Research ArticleSome Remarks on the Mathieu Series

Robert Frontczak

Landesbank Baden-Wurttemberg (LBBW) Am Hauptbahnhof 2 70173 Stuttgart Germany

Correspondence should be addressed to Robert Frontczak robertfrontczaklbbwde

Received 11 December 2013 Accepted 29 January 2014 Published 13 March 2014

Academic Editors G Fikioris and C-H Lien

Copyright copy 2014 Robert Frontczak This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The object of this note is to present new expressions for the classical Mathieu series in terms of hyperbolic functionsThe derivationis based on elementary arguments concerning the integral representation of the series The results are used afterwards to proveamong others a new relationship between the Mathieu series and its alternating companion A recursion formula for the Mathieuseries is also presented As a byproduct some closed-form evaluations of integrals involving hyperbolic functions are inferred

1 Introduction

The infinite series

119878 (119903) =

infin

sum

119899=1

2119899

(1198992 + 1199032)2 119903 gt 0 (1)

is called a Mathieu series It was introduced in 1890 by EL Mathieu (1835ndash1890) who studied various problems inmathematical physics Since its introduction the series 119878(119903)has been studied intensively Mathieu himself conjecturedthat 119878(119903) lt 1119903

2 The conjecture was proved in 1952 by Bergin [1] Nowadays themathematical literature provides a rangeof papers dealing with inequalities for the series 119878(119903) In 1957Makai [2] derived the double inequality

1

1199032 + 12lt 119878 (119903) lt

1

1199032 (2)

More recently Alzer et al proved in [3] that

1

1199032 + 1 (2120577 (3))lt 119878 (119903) lt

1

1199032 + 16 (3)

Here as usual 120577(119904) denotes the Riemann zeta functiondefined by 120577(119904) = sum

infin

119899=11119899119904 Re(119904) gt 1 The constants

1(2120577(3)) and 16 are the best possibleOther lower andupperbound estimates for the Mathieu series can be found in thearticles of Qi et al [4] and Hoorfar and Qi [5]

An integral representation for the Mathieu series (1) isgiven by

119878 (119903) =1

119903int

infin

0

119909

119890119909 minus 1sin (119903119909) 119889119909 (4)

The integral representationwas used by Elbert in [6] to derivethe asymptotic expansion of 119878(119903)

119878 (119903) =

infin

sum

119896=0

(minus1)1198961198612119896

1199032119896+2=

1

1199032minus

1

61199034plusmn sdot sdot sdot (119903 997888rarr infin) (5)

where 1198612119896

denote the even indexed Bernoulli numbersdefined by the generating function

119909

119890119909 minus 1=

infin

sum

119896=0

119861119896

119909119896

119896 |119909| lt 2120587 (6)

See also [7] for a derivationThe Mathieu series admits various generalizations that

have been introduced and investigated intensively in recentyears The generalizations include the alternating Mathieuseries the 119898-fold generalized Mathieu series Mathieu 119886-series and Mathieu (119886 120582)-series [8ndash11] The generalizationsrecapture the classical Mathieu series as a special case Onthe other hand the alternating Mathieu series althoughconnected to its classical companion is a variant that allows

Hindawi Publishing CorporationISRN Applied MathematicsVolume 2014 Article ID 985782 8 pageshttpdxdoiorg1011552014985782

2 ISRN Applied Mathematics

a separate study It was introduced by Pogany et al in [11] bythe equation

119878lowast

(119903) =

infin

sum

119899=1

(minus1)119899minus1

2119899

(1198992 + 1199032)2 119903 gt 0 (7)

It possesses the integral representation

119878lowast

(119903) =1

119903int

infin

0

119909

119890119909 + 1sin (119903119909) 119889119909 (8)

Recently derived bounding inequalities for alternatingMathieu-type series can be found in the paper of Poganyand Tomovski [12] The latest research article on integralforms for Mathieu-type series is the paper of Milovanovicand Pogany [13] The authors present a systematic treatmentof the subject based on contour integration Among othersthe following new integral representation for 119878(119903) is derived[13 Corollary 22]

119878 (119903) = 120587int

infin

0

1199032

minus 1199092

+ (14)

(1199092 minus 1199032 + (14))2

+ 1199032sdot

119889119909

cosh2 (120587119909) (9)

In this note we restrict the attention to the classical Mathieuseries Starting with the integral form (4) new representationsfor 119878(119903) are derived The derivation is based on elementaryarguments concerning the integrand combined with relatedintegral identitiesThe results are used afterwards to establishinteresting properties of the series Among others a newrelationship between the Mathieu series and its alternatingvariant is derived A recursion formula for the Mathieuseries is also presented As a byproduct some closed-formevaluations of integrals involving hyperbolic functions areinferred Finally a new proof is given for an exact evaluationof an infinite series related to 119878(119903)

2 Main Results

In what follows we will use the following integral identitiesThe identities are well known and are stated without proof(see [14] for a reference)

Lemma 1 For 119903 gt 0 and Re(120573) gt 0 it holds that

int

infin

0

119909

119890120573119909 minus 1cos (119903119909) 119889119909 = 1

21199032minus

1205872

21205732csch2 (119903120587

120573) (10)

where csch(119909) denotes the hyperbolic cosecant of 119909 defined by

csch (119909) = 1

sinh (119909)=

2

119890119909 minus 119890minus119909 (11)

Similarly for 119903 gt 0 and 119899 ⩾ 0 an integer

int

infin

0

1199092119899

119890119909 minus 1sin (119903119909) 119889119909 = (minus1)

1198991205972119899

1205971199032119899(120587

2coth (119903120587) minus 1

2119903)

(12)

where coth(119909) denotes the hyperbolic cotangent of 119909 defined by

coth (119909) = cosh (119909)sinh (119909)

=1198902119909

+ 1

1198902119909 minus 1 (13)

Finally for 119903 gt 0 Re(120573) gt 0 and 119899 ⩾ 0 an integer

int

infin

0

119909119899

119890minus120573119909 cos (119903119909) 119889119909 = (minus1)

119899120597119899

120597120573119899(

120573

1205732 + 1199032) (14)

Themain theorem of this note is stated next It gives threeexpressions for the Mathieu series in a semi-integral form

Theorem 2 The Mathieu series 119878(119903) has the following repre-sentations

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909(15)

119878 (119903) =1

1199032minus

1

41199034+ (

120587

2119903 sinh (120587119903))

2

+1

1199032int

infin

0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) 119889119909

(16)

119878 (119903) =1

1199032minus1

1199034+

120587

21199033coth (120587119903)

+1

2(

120587

119903 sinh (120587119903))

2

+2

1199033int

infin

0

1198921015840

(119909)

119890119909 minus 1sin (119903119909) 119889119909

(17)

where sinh(119909) and cosh(119909) denote the hyperbolic sine andcosine functions respectively 119892(119909) = 119909(119890

119909

minus 1) and 1198921015840

(119909)

denotes its first derivative

Proof Let 119892(119909) = 119909(119890119909

minus1) Themain argument in the proofis the observation that 119892(119909) satisfies the nonlinear differentialequation

1198921015840

(119909) = minus119892 (119909) +119892 (119909)

119909minus1198922

(119909)

119909 (18)

which may be also written in the form

119892 (119909) =119892 (119909)

119909minus1198922

(119909)

119909minus 1198921015840

(119909) (19)

Inserting the relation in (4) and using (12) with 119899 = 0 give

119878 (119903) = minus1

21199032+120587

2119903coth (120587119903)

minus1

119903int

infin

0

(1198922

(119909)

119909+ 1198921015840

(119909)) sin (119903119909) 119889119909(20)

Since

1198922

(119909)

119909= minus

1

2(119892 (119909) minus 119892 (119909) coth(119909

2)) (21)

ISRN Applied Mathematics 3

we see that

119878 (119903) = minus1

1199032+120587

119903coth (120587119903)

minus1

119903int

infin

0

(119892 (119909) coth(1199092) + 2119892

1015840

(119909)) sin (119903119909) 119889119909

(22)

Integration by parts gives

int

infin

0

1198921015840

(119909) sin (119903119909) 119889119909 = minus119903int

infin

0

119892 (119909) cos (119903119909) 119889119909 (23)

which is easily evaluated using (10) with 120573 = 1 Finally theelementary relation

coth(1199092) =

1

tanh (1199092)=cosh (119909) + 1sinh (119909)

(24)

establishes (15) Integrating (4) by parts results in

119878 (119903) =1

1199032+1

1199032int

infin

0

1198921015840

(119909) cos (119903119909) 119889119909 (25)

Equations (18) and (10) give

119878 (119903) =1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909

(26)

Let 119866(119903) denote the last component of (26) that is

119866 (119903) =1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 (27)

Then 119866(119903) can be evaluated in two different ways First therelation from (21) in combination with (10) shows that

119866 (119903) =1

41199034minus (

120587

2119903 sinh (120587119903))

2

+1

1199032int

infin

0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) 119889119909

(28)

This proves (16) Finally by the rule of lrsquoHospital

lim119909rarr0

(119892 (119909)

119909minus1198922

(119909)

119909) = lim119909rarr0

119890119909

minus 1 minus 119909

(119890119909 minus 1)2=1

2

lim119909rarrinfin

(119892 (119909)

119909minus1198922

(119909)

119909) = 0

(29)

Splitting the integrand of 119866(119903) into two parts and integratingby parts in each case we obtain

119866 (119903) =1

1199033int

infin

0

(119890119909

+ 1

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909 (30)

In the last equation we have used the fact that

lim119909rarr0

sin (119903119909)119890119909 minus 1

= lim119909rarr0

119909 sin (119903119909)(119890119909 minus 1)

2= 119903 (31)

Since119890119909

+ 1

(119890119909 minus 1)2=

1

119890119909 minus 1+

2

(119890119909 minus 1)2 (32)

119866(119903) is equal to

119866 (119903) =1

1199033int

infin

0

sin (119903119909)119890119909 minus 1

119889119909

+1

1199033int

infin

0

(2

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909

(33)

For the first integral we apply once more the result from (12)with 119899 = 0 Direct calculation shows that the expressionin brackets under the second integral sign is equal to21198921015840

(119909)(119890119909

minus 1) This completes the proof

The elementary relation in (18) seems to have beenoverseen in the literature It allows to express 119878(119903) in termsof the hyperbolic sine and cosine functions respectivelyThe representations may turn out to be useful to study theproperties of 119878(119903) For the remainder of this note howeverwe will mainly work with (15) The equation will be used toinfer interesting consequences which we are going to presentimmediately Additionally in Section 3 it will be outlined that(18) is also useful to study some topics that are related to theevaluation of 119878(119903)

Concerning the integrands in Theorem 2 we make thefollowing observations Firstly standard arguments show that(cosh(119909) + 1) sinh(119909)119892(119909) gt 0 for 119909 gt 0 and

lim119909rarr0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) = 2119903 (34)

Secondly 1119909 minus 12(cosh(119909) + 1) sinh(119909)119892(119909) lt 0 for 119909 gt 0

and

lim119909rarr0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) = 0 (35)

Finally since 119890119909minus1 lt 119909119890119909 for 119909 gt 0 it is clear that 1198921015840(119909)(119890119909minus

1) lt 0 and

lim119909rarr0

1198921015840

(119909) sin (119903119909)119890119909 minus 1

= minus119903

2 (36)

Figure 1 illustrates the results from Theorem 2 numerically119878(119903) is plotted together with (15)ndash(17) where the new rep-resentations are decomposed into nonintegral and integralparts (termed first and second parts resp) More preciselythis means that (15) is decomposed in the following manner

First Part = 120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

Second Part = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909(37)

Equations (16) and (17) are decomposed analogously

4 ISRN Applied Mathematics

0 1 2 3 4 5

0

2

4

6

8Equation (15)

First partSecond part

Equation (16) Equation (17)

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

S(r)

First partSecond part

S(r)

First partSecond part

S(r)

Figure 1 Numerical (graphical) illustrations of the results fromTheorem 2

A first interesting consequence of the theorem is thefollowing closed-form evaluation of an integral

Corollary 3 Let 119892(119909) = 119909(119890119909

minus 1) Then it holds that

int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) 119909 119889119909 = 2 (2120577 (2) minus 120577 (3)) asymp 41756

(38)

Proof From (1) it is obvious that lim119903rarr0

119878(119903) = 2120577(3)Further from the identity for Bernoulli numbers (see [14])

int

infin

0

1199092119899minus1

119890119901119909 minus 1119889119909 = (minus1)

119899minus1

(2120587

119901)

2119899

1198612119899

4119899 119899 = 1 2 3

(39)

and the fact that 1198612= 16 we easily deduce that

lim119903rarr0

2int

infin

0

119892 (119909) cos (119903119909) 119889119909

= lim119903rarr0

(1

1199032minus (

120587

sinh (120587119903))

2

) = 2120577 (2)

lim119903rarr0

(120587

119903coth (120587119903) minus 1

1199032) = 2120577 (2)

(40)

Hence

lim119903rarr0

(120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

) = 4120577 (2) (41)

The assertion follows from (15)

It is clear that a successive repetition of integration byparts in Theorem 2 will produce a range of other expressionsfor 119878(119903) Among others from (15) and the fact that

tanh(1199092) =

cosh (119909) minus 1sinh (119909)

(42)

it is fairly easy to produce the following characterization

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

+ int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) cos (119903119909) 119889119909

+1

119903int

infin

0

ln (cosh (119909) minus 1) 1198921015840 (119909) sin (119903119909) 119889119909

(43)

In the above characterization the sine and cosine functionsappear simultaneously as integrands Now in view of theprevious proof and (18) it is straightforward to get

int

infin

0

ln (cosh (119909) minus 1) (2119892 (119909) minus 119892 (119909) 119892 (minus119909)) 119889119909

= 2 (120577 (3) minus 2120577 (2))

(44)

where we have used the relation 119909 + 119892(119909) = 119892(minus119909) Further-more since direct computation verifies that

119892 (119909) 119892 (minus119909) =1

21199092

1

cosh (119909) minus 1 (45)

ISRN Applied Mathematics 5

the last integral may be written in the form

int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) 119889119909

= 120577 (3) minus 2120577 (2) + int

infin

0

(119909

2)

2 ln (cosh (119909) minus 1)cosh (119909) minus 1

119889119909

(46)

A further interesting consequence of the previous theoremis the following new integral-type representation for thealternating Mathieu series 119878lowast(119903)

Proposition 4 The alternating Mathieu series 119878lowast(119903) may berepresented as

119878lowast

(119903) = minus119878 (119903) +120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903)

(47)

with

119869 (119903) =1

119903int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(48)

Proof We start with the identity

119878lowast

(119903) = 119878 (119903) minus1

4119878 (

119903

2) (49)

In view of (15) and applying the relation

coth(1205871199032) = coth (120587119903) + 1

sinh (120587119903) (50)

we obtain

119878lowast

(119903) =120587

2119903coth (120587119903) minus 120587

2119903

1

sinh (120587119903)

+ (120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

+ 119868 (119903)

(51)

where

119868 (119903) = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909

+1

2119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin(12119903119909) 119889119909

(52)

For the second integral a change of variable (119910 = 12119909) isused in combination with themultiple-angle formulas for thehyperbolic sine and cosine functions respectively to get

cosh (2119909) + 1sinh (2119909)

= coth (119909) (53)

and hence

119868 (119903) = minus1

119903int

infin

0

( (119892 (119909) minus 119892 (2119909)) coth (119909)

+119892 (119909)

sinh (119909)) sin (119903119909) 119889119909

(54)

Next since

119892 (2119909) =2119892 (119909)

119890119909 + 1 (55)

119868(119903) can be simplified to

119868 (119903) = minus1

119903int

infin

0

( coth (119909) coth(1199092)

+1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(56)

The relation

coth (119909) coth(1199092) =

cosh (119909)cosh (119909) + 1

= 1 minus1

cosh (119909) + 1(57)

shows that 119868(119903) = minus119878(119903)+119869(119903)The final formula follows from

(120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

=1205872

sinh2 (120587119903)(cosh2 (1

2120587119903) minus 1)

(58)

and the half-angle formula

cosh (12119909) = radic

cosh (119909) + 12

(59)

Corollary 5 It holds that

int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) 119909 119889119909

=7

2120577 (3) minus 3120577 (2) asymp minus07276

(60)

int

infin

0

(1

cosh (119909) + 1+ coth (119909)) 119892 (119909) 119909 119889119909

=3

2120577 (3) + 120577 (2) asymp 34480

(61)

Proof We have lim119903rarr0

(119878lowast

(119903)+119878(119903)) = 72120577(3) Additionallyusing similar arguments as in Corollary 3 it can be shownthat

lim119903rarr0

(120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903))) = 3120577 (2) (62)

This proves the first expression The second follows fromcombining (38) and (60)

The next assertion provides a recursion formula for theMathieu series Surprisingly it is also a consequence ofProposition 4

Corollary 6 For 119903 gt 0 define the function 119891(119903) as

119891 (119903) =120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903) (63)

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 ISRN Applied Mathematics

a separate study It was introduced by Pogany et al in [11] bythe equation

119878lowast

(119903) =

infin

sum

119899=1

(minus1)119899minus1

2119899

(1198992 + 1199032)2 119903 gt 0 (7)

It possesses the integral representation

119878lowast

(119903) =1

119903int

infin

0

119909

119890119909 + 1sin (119903119909) 119889119909 (8)

Recently derived bounding inequalities for alternatingMathieu-type series can be found in the paper of Poganyand Tomovski [12] The latest research article on integralforms for Mathieu-type series is the paper of Milovanovicand Pogany [13] The authors present a systematic treatmentof the subject based on contour integration Among othersthe following new integral representation for 119878(119903) is derived[13 Corollary 22]

119878 (119903) = 120587int

infin

0

1199032

minus 1199092

+ (14)

(1199092 minus 1199032 + (14))2

+ 1199032sdot

119889119909

cosh2 (120587119909) (9)

In this note we restrict the attention to the classical Mathieuseries Starting with the integral form (4) new representationsfor 119878(119903) are derived The derivation is based on elementaryarguments concerning the integrand combined with relatedintegral identitiesThe results are used afterwards to establishinteresting properties of the series Among others a newrelationship between the Mathieu series and its alternatingvariant is derived A recursion formula for the Mathieuseries is also presented As a byproduct some closed-formevaluations of integrals involving hyperbolic functions areinferred Finally a new proof is given for an exact evaluationof an infinite series related to 119878(119903)

2 Main Results

In what follows we will use the following integral identitiesThe identities are well known and are stated without proof(see [14] for a reference)

Lemma 1 For 119903 gt 0 and Re(120573) gt 0 it holds that

int

infin

0

119909

119890120573119909 minus 1cos (119903119909) 119889119909 = 1

21199032minus

1205872

21205732csch2 (119903120587

120573) (10)

where csch(119909) denotes the hyperbolic cosecant of 119909 defined by

csch (119909) = 1

sinh (119909)=

2

119890119909 minus 119890minus119909 (11)

Similarly for 119903 gt 0 and 119899 ⩾ 0 an integer

int

infin

0

1199092119899

119890119909 minus 1sin (119903119909) 119889119909 = (minus1)

1198991205972119899

1205971199032119899(120587

2coth (119903120587) minus 1

2119903)

(12)

where coth(119909) denotes the hyperbolic cotangent of 119909 defined by

coth (119909) = cosh (119909)sinh (119909)

=1198902119909

+ 1

1198902119909 minus 1 (13)

Finally for 119903 gt 0 Re(120573) gt 0 and 119899 ⩾ 0 an integer

int

infin

0

119909119899

119890minus120573119909 cos (119903119909) 119889119909 = (minus1)

119899120597119899

120597120573119899(

120573

1205732 + 1199032) (14)

Themain theorem of this note is stated next It gives threeexpressions for the Mathieu series in a semi-integral form

Theorem 2 The Mathieu series 119878(119903) has the following repre-sentations

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909(15)

119878 (119903) =1

1199032minus

1

41199034+ (

120587

2119903 sinh (120587119903))

2

+1

1199032int

infin

0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) 119889119909

(16)

119878 (119903) =1

1199032minus1

1199034+

120587

21199033coth (120587119903)

+1

2(

120587

119903 sinh (120587119903))

2

+2

1199033int

infin

0

1198921015840

(119909)

119890119909 minus 1sin (119903119909) 119889119909

(17)

where sinh(119909) and cosh(119909) denote the hyperbolic sine andcosine functions respectively 119892(119909) = 119909(119890

119909

minus 1) and 1198921015840

(119909)

denotes its first derivative

Proof Let 119892(119909) = 119909(119890119909

minus1) Themain argument in the proofis the observation that 119892(119909) satisfies the nonlinear differentialequation

1198921015840

(119909) = minus119892 (119909) +119892 (119909)

119909minus1198922

(119909)

119909 (18)

which may be also written in the form

119892 (119909) =119892 (119909)

119909minus1198922

(119909)

119909minus 1198921015840

(119909) (19)

Inserting the relation in (4) and using (12) with 119899 = 0 give

119878 (119903) = minus1

21199032+120587

2119903coth (120587119903)

minus1

119903int

infin

0

(1198922

(119909)

119909+ 1198921015840

(119909)) sin (119903119909) 119889119909(20)

Since

1198922

(119909)

119909= minus

1

2(119892 (119909) minus 119892 (119909) coth(119909

2)) (21)

ISRN Applied Mathematics 3

we see that

119878 (119903) = minus1

1199032+120587

119903coth (120587119903)

minus1

119903int

infin

0

(119892 (119909) coth(1199092) + 2119892

1015840

(119909)) sin (119903119909) 119889119909

(22)

Integration by parts gives

int

infin

0

1198921015840

(119909) sin (119903119909) 119889119909 = minus119903int

infin

0

119892 (119909) cos (119903119909) 119889119909 (23)

which is easily evaluated using (10) with 120573 = 1 Finally theelementary relation

coth(1199092) =

1

tanh (1199092)=cosh (119909) + 1sinh (119909)

(24)

establishes (15) Integrating (4) by parts results in

119878 (119903) =1

1199032+1

1199032int

infin

0

1198921015840

(119909) cos (119903119909) 119889119909 (25)

Equations (18) and (10) give

119878 (119903) =1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909

(26)

Let 119866(119903) denote the last component of (26) that is

119866 (119903) =1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 (27)

Then 119866(119903) can be evaluated in two different ways First therelation from (21) in combination with (10) shows that

119866 (119903) =1

41199034minus (

120587

2119903 sinh (120587119903))

2

+1

1199032int

infin

0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) 119889119909

(28)

This proves (16) Finally by the rule of lrsquoHospital

lim119909rarr0

(119892 (119909)

119909minus1198922

(119909)

119909) = lim119909rarr0

119890119909

minus 1 minus 119909

(119890119909 minus 1)2=1

2

lim119909rarrinfin

(119892 (119909)

119909minus1198922

(119909)

119909) = 0

(29)

Splitting the integrand of 119866(119903) into two parts and integratingby parts in each case we obtain

119866 (119903) =1

1199033int

infin

0

(119890119909

+ 1

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909 (30)

In the last equation we have used the fact that

lim119909rarr0

sin (119903119909)119890119909 minus 1

= lim119909rarr0

119909 sin (119903119909)(119890119909 minus 1)

2= 119903 (31)

Since119890119909

+ 1

(119890119909 minus 1)2=

1

119890119909 minus 1+

2

(119890119909 minus 1)2 (32)

119866(119903) is equal to

119866 (119903) =1

1199033int

infin

0

sin (119903119909)119890119909 minus 1

119889119909

+1

1199033int

infin

0

(2

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909

(33)

For the first integral we apply once more the result from (12)with 119899 = 0 Direct calculation shows that the expressionin brackets under the second integral sign is equal to21198921015840

(119909)(119890119909

minus 1) This completes the proof

The elementary relation in (18) seems to have beenoverseen in the literature It allows to express 119878(119903) in termsof the hyperbolic sine and cosine functions respectivelyThe representations may turn out to be useful to study theproperties of 119878(119903) For the remainder of this note howeverwe will mainly work with (15) The equation will be used toinfer interesting consequences which we are going to presentimmediately Additionally in Section 3 it will be outlined that(18) is also useful to study some topics that are related to theevaluation of 119878(119903)

Concerning the integrands in Theorem 2 we make thefollowing observations Firstly standard arguments show that(cosh(119909) + 1) sinh(119909)119892(119909) gt 0 for 119909 gt 0 and

lim119909rarr0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) = 2119903 (34)

Secondly 1119909 minus 12(cosh(119909) + 1) sinh(119909)119892(119909) lt 0 for 119909 gt 0

and

lim119909rarr0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) = 0 (35)

Finally since 119890119909minus1 lt 119909119890119909 for 119909 gt 0 it is clear that 1198921015840(119909)(119890119909minus

1) lt 0 and

lim119909rarr0

1198921015840

(119909) sin (119903119909)119890119909 minus 1

= minus119903

2 (36)

Figure 1 illustrates the results from Theorem 2 numerically119878(119903) is plotted together with (15)ndash(17) where the new rep-resentations are decomposed into nonintegral and integralparts (termed first and second parts resp) More preciselythis means that (15) is decomposed in the following manner

First Part = 120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

Second Part = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909(37)

Equations (16) and (17) are decomposed analogously

4 ISRN Applied Mathematics

0 1 2 3 4 5

0

2

4

6

8Equation (15)

First partSecond part

Equation (16) Equation (17)

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

S(r)

First partSecond part

S(r)

First partSecond part

S(r)

Figure 1 Numerical (graphical) illustrations of the results fromTheorem 2

A first interesting consequence of the theorem is thefollowing closed-form evaluation of an integral

Corollary 3 Let 119892(119909) = 119909(119890119909

minus 1) Then it holds that

int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) 119909 119889119909 = 2 (2120577 (2) minus 120577 (3)) asymp 41756

(38)

Proof From (1) it is obvious that lim119903rarr0

119878(119903) = 2120577(3)Further from the identity for Bernoulli numbers (see [14])

int

infin

0

1199092119899minus1

119890119901119909 minus 1119889119909 = (minus1)

119899minus1

(2120587

119901)

2119899

1198612119899

4119899 119899 = 1 2 3

(39)

and the fact that 1198612= 16 we easily deduce that

lim119903rarr0

2int

infin

0

119892 (119909) cos (119903119909) 119889119909

= lim119903rarr0

(1

1199032minus (

120587

sinh (120587119903))

2

) = 2120577 (2)

lim119903rarr0

(120587

119903coth (120587119903) minus 1

1199032) = 2120577 (2)

(40)

Hence

lim119903rarr0

(120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

) = 4120577 (2) (41)

The assertion follows from (15)

It is clear that a successive repetition of integration byparts in Theorem 2 will produce a range of other expressionsfor 119878(119903) Among others from (15) and the fact that

tanh(1199092) =

cosh (119909) minus 1sinh (119909)

(42)

it is fairly easy to produce the following characterization

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

+ int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) cos (119903119909) 119889119909

+1

119903int

infin

0

ln (cosh (119909) minus 1) 1198921015840 (119909) sin (119903119909) 119889119909

(43)

In the above characterization the sine and cosine functionsappear simultaneously as integrands Now in view of theprevious proof and (18) it is straightforward to get

int

infin

0

ln (cosh (119909) minus 1) (2119892 (119909) minus 119892 (119909) 119892 (minus119909)) 119889119909

= 2 (120577 (3) minus 2120577 (2))

(44)

where we have used the relation 119909 + 119892(119909) = 119892(minus119909) Further-more since direct computation verifies that

119892 (119909) 119892 (minus119909) =1

21199092

1

cosh (119909) minus 1 (45)

ISRN Applied Mathematics 5

the last integral may be written in the form

int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) 119889119909

= 120577 (3) minus 2120577 (2) + int

infin

0

(119909

2)

2 ln (cosh (119909) minus 1)cosh (119909) minus 1

119889119909

(46)

A further interesting consequence of the previous theoremis the following new integral-type representation for thealternating Mathieu series 119878lowast(119903)

Proposition 4 The alternating Mathieu series 119878lowast(119903) may berepresented as

119878lowast

(119903) = minus119878 (119903) +120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903)

(47)

with

119869 (119903) =1

119903int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(48)

Proof We start with the identity

119878lowast

(119903) = 119878 (119903) minus1

4119878 (

119903

2) (49)

In view of (15) and applying the relation

coth(1205871199032) = coth (120587119903) + 1

sinh (120587119903) (50)

we obtain

119878lowast

(119903) =120587

2119903coth (120587119903) minus 120587

2119903

1

sinh (120587119903)

+ (120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

+ 119868 (119903)

(51)

where

119868 (119903) = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909

+1

2119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin(12119903119909) 119889119909

(52)

For the second integral a change of variable (119910 = 12119909) isused in combination with themultiple-angle formulas for thehyperbolic sine and cosine functions respectively to get

cosh (2119909) + 1sinh (2119909)

= coth (119909) (53)

and hence

119868 (119903) = minus1

119903int

infin

0

( (119892 (119909) minus 119892 (2119909)) coth (119909)

+119892 (119909)

sinh (119909)) sin (119903119909) 119889119909

(54)

Next since

119892 (2119909) =2119892 (119909)

119890119909 + 1 (55)

119868(119903) can be simplified to

119868 (119903) = minus1

119903int

infin

0

( coth (119909) coth(1199092)

+1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(56)

The relation

coth (119909) coth(1199092) =

cosh (119909)cosh (119909) + 1

= 1 minus1

cosh (119909) + 1(57)

shows that 119868(119903) = minus119878(119903)+119869(119903)The final formula follows from

(120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

=1205872

sinh2 (120587119903)(cosh2 (1

2120587119903) minus 1)

(58)

and the half-angle formula

cosh (12119909) = radic

cosh (119909) + 12

(59)

Corollary 5 It holds that

int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) 119909 119889119909

=7

2120577 (3) minus 3120577 (2) asymp minus07276

(60)

int

infin

0

(1

cosh (119909) + 1+ coth (119909)) 119892 (119909) 119909 119889119909

=3

2120577 (3) + 120577 (2) asymp 34480

(61)

Proof We have lim119903rarr0

(119878lowast

(119903)+119878(119903)) = 72120577(3) Additionallyusing similar arguments as in Corollary 3 it can be shownthat

lim119903rarr0

(120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903))) = 3120577 (2) (62)

This proves the first expression The second follows fromcombining (38) and (60)

The next assertion provides a recursion formula for theMathieu series Surprisingly it is also a consequence ofProposition 4

Corollary 6 For 119903 gt 0 define the function 119891(119903) as

119891 (119903) =120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903) (63)

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Applied Mathematics 3

we see that

119878 (119903) = minus1

1199032+120587

119903coth (120587119903)

minus1

119903int

infin

0

(119892 (119909) coth(1199092) + 2119892

1015840

(119909)) sin (119903119909) 119889119909

(22)

Integration by parts gives

int

infin

0

1198921015840

(119909) sin (119903119909) 119889119909 = minus119903int

infin

0

119892 (119909) cos (119903119909) 119889119909 (23)

which is easily evaluated using (10) with 120573 = 1 Finally theelementary relation

coth(1199092) =

1

tanh (1199092)=cosh (119909) + 1sinh (119909)

(24)

establishes (15) Integrating (4) by parts results in

119878 (119903) =1

1199032+1

1199032int

infin

0

1198921015840

(119909) cos (119903119909) 119889119909 (25)

Equations (18) and (10) give

119878 (119903) =1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909

(26)

Let 119866(119903) denote the last component of (26) that is

119866 (119903) =1

1199032int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 (27)

Then 119866(119903) can be evaluated in two different ways First therelation from (21) in combination with (10) shows that

119866 (119903) =1

41199034minus (

120587

2119903 sinh (120587119903))

2

+1

1199032int

infin

0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) 119889119909

(28)

This proves (16) Finally by the rule of lrsquoHospital

lim119909rarr0

(119892 (119909)

119909minus1198922

(119909)

119909) = lim119909rarr0

119890119909

minus 1 minus 119909

(119890119909 minus 1)2=1

2

lim119909rarrinfin

(119892 (119909)

119909minus1198922

(119909)

119909) = 0

(29)

Splitting the integrand of 119866(119903) into two parts and integratingby parts in each case we obtain

119866 (119903) =1

1199033int

infin

0

(119890119909

+ 1

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909 (30)

In the last equation we have used the fact that

lim119909rarr0

sin (119903119909)119890119909 minus 1

= lim119909rarr0

119909 sin (119903119909)(119890119909 minus 1)

2= 119903 (31)

Since119890119909

+ 1

(119890119909 minus 1)2=

1

119890119909 minus 1+

2

(119890119909 minus 1)2 (32)

119866(119903) is equal to

119866 (119903) =1

1199033int

infin

0

sin (119903119909)119890119909 minus 1

119889119909

+1

1199033int

infin

0

(2

(119890119909 minus 1)2minus

2119909119890119909

(119890119909 minus 1)3) sin (119903119909) 119889119909

(33)

For the first integral we apply once more the result from (12)with 119899 = 0 Direct calculation shows that the expressionin brackets under the second integral sign is equal to21198921015840

(119909)(119890119909

minus 1) This completes the proof

The elementary relation in (18) seems to have beenoverseen in the literature It allows to express 119878(119903) in termsof the hyperbolic sine and cosine functions respectivelyThe representations may turn out to be useful to study theproperties of 119878(119903) For the remainder of this note howeverwe will mainly work with (15) The equation will be used toinfer interesting consequences which we are going to presentimmediately Additionally in Section 3 it will be outlined that(18) is also useful to study some topics that are related to theevaluation of 119878(119903)

Concerning the integrands in Theorem 2 we make thefollowing observations Firstly standard arguments show that(cosh(119909) + 1) sinh(119909)119892(119909) gt 0 for 119909 gt 0 and

lim119909rarr0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) = 2119903 (34)

Secondly 1119909 minus 12(cosh(119909) + 1) sinh(119909)119892(119909) lt 0 for 119909 gt 0

and

lim119909rarr0

(1

119909minus1

2

cosh (119909) + 1sinh (119909)

) 119892 (119909) cos (119903119909) = 0 (35)

Finally since 119890119909minus1 lt 119909119890119909 for 119909 gt 0 it is clear that 1198921015840(119909)(119890119909minus

1) lt 0 and

lim119909rarr0

1198921015840

(119909) sin (119903119909)119890119909 minus 1

= minus119903

2 (36)

Figure 1 illustrates the results from Theorem 2 numerically119878(119903) is plotted together with (15)ndash(17) where the new rep-resentations are decomposed into nonintegral and integralparts (termed first and second parts resp) More preciselythis means that (15) is decomposed in the following manner

First Part = 120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

Second Part = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909(37)

Equations (16) and (17) are decomposed analogously

4 ISRN Applied Mathematics

0 1 2 3 4 5

0

2

4

6

8Equation (15)

First partSecond part

Equation (16) Equation (17)

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

S(r)

First partSecond part

S(r)

First partSecond part

S(r)

Figure 1 Numerical (graphical) illustrations of the results fromTheorem 2

A first interesting consequence of the theorem is thefollowing closed-form evaluation of an integral

Corollary 3 Let 119892(119909) = 119909(119890119909

minus 1) Then it holds that

int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) 119909 119889119909 = 2 (2120577 (2) minus 120577 (3)) asymp 41756

(38)

Proof From (1) it is obvious that lim119903rarr0

119878(119903) = 2120577(3)Further from the identity for Bernoulli numbers (see [14])

int

infin

0

1199092119899minus1

119890119901119909 minus 1119889119909 = (minus1)

119899minus1

(2120587

119901)

2119899

1198612119899

4119899 119899 = 1 2 3

(39)

and the fact that 1198612= 16 we easily deduce that

lim119903rarr0

2int

infin

0

119892 (119909) cos (119903119909) 119889119909

= lim119903rarr0

(1

1199032minus (

120587

sinh (120587119903))

2

) = 2120577 (2)

lim119903rarr0

(120587

119903coth (120587119903) minus 1

1199032) = 2120577 (2)

(40)

Hence

lim119903rarr0

(120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

) = 4120577 (2) (41)

The assertion follows from (15)

It is clear that a successive repetition of integration byparts in Theorem 2 will produce a range of other expressionsfor 119878(119903) Among others from (15) and the fact that

tanh(1199092) =

cosh (119909) minus 1sinh (119909)

(42)

it is fairly easy to produce the following characterization

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

+ int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) cos (119903119909) 119889119909

+1

119903int

infin

0

ln (cosh (119909) minus 1) 1198921015840 (119909) sin (119903119909) 119889119909

(43)

In the above characterization the sine and cosine functionsappear simultaneously as integrands Now in view of theprevious proof and (18) it is straightforward to get

int

infin

0

ln (cosh (119909) minus 1) (2119892 (119909) minus 119892 (119909) 119892 (minus119909)) 119889119909

= 2 (120577 (3) minus 2120577 (2))

(44)

where we have used the relation 119909 + 119892(119909) = 119892(minus119909) Further-more since direct computation verifies that

119892 (119909) 119892 (minus119909) =1

21199092

1

cosh (119909) minus 1 (45)

ISRN Applied Mathematics 5

the last integral may be written in the form

int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) 119889119909

= 120577 (3) minus 2120577 (2) + int

infin

0

(119909

2)

2 ln (cosh (119909) minus 1)cosh (119909) minus 1

119889119909

(46)

A further interesting consequence of the previous theoremis the following new integral-type representation for thealternating Mathieu series 119878lowast(119903)

Proposition 4 The alternating Mathieu series 119878lowast(119903) may berepresented as

119878lowast

(119903) = minus119878 (119903) +120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903)

(47)

with

119869 (119903) =1

119903int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(48)

Proof We start with the identity

119878lowast

(119903) = 119878 (119903) minus1

4119878 (

119903

2) (49)

In view of (15) and applying the relation

coth(1205871199032) = coth (120587119903) + 1

sinh (120587119903) (50)

we obtain

119878lowast

(119903) =120587

2119903coth (120587119903) minus 120587

2119903

1

sinh (120587119903)

+ (120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

+ 119868 (119903)

(51)

where

119868 (119903) = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909

+1

2119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin(12119903119909) 119889119909

(52)

For the second integral a change of variable (119910 = 12119909) isused in combination with themultiple-angle formulas for thehyperbolic sine and cosine functions respectively to get

cosh (2119909) + 1sinh (2119909)

= coth (119909) (53)

and hence

119868 (119903) = minus1

119903int

infin

0

( (119892 (119909) minus 119892 (2119909)) coth (119909)

+119892 (119909)

sinh (119909)) sin (119903119909) 119889119909

(54)

Next since

119892 (2119909) =2119892 (119909)

119890119909 + 1 (55)

119868(119903) can be simplified to

119868 (119903) = minus1

119903int

infin

0

( coth (119909) coth(1199092)

+1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(56)

The relation

coth (119909) coth(1199092) =

cosh (119909)cosh (119909) + 1

= 1 minus1

cosh (119909) + 1(57)

shows that 119868(119903) = minus119878(119903)+119869(119903)The final formula follows from

(120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

=1205872

sinh2 (120587119903)(cosh2 (1

2120587119903) minus 1)

(58)

and the half-angle formula

cosh (12119909) = radic

cosh (119909) + 12

(59)

Corollary 5 It holds that

int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) 119909 119889119909

=7

2120577 (3) minus 3120577 (2) asymp minus07276

(60)

int

infin

0

(1

cosh (119909) + 1+ coth (119909)) 119892 (119909) 119909 119889119909

=3

2120577 (3) + 120577 (2) asymp 34480

(61)

Proof We have lim119903rarr0

(119878lowast

(119903)+119878(119903)) = 72120577(3) Additionallyusing similar arguments as in Corollary 3 it can be shownthat

lim119903rarr0

(120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903))) = 3120577 (2) (62)

This proves the first expression The second follows fromcombining (38) and (60)

The next assertion provides a recursion formula for theMathieu series Surprisingly it is also a consequence ofProposition 4

Corollary 6 For 119903 gt 0 define the function 119891(119903) as

119891 (119903) =120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903) (63)

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 ISRN Applied Mathematics

0 1 2 3 4 5

0

2

4

6

8Equation (15)

First partSecond part

Equation (16) Equation (17)

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

0 1 2 3 4 5

0

2

4

6

8

minus2

minus4

minus6

r

S(r)

First partSecond part

S(r)

First partSecond part

S(r)

Figure 1 Numerical (graphical) illustrations of the results fromTheorem 2

A first interesting consequence of the theorem is thefollowing closed-form evaluation of an integral

Corollary 3 Let 119892(119909) = 119909(119890119909

minus 1) Then it holds that

int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) 119909 119889119909 = 2 (2120577 (2) minus 120577 (3)) asymp 41756

(38)

Proof From (1) it is obvious that lim119903rarr0

119878(119903) = 2120577(3)Further from the identity for Bernoulli numbers (see [14])

int

infin

0

1199092119899minus1

119890119901119909 minus 1119889119909 = (minus1)

119899minus1

(2120587

119901)

2119899

1198612119899

4119899 119899 = 1 2 3

(39)

and the fact that 1198612= 16 we easily deduce that

lim119903rarr0

2int

infin

0

119892 (119909) cos (119903119909) 119889119909

= lim119903rarr0

(1

1199032minus (

120587

sinh (120587119903))

2

) = 2120577 (2)

lim119903rarr0

(120587

119903coth (120587119903) minus 1

1199032) = 2120577 (2)

(40)

Hence

lim119903rarr0

(120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

) = 4120577 (2) (41)

The assertion follows from (15)

It is clear that a successive repetition of integration byparts in Theorem 2 will produce a range of other expressionsfor 119878(119903) Among others from (15) and the fact that

tanh(1199092) =

cosh (119909) minus 1sinh (119909)

(42)

it is fairly easy to produce the following characterization

119878 (119903) =120587

119903coth (120587119903) minus ( 120587

sinh (120587119903))

2

+ int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) cos (119903119909) 119889119909

+1

119903int

infin

0

ln (cosh (119909) minus 1) 1198921015840 (119909) sin (119903119909) 119889119909

(43)

In the above characterization the sine and cosine functionsappear simultaneously as integrands Now in view of theprevious proof and (18) it is straightforward to get

int

infin

0

ln (cosh (119909) minus 1) (2119892 (119909) minus 119892 (119909) 119892 (minus119909)) 119889119909

= 2 (120577 (3) minus 2120577 (2))

(44)

where we have used the relation 119909 + 119892(119909) = 119892(minus119909) Further-more since direct computation verifies that

119892 (119909) 119892 (minus119909) =1

21199092

1

cosh (119909) minus 1 (45)

ISRN Applied Mathematics 5

the last integral may be written in the form

int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) 119889119909

= 120577 (3) minus 2120577 (2) + int

infin

0

(119909

2)

2 ln (cosh (119909) minus 1)cosh (119909) minus 1

119889119909

(46)

A further interesting consequence of the previous theoremis the following new integral-type representation for thealternating Mathieu series 119878lowast(119903)

Proposition 4 The alternating Mathieu series 119878lowast(119903) may berepresented as

119878lowast

(119903) = minus119878 (119903) +120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903)

(47)

with

119869 (119903) =1

119903int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(48)

Proof We start with the identity

119878lowast

(119903) = 119878 (119903) minus1

4119878 (

119903

2) (49)

In view of (15) and applying the relation

coth(1205871199032) = coth (120587119903) + 1

sinh (120587119903) (50)

we obtain

119878lowast

(119903) =120587

2119903coth (120587119903) minus 120587

2119903

1

sinh (120587119903)

+ (120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

+ 119868 (119903)

(51)

where

119868 (119903) = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909

+1

2119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin(12119903119909) 119889119909

(52)

For the second integral a change of variable (119910 = 12119909) isused in combination with themultiple-angle formulas for thehyperbolic sine and cosine functions respectively to get

cosh (2119909) + 1sinh (2119909)

= coth (119909) (53)

and hence

119868 (119903) = minus1

119903int

infin

0

( (119892 (119909) minus 119892 (2119909)) coth (119909)

+119892 (119909)

sinh (119909)) sin (119903119909) 119889119909

(54)

Next since

119892 (2119909) =2119892 (119909)

119890119909 + 1 (55)

119868(119903) can be simplified to

119868 (119903) = minus1

119903int

infin

0

( coth (119909) coth(1199092)

+1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(56)

The relation

coth (119909) coth(1199092) =

cosh (119909)cosh (119909) + 1

= 1 minus1

cosh (119909) + 1(57)

shows that 119868(119903) = minus119878(119903)+119869(119903)The final formula follows from

(120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

=1205872

sinh2 (120587119903)(cosh2 (1

2120587119903) minus 1)

(58)

and the half-angle formula

cosh (12119909) = radic

cosh (119909) + 12

(59)

Corollary 5 It holds that

int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) 119909 119889119909

=7

2120577 (3) minus 3120577 (2) asymp minus07276

(60)

int

infin

0

(1

cosh (119909) + 1+ coth (119909)) 119892 (119909) 119909 119889119909

=3

2120577 (3) + 120577 (2) asymp 34480

(61)

Proof We have lim119903rarr0

(119878lowast

(119903)+119878(119903)) = 72120577(3) Additionallyusing similar arguments as in Corollary 3 it can be shownthat

lim119903rarr0

(120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903))) = 3120577 (2) (62)

This proves the first expression The second follows fromcombining (38) and (60)

The next assertion provides a recursion formula for theMathieu series Surprisingly it is also a consequence ofProposition 4

Corollary 6 For 119903 gt 0 define the function 119891(119903) as

119891 (119903) =120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903) (63)

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Applied Mathematics 5

the last integral may be written in the form

int

infin

0

ln (cosh (119909) minus 1) 119892 (119909) 119889119909

= 120577 (3) minus 2120577 (2) + int

infin

0

(119909

2)

2 ln (cosh (119909) minus 1)cosh (119909) minus 1

119889119909

(46)

A further interesting consequence of the previous theoremis the following new integral-type representation for thealternating Mathieu series 119878lowast(119903)

Proposition 4 The alternating Mathieu series 119878lowast(119903) may berepresented as

119878lowast

(119903) = minus119878 (119903) +120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903)

(47)

with

119869 (119903) =1

119903int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(48)

Proof We start with the identity

119878lowast

(119903) = 119878 (119903) minus1

4119878 (

119903

2) (49)

In view of (15) and applying the relation

coth(1205871199032) = coth (120587119903) + 1

sinh (120587119903) (50)

we obtain

119878lowast

(119903) =120587

2119903coth (120587119903) minus 120587

2119903

1

sinh (120587119903)

+ (120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

+ 119868 (119903)

(51)

where

119868 (119903) = minus1

119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin (119903119909) 119889119909

+1

2119903int

infin

0

cosh (119909) + 1sinh (119909)

119892 (119909) sin(12119903119909) 119889119909

(52)

For the second integral a change of variable (119910 = 12119909) isused in combination with themultiple-angle formulas for thehyperbolic sine and cosine functions respectively to get

cosh (2119909) + 1sinh (2119909)

= coth (119909) (53)

and hence

119868 (119903) = minus1

119903int

infin

0

( (119892 (119909) minus 119892 (2119909)) coth (119909)

+119892 (119909)

sinh (119909)) sin (119903119909) 119889119909

(54)

Next since

119892 (2119909) =2119892 (119909)

119890119909 + 1 (55)

119868(119903) can be simplified to

119868 (119903) = minus1

119903int

infin

0

( coth (119909) coth(1199092)

+1

sinh (119909)) 119892 (119909) sin (119903119909) 119889119909

(56)

The relation

coth (119909) coth(1199092) =

cosh (119909)cosh (119909) + 1

= 1 minus1

cosh (119909) + 1(57)

shows that 119868(119903) = minus119878(119903)+119869(119903)The final formula follows from

(120587

2 sinh ((12)120587119903))

2

minus (120587

sinh (120587119903))

2

=1205872

sinh2 (120587119903)(cosh2 (1

2120587119903) minus 1)

(58)

and the half-angle formula

cosh (12119909) = radic

cosh (119909) + 12

(59)

Corollary 5 It holds that

int

infin

0

(1

cosh (119909) + 1minus

1

sinh (119909)) 119892 (119909) 119909 119889119909

=7

2120577 (3) minus 3120577 (2) asymp minus07276

(60)

int

infin

0

(1

cosh (119909) + 1+ coth (119909)) 119892 (119909) 119909 119889119909

=3

2120577 (3) + 120577 (2) asymp 34480

(61)

Proof We have lim119903rarr0

(119878lowast

(119903)+119878(119903)) = 72120577(3) Additionallyusing similar arguments as in Corollary 3 it can be shownthat

lim119903rarr0

(120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903))) = 3120577 (2) (62)

This proves the first expression The second follows fromcombining (38) and (60)

The next assertion provides a recursion formula for theMathieu series Surprisingly it is also a consequence ofProposition 4

Corollary 6 For 119903 gt 0 define the function 119891(119903) as

119891 (119903) =120587

2

cosh (120587119903) minus 1sinh (120587119903)

(1

119903+

120587

sinh (120587119903)) + 119869 (119903) (63)

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 ISRN Applied Mathematics

where 119869(119903) is given in (48)Then for 119899 ⩾ 1 one has the recursionformula

119878 (119903

2119899) = 23119899

119878 (119903) minus

119899minus1

sum

119894=0

23119894+2

119891(119903

2119899minus(119894+1)) (64)

Proof Combine (47) and (49) to get

119878 (119903

2) = 8119878 (119903) minus 4119891 (119903) (65)

Successive repetition of the identity establishes the statedformula

3 Evaluation of a Related Series

It is interesting to mention that (18) may be used to prove aresult for a closed-form evaluation of an infinite series that isrelated to 119878(119903) More precisely if we define the function 119879(119903)for 119903 isin R by the infinite series

119879 (119903) =

infin

sum

119899=1

1198992

minus 1199032

(1198992 + 1199032)2 119879 (0) = 120577 (2) =

1205872

6 (66)

then 119878(119903) and 119879(119903) are connected via the following relation-ship Consider the (complex) function

119892 (119911) =

infin

sum

119899=0

1

(119899 + 119911)2= (

119889

119889119911)

2

ln Γ (119911) 119911 = 0 minus1 minus2

(67)

with Γ(119911) being the Gamma function Note first that

infin

sum

119899=1

1

(119899 + 119894119903)2=

infin

sum

119899=1

(119899 minus 119894119903)2

(1198992 + 1199032)2= 119879 (119903) minus 119894119903119878 (119903) (68)

Comparing the two equations we see that

119879 (119903) minus 119894119903119878 (119903) = 119892 (1 + 119894119903) = 119892 (119894119903) +1

1199032 (69)

The function 119879(119903) admits an exact evaluation (see [7]) forwhich we can provide a new elementary proof applyingthe analytical structure of 119878(119903) However in contrary to thepervious section where we have focused on integral-typeexpressions we change the point of view and work withsummation-type representations of 119878(119903)

Proposition 7 The function 119879(119903) can be evaluated exactly as

119879 (119903) =1

21199032(1 minus (

120587119903

sinh (120587119903))

2

) 119903 = 0 (70)

Proof Once more let 119892(119909) = 119909(119890119909

minus 1) We start with (18)and (26) Since

1

119890119909 minus 1=

infin

sum

119899=1

119890minus119899119909

119909 gt 0 (71)

we have1

(119890119909 minus 1)2=

infin

sum

119899=1

119899119890minus(119899+1)119909

119909 gt 0

119892 (119909)

119909minus1198922

(119909)

119909=

infin

sum

119899=1

119899119890minus119899119909

minus

infin

sum

119899=1

119899119890minus(119899+1)119909

minus 119909

infin

sum

119899=1

119899119890minus(119899+1)119909

(72)

Thus after interchanging summation and integration andapplying (14)

int

infin

0

(119892 (119909)

119909minus1198922

(119909)

119909) cos (119903119909) 119889119909 = 119862 (73)

where

119862 =

infin

sum

119899=1

1198992

1198992 + 1199032minus

infin

sum

119899=1

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus

infin

sum

119899=1

119899(119899 + 1)

2

minus 1199032

((119899 + 1)2

+ 1199032)2

(74)

This gives

119878 (119903) =

infin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2+1

1199032minus

1

21199034+1

2(

120587

119903 sinh (120587119903))

2

+1

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(75)

From the identityinfin

sum

119899=1

119899

((119899 + 1)2

+ 1199032)2=1

2119878 (119903) minus

infin

sum

119899=1

1

(1198992 + 1199032)2 (76)

we arrive at

119878 (119903) = minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032

minus119899(119899 + 1)

2

((119899 + 1)2

+ 1199032)2)

(77)

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Applied Mathematics 7

Next we simplify the sums in the above equation for 119878(119903)further Using the fact that 119899(119899 + 1) = (119899 + 1)

2

minus (119899 + 1) itis straightforward to show that

infin

sum

119899=1

(1198992

1198992 + 1199032minus

119899 (119899 + 1)

(119899 + 1)2

+ 1199032)

= 1199032

infin

sum

119899=1

(1

(119899 + 1)2

+ 1199032minus

1

1198992 + 1199032)

+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

= minus1199032

1 + 1199032+

infin

sum

119899=1

119899 + 1

(119899 + 1)2

+ 1199032

(78)

The last identity is true since the first sum on the right-handside of the equality telescopes This gives

119878 (119903) = minus2

1 + 1199032minus

infin

sum

119899=1

2

(1198992 + 1199032)2+2

1199032minus1

1199034

+ (120587

119903 sinh (120587119903))

2

minus2

1199032 (1 + 1199032)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(79)

which may be written as

119878 (119903) =

infin

sum

119899=1

minus2

(1198992 + 1199032)2+2

1199032(minus

1

21199032+1

2(

120587

sinh (120587119903))

2

)

+2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2)

(80)

Finally notice that

2

1199032

infin

sum

119899=1

(119899

1198992 + 1199032minus

119899(119899 + 1)2

((119899 + 1)2

+ 1199032)2) =

2

1199032

infin

sum

119899=1

1198991199032

+ 1198992

(1198992 + 1199032)2

(81)

This leads to canceling out 119878(119903) in (80) and the proof iscomplete

Remark 8 Using the further observation that1198922(119909)119909may beexpressed as

1198922

(119909)

119909= 119909119890minus2119909

(1 +119892 (119909)

119909+ 119890119909

(119892 (119909)

119909)

2

)

=119909119890minus119909

119890119909 minus 1+

119909119890minus119909

(119890119909 minus 1)2

(82)

it is possible to derive the following summation-type form of119878(119903)

119878 (119903) =

minus (1 + 1199034

)

(1199032 (1 + 1199032))2+ (

120587

119903 sinh (120587119903))

2

+2

1199032

infin

sum

119899=1

(119899 minus 1

1198992 + 1199032minus

119899(119899 + 2)2

((119899 + 2)2

+ 1199032)2)

(83)

4 Conclusion

In this paper new expressions for the Mathieu series werederived in terms of hyperbolic functions To derive thenew identities a basic property of the function 119909(119890

119909

minus 1)

was utilized Using a particular identity it was possible toprove a new relationship between the Mathieu series andthe alternating variant Also a new recursion formula andsome interesting closed-form evaluations of definite integralsinvolving hyperbolic functions were established Finally anew elementary proof for an evaluation of an infinite seriesrelated to the Mathieu series was presented

It would be interesting to knowwhether the new identitiescan be used to derive new (double) inequalities for the series

Disclaimer

The statements and conclusions made in this paper areentirely those of the author They do not necessarily reflectthe views of LBBW

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author thanks two anonymous referees for a carefulreading of the paper and valuable comments that helped toimprove the content of the paper

References

[1] L Berg ldquoUber eine Abschatzung von MathieurdquoMathematischeNachrichten vol 7 pp 257ndash259 1952 (German)

[2] E Makai ldquoOn the inequality of Mathieurdquo Publicationes Mathe-maticae Debrecen vol 5 pp 204ndash205 1957

[3] H Alzer J L Brenner and O G Ruehr ldquoOn Mathieursquosinequalityrdquo Journal of Mathematical Analysis and Applicationsvol 218 no 2 pp 607ndash610 1998

[4] F Qi C-P Chen and B-N Guo ldquoNotes on double inequalitiesof Mathieursquos seriesrdquo International Journal of Mathematics andMathematical Sciences no 16 pp 2547ndash2554 2005

[5] A Hoorfar and F Qi ldquoSome new bounds for Mathieursquos seriesrdquoAbstract and Applied Analysis vol 2007 Article ID 94854 10pages 2007

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 ISRN Applied Mathematics

[6] A Elbert ldquoAsymptotic expansion and continued fraction forMathieursquos seriesrdquo Periodica Mathematica Hungarica vol 13 no1 pp 1ndash8 1982

[7] D C Russell ldquoA note on Mathieursquos inequalityrdquo AequationesMathematicae vol 36 no 2-3 pp 294ndash302 1988

[8] P Cerone and C T Lenard ldquoOn integral forms of generalisedMathieu seriesrdquo Journal of Inequalities in Pure and AppliedMathematics vol 4 no 5 2003

[9] H M Srivastava and Z Tomovski ldquoSome problems andsolutions involving Mathieursquos series and its generalizationsrdquoJournal of Inequalities in Pure and Applied Mathematics vol 5no 2 2004

[10] T K Pogany and Z Tomovski ldquoOn multiple generalizedMathieu seriesrdquo Integral Transforms and Special Functions vol17 no 4 pp 285ndash293 2006

[11] T K Pogany HM Srivastava and Z Tomovski ldquoSome familiesof Mathieu a-series and alternating Mathieu a-seriesrdquo AppliedMathematics and Computation vol 173 no 1 pp 69ndash108 2006

[12] T K Pogany and Z Tomovski ldquoBounds improvement for alter-nating Mathieu type seriesrdquo Journal of Mathematical Inequali-ties vol 4 no 3 pp 315ndash324 2010

[13] G V Milovanovic and T K Pogany ldquoNew integral forms ofgeneralized Mathieu series and related applicationsrdquo ApplicableAnalysis and Discrete Mathematics vol 7 no 1 pp 180ndash1922013

[14] I S Gradshteyn and I M Ryzhik Table of Integrals Series andProducts Elsevier Academic Press Amsterdam The Nether-lands 7th edition 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of