ROMANIAN MATHEMATICAL MAGAZINE · R M M ROMANIAN MATHEMATICAL MAGAZINE Founding Editor DANIEL SITARU Available online ISSN-L 2501-0099 RMM - Calculus Marathon 801 - 900

R M MROMANIAN MATHEMATICAL MAGAZINE

Founding EditorFounding EditorDANIEL SITARUDANIEL SITARU

Available onlineAvailable onlinewww.ssmrmh.rowww.ssmrmh.ro

ISSN-L 2501-0099ISSN-L 2501-0099

RMM - Calculus Marathon 801 - 900RMM - Calculus Marathon 801 - 900

www.ssmrmh.ro

1

RMM

CALCULUS

MARATHON

801-900

www.ssmrmh.ro

2

Proposed by

Daniel Sitaru – Romania, Srinivasa Raghava-AIRMC-India

Mokhtar Khassani-Mostaganem-Algerie

Naren Bhandari-Bajura-Nepal,Rahim Shahbazov-Baku-Azerbaijan

Mohammed Bouras-Morocco,Abdul Mukhtar-Nigeria

Vasile Mircea Popa-Romania,Ekpo Samuel-Nigeria

Jalil Hajimir-Toronto-Canada,Florică Anastase-Romania

Ahmed Hegazi-Cairo-Egypt,Radu Diaconu-Romania

Igor Soposki-Skopje-Macedonia, Dinu Șerbănescu-Romania

Abdul Hafeez Ayinde-Nigeria, Prem Kumar-India

www.ssmrmh.ro

3

Solutions by

Daniel Sitaru – Romania,Nassim Nicholas Taleb-USA,Soumava Chakraborty-Kolkata-India

Abdul Hafeez Ayinde-Nigeria, Mokhtar Khassani-Mostaganem-Algerie, Ahmed Salama

Hegazy-Cairo-Egypt, Sergio Esteban-Argentina,Gabriel Ruddy Cruz Mendez-Lima-Peru

Gerald Bernabel-Lima-Peru, Dawid Bialek-Poland, Kamel Benaicha-Algiers-Algerie, Pedro

Nagasava-Brazil,Ravi Prakash-New Delhi-India, Timson Azeez Folorunsho-Nigeria

Surjeet Singhania-India, Remus Florin Stanca-Romania,Igor Soposki-Skopje-Macedonia,

Kartick Chandra Betal-India,Yen Tung Chung-Taichung-Taiwan, Kartick Chandra Betal-

India,Florentin Vișescu-Romania , George Florin Șerban-Romania,Florică Anastase-

Romania, Avishek Mitra-West Bengal-India,Ovwie Edafe-Nigeria, Samir HajAli-Damascus-

Syria, Zaharia Burghelea-Romania, Tobi Joshua-Nigeria,Jalil Hajimir-Canada, Sanong

Huayrerai-Nakon Pathom-Thailand, Ali Jaffal-Lebanon, Lazaros Zachariadis-Thessaloniki-

Greece,Yuval Peres-USA, Chris Kyriazis-Athens-Greece, Adrian Popa – Romania, Khaled Abd

Imouti-Damascus-Syria, Soumitra Mandal-Chandar Nagore-India,Tran Hong-Dong Thap-

Vietnam, Sujit Bhowmick-North Bengal-India, Nelson Javier Villaherrera Lopez-El

Salvador,Marian Ursărescu – Romania, Naren Bhandari-Bajura-Nepal, Artan Ajredini-

Presheva-Serbie

www.ssmrmh.ro

4

801. Prove that:

(

*

( )

G: Catalan’s constant

Proposed by Mokhtar Khassani-Mostaganem-Algerie

Solution by Abdul Hafeez Ayinde-Nigeria

Let .

/

. / .

/ .

/

. / .

/ .

/∑ ( ) .

/ .

/ .

/

. / .

/ .

/

( )

∑ ( ) .

/ .

/ .

/

. / .

/ .

/

( )

∑ ( ) .

/ .

/ .

/

. / .

/ .

/

( )

∑

. / .

/ .

/

. / .

/ .

/

( )

∑

. /

. / .

/ .

/

( )

∑

( )

. / .

/.

/ .

/ .

/

∑( )

( )( )( )( )( )

∑( )

( ) ( )( )( )

www.ssmrmh.ro

5

(

∑

( )

∑

( )

( )

∑

( )

∑

( )

∑

( )

+

{

(∑

( )

( )

+

∑

( )

(∑

( )

+

(∑

( )

+}

{

(

*}

{

}

(

*

(

*

Catalan’s constant.

802. Let’s define the function

( ) ∫

( )

Prove that:

∫( (

* ( )* ( )

Proposed by Srinivasa Raghava-AIRMC-India

Solution by Mokhtar Khassani-Mostaganem-Algerie

( ) ∫

∫

∫

( ) ∫ ∑

www.ssmrmh.ro

6

( ) ∑∫ ( )

( ) ∑8 ( )

( ) ( )

9

( ) ∑(

( )

( )

( ) *

(

* (

* (

* ( ) (

*

(

* ( ) ( )

( ) ( ) (

* ( ) ( )

Now:

∫( (

* * ( )

∫( (

* * ( )

∫( ( ) ( ) ( )

( ) ( )

( )

( ) ( )+

( )

∫ ( ) ( )

∫ ( )∑( )

∑

∫( )

∑

( )

∑(

*

∑(

( )

( )

*

www.ssmrmh.ro

7

∑(

*

( ) ( )

∫ ( )

∫ ( )

⏞

{ ( )}

∫ ( )

( ) ( )

( ) ( ) ( )

∫ ( ) ( )

⏞

{ ( )( )}

⏞

∫(

( )+

∑∫ ( )

∑(

( )

( )

( ) *

( ) ( ) ( ) ( )

(Answer)

803.

(√ √ √ √ √ +

:√ √√ (√ √ ) √ √ ;

Proposed by Naren Bhandari-Bajura-Nepal

Solution 1 by Ahmed Salama Hegazy-Cairo-Egypt

( ) ( ) ( ) ( ) ( )

Now we find ( ), let

( ) ( ) ( )

( )

www.ssmrmh.ro

8

( ) ( )

( ( ) ) ( )

√

√

, and √

√

since

√ √

√ √ , put

√

√

√

√ , by substitution we have

√ √ √ √ √

Solution 2 by Sergio Esteban-Argentina

Calculo del geometricamente. Por triangulos notables sabemos

i) Por Angulo inscrito sabemos que

www.ssmrmh.ro

9

y

ii) Sabemos que:

√ √ √ √ √ √ √ y

Usando el teorema de Ptolomeo

(√ √ )(√ ) (√ √ ) 4√ √ 5

(√ √ )(√ ) (√ √ ).√ √ /

Entonces:

√ √ √ √ √

Preliminares

i) Sabemos que el triangulo equilatero

www.ssmrmh.ro

10

Aplicando Pitagoras: ( ) √

ii) Trabajando i)

Prolongamos hasta tai que . Si damos (√ √ )

Entonces

iii) en el decagon regular

www.ssmrmh.ro

11

medida de

Se traza la bisectriz interior

Por el teorema de la bisectriz interior

. Operando .

√

/. Entonces:

: 4√

5;

√ √ si damos

www.ssmrmh.ro

12

804. Find:

Proposed by Rahim Shahbazov-Baku-Azerbaijan

Solution 1 by Gabriel Ruddy Cruz Mendez-Lima-Peru

Sabemos: ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) }

Sumando verticalmente: ∑ ( ( )) ∑ ( ( ))

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

∑ , ( )-

4 ⏟

5

∑ , ( )-

∑ ( ( ))

∑ ( ( ))

www.ssmrmh.ro

13

Solution 2 by Gerald Bernabel-Lima-Peru

∑ ( )

∑ ( )

( ) . / ( ) .

/ ( ) .

/ ( ) .

/ ( ) .

/ ( )

. / ( ) .

/ ( ) .

/ ( )

Si: ( )

( ) 4. / ( ) .

/ ( ) .

/ ( ) .

/5

. / .

/ ( ) .

/ ( ) .

/ ( ) .

/ ( )

Si: ( ) entonces la ecuacion: . / .

/ .

/ .

/

Presenta 22 raices de la forma ( ) donde * +

∑ ( )

. /

. /

805. Prove that:

∑ ( )

( )( )( )

4 ( )

5 .

/√ :

. /

√ ; 4

( )

5

4 ( )

5 ( )

Euler-Mascheroni constant; exponential integral

imaginary error function


www.ssmrmh.ro

14

Solution by Dawid Bialek-Poland

∑ ( )

( )( )( )

( )

∑(

*

( )( )

∑(

*

⏟

∑.

/

( )

⏟

∑.

/

( )

⏟

(1)

∑.

/

( ) .

/ .

/ .

( )

/ .

( )

/ (2)

Where (1) ( ) ( ) ∑

– exponential integral

∑.

/

( )

∑

.

/

( )

∑

.

/

( )

( ) (

*

. ( )

/ ( ) (3)

Where (II)

∑

∑

∑

( )

∑(

*

( )

∑

4√

5

( )

√

∑

4√

5

( )

√

[

∑

4√

5

( )

√

]

√ ∑

4√

5

( )

( )

√ √

4√

5

.

/ √ 4

. /

√ 5 .

/√ 4

.

/

√ 5 (4)

Where (III) ( )

√ ∑

( )

( )

( ) - imaginary function

Rewriting (1) with (2), (3) and (4) we get:

∑ ( )

( )( )( )

4 ( )

5 4

( )

5

4 ( )

5 ( ) .

/√ :

. /

√ ;

www.ssmrmh.ro

15

4 ( )

5 .

/√ :

. /

√ ; 4

( )

5

4 ( )

5 ( )

806.

∫√ √

√ √ √ √ √ √ √ √

⏟

– fibonacci number;

Prove that

is faster to give golden number then

Proposed by Mohammed Bouras-Morocco

Solution by Kamel Benaicha-Algiers-Algerie

∫√ √

√ √ √ √ √ √ √

Put: ( ) √ √ √ √ (1)

and ( ) √ √ √ √ √ √ √ √ √

So, ( ) ( ) ( ) (2)

We have: ( ) √ ( ) (a)

( ) ( ) √ ( ) ( ) ( )

( )

√ ( ) ( ) ( ( ))4

( )5

√ ( ) ( ) (3)

Put ( ) √ √ ( )

increasing and we have: , - ( )

www.ssmrmh.ro

16

√ √ ( ) √ √ √ √ √ ( √ )

( ) ( ) and then ( ( )) is increasing, ( ) ( ) (4)

( ) , by using (3) and (4), then

So, ( ( )) and ( )

is solution of equation (result (a)) with condion

So, ( ) (b)

∫ ( )

( )

∫ ( )

(using result (2))

, - ( ) then by using second Middle value theory:

- , such that ∫ ( )

( ) ( ) ∫ ( )

( )∫ ( )

∫ ( )

( ) (using result (b))

. We have

( ) - ,⁄

Put

where ( ) is the Fibonacci sequence.

Put ( )

is decreasing ( ) is fluctuate sequence.

By comparison of fluctuate sequence ( ) and increasing sequence ( ( ))

then

is faster than

to . denote the golden ration:

√

807.

∫


Solution by Pedro Nagasava-Brazil

∫

( )

( )

www.ssmrmh.ro

17

∫

( )

( ) ( ) ∫

∫

( )( )( )

( )( )

∫

6. /

7

6. /

. /

7

(

*

∫

∫

< .

/

=

∫

808. Define for all ( ) √ √ √ √ ( )

⏟

then prove that

∫ ( )

( )

√ √ √ √ √

√ √


Solution by Sergio Esteban-Argentina

www.ssmrmh.ro

18

i) Probamos que si , -, se cumpie que

√ √ √ √ (

*

Demonstration:

√ √ ( ) √

| (

*|

Lado que

.

/

√ √ √ (

*

Reemplarando lo intenso:

√ √ √ √ (

* (

*

Analogamente

√ √ √ √

⏟

(

*

en el problema ( ) (1)

ii) Probamos que (2)

Es facel probar que

Entonces: ⏟

⏟

⏟

Nota:

www.ssmrmh.ro

19

Usando (1) y (2) en la integral

∫

∫

∫ ( )

(

∫

∫

)

( |

|

,

(

(

*

*

(

*

Resolvemos .

/. Consideramos el triangulo notable de y

i)

ii)

√[ (√ √ )] (√ √ )

www.ssmrmh.ro

20

√ (√ √ ) (√ √ ) (√ √ )

√ √ √

SABEMOS que √ √ √ √ esto se deduce de (√ √ )

. Entonces

(

* (

*

(√ √ )

√ √ √ √ √ √ √

√ √

Por lo tanto

.

/

√ √ √ √ √

√ √

(√ √ √ √ √

√ √ +

809. Prove that Catalan’s constant:

∑ ∑( )

( )

∑∑( )

( ) ∑ ∑

( )

( )



∑∑( ) ( ) ( )

( )

∑ ∑( )

( )

We calculate

∑∑( ) ( ( ) )∫ ( )( )

∫∑∑( ) ( ( ) ) ( )

∫ ( )

∑( ) [∑ ∑( )

]

∫ ( )

4

5 [

]

www.ssmrmh.ro

21

∫ ( )

( )( )( )

∫ ( ) [

( )

( )]

∫ ( )

∫ ( )

∫ ( )

Now, we calculate

∑∑( )

( )

∑∑( )

∫ ( )

∫ ( )

∑

∑( )

∫ ( )

.

/ (

*

∫ ( ) [

( )

( )]

∫ ( )

∫ ( )

∫ ( )

810. Prove that:

∫ .

/ ( )

( )

Proposed by Abdul Mukhtar-Nigeria


∫ .

/ ( )

∫ ( )

( )

∫ ( )

( )

( )

( ) .

/ ( )

( )

www.ssmrmh.ro

22

∫ ( )

( )

∑

∫ ( )

⏞

∑

:[

( )] |

∫

; ∑

[

] |

∑

⏞

( ) ( )

∫ ( )

( )

⏞

∫ ( )

( )

⏞( )

∫

( )

∑

∫ ( )

⏞

∑

:6

( )7 |

∫ ( )

;

∑

∫ ( )

⏞

∑

:6

( )7 |

∫

;

∑

( ) 46

( )7 |

5

∑ ( )

⏞

∑ ( )

∑

( )

∑

∑

⏞

( )

( ) ( )

( ) ( ) Where:

( )∑

( ) ∑

( )

( ) ∑

( )

Rewriting (1) with (2) and (3), we get

∫ .

/ ( )

( )

www.ssmrmh.ro

23

811. Solve for :

∫. ( )/

Proposed by Daniel Sitaru – Romania

Solution 1 by Kamel Benaicha-Algiers-Algerie

∫ ( )( ( ))

(E); for

( ) ∫ ( )( ( ))

∫ ( )( ( ))

∫ ( ) ( )

( ( ))

∫

( )

( )

|

( )

( ) ( )

(E) ( ) ( )

( ) ( ) ( )

bijection function

So: (E) ( ) ( ) ( ) ( )

Solution 2 by Ravi Prakash-New Delhi-India

Let ∫

( )dt ∫

.

/ ∫

Let ( )( )

( )

∫

( )

∫ . Thus,

]

∫

∫

Thus, the given equation becomes

or or

Solution 3 by Timson Azeez Folorunsho-Nigeria

∫ . ( ) ( )/

put

www.ssmrmh.ro

24

∫ . ( )/

∫ ( ( ))

∫ ( ( ) )

∫ ( ( ) )

∫ ( ( ) )

Put

∫ ( ( ))

∫

|

|

( )|

( )

( ) is also a solution

812. Solve for real numbers:

∫(

*

(

*



∫ ( )

( )

∫ ( )

( ( ) *

Put: ( )

, so

( )

∫

( )

|

( )

( ) |

Then, if ∫ ( )

( )

.

/ (E)

(E) ( )

( )

( )

( )

( ) ( )

www.ssmrmh.ro

25

Put: ( ) ( )

- , ( )

( )

. So: ( )

( )

for and for: . So: ( )

- ,

is the maximum of

( )

; ( ) unique value for real numbers

( ) * +

Solution 2 by Surjeet Singhania-India

∫ ( )

( )

(

* ∫

( )

( )

Put ( )

( )

∫

( )

∫

( )( )

(

*

(

*

(

*

Comparing with other side we get

Solution 3 by Remus Florin Stanca-Romania

∫ ( )

( ) ∫

( )

( ) 4(

( )*

5

∫ ( )

( )

(

( )*

Let

( )

( )

( ) ∫

( )

( ) ∫

∫ ( )

( )( )

∫(

*

(|

|*

(|

( )

( )

|,

As we know that ( )

∫ ( )

( )

4

( )

( )5

4

( )

( )5

(

*

( )

( )

( ) ( ) ( ) ( )

www.ssmrmh.ro

26

( ) ( )

Let ( ) ( )

( )

( )

and we have that for ( ) and for

( ) and ( )

on ( ) the function is increasing ( )

and for , ) the function is

decreasing with

.

813. ( )

Find:

( ) ∫4( ) ( ) ( ) ( )

( ( ))( ( ))5


Solution by Igor Soposki-Skopje-Macedonia

( ) ∫( ) ( )

( )( )

( )( ) ( )( )

( )

( )

( ) ( ) ( )

( )

( ) ∫

⏟

( ) ( )

( ) ⏟

( ) ∫

|

|

( ) |

| |

|

www.ssmrmh.ro

27

( )

( ) ( )

( )

2

( )

( ) | ( ) ( )

( ) ( ) |

814. Find:

∫

(

( )√( )( ) 4 4√

5 5

)


Solution by Igor Soposki-Skopje-Macedonia

:√

; ∫

( )√( )( ) 4 √ 5

4√ 5

√

( )

( )

www.ssmrmh.ro

28

√

( )

( )

√

( )

( )

√( ) . /

( )

√( )( )

( )√( )( ) ∫

( ) ∫

( )

2

3 ∫

∫

∫

( )

4 √ 5

815. Given that:

( ) ∫4( )√

( )√ 5

Prove or disprove that:

( ) ∫ ( ) √ 4 4√

√ 55

Proposed by Ekpo Samuel-Nigeria


( ) ∫

( )√ √ ∫

√

√ √

√

(√ )

√

√

( ) ∫ ( ) ∫√

√

⏞ √ √

√ ∫ √

√

( )

√ 4√

√ 5 √ 4

√

√ 5

Note: ( )

√ ∫

and ( ) ( )

www.ssmrmh.ro

29

816. Prove that for:

∫ .

/

√

( )


Solution 1 by Kartick Chandra Betal-India

∫ .

/

√ ( )

( ) ∫ .

/

√ ( )

( ) ∫

√ ( )

(

( ) *

( )

∫

√

( )

∫( )

( ) ∫

( ) ∫( )

( )

∫ ( )

( )

Solution 2 by Yen Tung Chung-Taichung-Taiwan

∫ .

/

√

⏟

∫ .

/

√

∫ (

*

∫ ( )

⏟ ( )

∫ ( )

∫

( )

www.ssmrmh.ro

30

817. Find:

( ) ∫ (

( )( )*

Proposed by Vasile Mircea Popa-Romania


∫

( )( )

∫4 ( )

( )( )

( )

( )( )

( )( )5

( )

( )

( )

∫ ( )

∫:

;

( ) ∫

. /

(

*

( ) (

√ * (

√ *

∫ ( )

∫:

( )

;

( ) ∫

. /

(

*

( ) (

√ * (

√ *

{

( )

( )( ( ) (

√ * (

√ **

( )( ( ) (

√ * (

√ **

}

www.ssmrmh.ro

31

( )

√ ( )

( )

√ ( )

(

√ *4

( )

√ ( )

( )

√ ( )5

818. Show that:

∫ ( ) ( ( )) ( ( ( )) ( ))



∫ ( ) ( ( ) )

∫ ( ) ( )

∫

∫

∫

now let:

So: ∫

| |

( (

*+ (

( )

*

819. Find:

∫4 ( )

√ 5


Solution 1 by Abdul Hafeez Ayinde-Nigeria

∫ ( )

√ ∫ ( ) ( √ ) √ ( )

√ ∫√

√ ( )

√ ∑( )

∫√

www.ssmrmh.ro

32

√ ( )

√ ∑( )

∫

√ ( ) √

√ ∑

( )

. /

√ ( ) √

√ ∑

( )

. /

√ ( ) √

√ ∑( ) .

/

. /

√ ( ) √

√ . /

. /∑( ) .

/

. /

√ ( ) √

√ ∑( ) .

/

. /

√ ( ) √

√ (

*

Where ( ) is the classic hypergeometric function.

Solution 2 by Mokhtar Khassani-Mostaganem-Algerie

∫ ( )

√

⏞

{ √ ( )}

√ ∫√

⏞

√ ( )

√ ∫

√ √ ( )

√ (

* √ ( )

√ (

*

820. Find a closed form:

∫4 ( )√

√ 5


www.ssmrmh.ro

33


∫ ( )√

√

. Let , then:

∫

( )

( )

∫

( )

( )

( ) ∫

( )

( )

Integration by parts gives:

∫

( )

∫

( )

(

*

∫

∫

(

*

∫(

* ( )

∫

∫

: ∫

∫

;

(

*

(

*

(

*

( (

* (

*+

By using Gauss theorem:

(

* (

* √ 4

√

√ 5

:√ √

√ √

√

√ ;

(

* (

* √ (√ )

:√ √

√ √

√

√ ;

www.ssmrmh.ro

34

√

(√ )

:√ √

√ √

√

√ ;

√ (√ )

√

So: ∫ ( )√

√

√ (√ )

√


∫ ( )√

√

∫( )

√

∫ ( )

( )

∫ (

*

∫(

*

{

( ) } ∫

( )

( )

∑ ( )

( ) ∫(

*

∑ ( )

( ) <

=

∑( )

( )

. / .

/

∑( ) ( )

( ) ( )

∑( ) ( )

∑( )

8

9

∑( )

∑( )

∑( )

( )( )

∑( )

>

?

∑( )

∑( )

www.ssmrmh.ro

35

∑( )

∑( )

∑( )

∫

∫

∑( )

∫

( )

∫

∫

∫>

.

/

?

∫

.

/

. /

(√ )

∫

. /

. /

(√ )

√ 2

3

√ 6 |

√

√ |7

√

√ | √

√ |

√

√ (√ )

821. Find without softs:

∫∫

( )

Proposed by Jalil Hajimir-Toronto-Canada

Solution by Daniel Sitaru-Romania

∫(

( ) *

⏞

∫4( )

( ) 5

∫4( )

( ) 5

∫4( )

( ) 5

www.ssmrmh.ro

36

∫4 ( )

( ) 5 ∫

∫:∫(

( ) *

;

∫.

/

822. For , we have:

∫

( ) √ ( )


Solution 1 by Florentin Vișescu-Romania

∫

√

∫

√

( ) ∫

√

∫

√

∫

√

√

∫ ( √ )

( √ )

∫ ( √ )

√

∫

|

Solution 2 by Tobi Joshua-Nigeria

∫

√

www.ssmrmh.ro

37

∫ ( √ )

( √ )( √ )

∫ ( √ )

∫

∫

∫ √

(odd fxn) ∫

(odd fxn)

∫

(even function);

1

823. Prove that:

∫∫(( ) ( ) ( ) ( ))

( ( ))

( )



∫∫( ) ( ) ( ) ( )

( ( ))

∫∫( ) ( ) ( )( ( ) ( ))

( ( ) ( ))

∫∫

( ) ( ( ) ( ))

∫∫( ) ( )

( )

( ( ) ( ))

∫

( )4 ( ) ( )

5 ∫∫

( ) ( )

( )

( ( ) ( ))

www.ssmrmh.ro

38

∫ ∫( ) ( )

( )

( ( ) ( ))

∫

( ) 4 ( ) ( )

5

∫∫( ) ( )

( )

( ( ) ( ))

∫ ∫( ) ( )

( )

( ( ) ( ))

(1)

( {( ) ⁄

}

*( ) + {( )

}

,

∫ ( ) ∫

( )

, ( ) ( )-

∫ ( )

[

( )

( )]

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫

( )

∫

( )

( ) (2)

∫ ∫( ) ( )

( )

( ( ) ( ))

∫ ∫ ( )

( )

( ( ) ( ))

www.ssmrmh.ro

39

∫ ( )

∫( ( ) ( ( )))

∫ ( )

( )

( ( ( ) ) ( ))=

∫ ( )

[ ( )

( ( ))]

∫ : ( )

( ) ( )

( )

∫

( )

( )

;

∫ ( )

( )

∫

( )

( )

(

* ∫

( )

∫ ( )

( ) (3)

(1), (2) and (3)

( )

∫∫( ) ( ) ( ) ( )

( ( ))

( )

824. Evaluate the integral

If ∫ ∫ 4

. /

. /5

Proposed by Abdul Hafeez Ayinde-Nigeria

Solution by Pedro Nagasava-Brazil

1. Separating variables:

∫ .

/

∫

2. For the variable, let’s apply Frullani integral:

www.ssmrmh.ro

40

∫

.

/ .

/ ( )

3. For the variable, let :

∫ .

/

Now, let’s apply Differentiation Under the Integral Sign:

( ) ∫ (

*

( )

∫ .

/

(1)

Let

:

( )

∫

.

/

( ) ∫

.

/

(2)

With (1) (2):

( ) ∫ (

* .

/

√ ( ) √

( )

∫

4

5

√

6

7

∫ 4

5

√

6

7

Hence: ∫ .

/

√

0

1

√

Finally:

∫ ∫

4

. /

. /5

√

( )

825. If , then 0

1 such that:

www.ssmrmh.ro

41

∫

( )( )

( )(

*

Proposed by Florică Anastase-Romania

Solution 1 by George Florin Șerban-Romania

From Osiann-Bonet formula we have:

, - , - , -

∫ ( ) ( ) ( )∫ ( ) ( )∫ ( )

∫

( )( )

∫

( )( )

, - ( )

( )

( )

, -

, - ( )

, -

( )∫ ( ) ( ) ∫ ( )

∫ ( ) ∫

∫

. /

(

*

www.ssmrmh.ro

42

∫ ( )

∫ ( )

(

*

( )(

*

Solution 2 by proposer

Theorem (Bonnet-Weierstrass):

If , - decreasing function of class and , - continuous function,

then , -

∫ ( ) ( ) ( )∫ ( ) ( )∫ ( )

Demonstration:

Let , - ( ) ( ) ( ) ( ) , -

From theorem 2 of means , - such that:

∫ ( ) ( ) ( )∫ ( )

∫ ( )( ( ) ( )) ( ( ) ( ))∫ ( )

∫ ( ) ( ) ( )∫ ( ) ( ( ) ( ))∫ ( )

( )∫ ( ) ( )∫ ( )

q.e.d. Let 0

1 ( )

( )

( )

( )

( ) ∫

∫

∫

. /

0

1

www.ssmrmh.ro

43

∫

( )( ) ( )( ( ) ( )) .

/ ( ( ) ( ))

(

*

( )(

*

826. Find:

∫ ( ( ))


Solution by Avishek Mitra-West Bengal-India

∫

( ) ∫ [ ∑ ( )

]

∫

∑

∫

( )

∫

( ) 6 ( )

7

∫

( )

∫

( )

[

8 ( )

9

∫

( )

]

www.ssmrmh.ro

44

[

( )

∫

( )

]

( )

[

8 ( )

9

∫

( )

]

( )

[

∫ ( )

]

( )

[

8 ( )

9

∫ ( )

]

( )

6

( )

* ( )+

7

( )

( )

6

7

∑( )

∑( )

( )

( )

( ) ( )

( ) ( )

( )

( )

( )

( )

827. Evaluate the following integral:

www.ssmrmh.ro

45

∫ ( ) ( )

( )

Proposed by Ahmed Hegazi-Cairo-Egypt


∫ ( )

( )

[

]

∫ ( )

( )

<∫ ( )

( )

∫ ( )

=

∫ ( )

, ( ) ( ) ( )- ( ) ( ) ( ) (Answer)

828. Prove that:

∫ (( )

( ) *



∫ (( ) ( ) *

∫ ( )

( )∫

( )

Put:

∑

∫

( )

( )

∑

( )

( ( ) )|

www.ssmrmh.ro

46

( )

( )

∫ (( )

( ) *

829. Let: ( ) ∑

Prove that:

∫ 4

√( ( ) ( ))5



∑

∑ . /

(

*

( ) | |

Now: ∫ .

√ ( ) ( )/

∫ 4√

5

⏞

8 4√

59

√

∫

√ ( )

√

:∫

√

∫

√ ( )

;

8√ 4√

59

830. Find:

∫ ( ) ( )

Proposed by Abdul Mukhtar-Nigeria

Solution 1 by Avishek Mitra-West Bengal-India

∫ ( )

www.ssmrmh.ro

47

6 ( ) ∫

7

∫ ( )

( )

6∫

7

[

( ) ]

∫ ( )

( )

∫ ( )

( )

∑

∫

∑

( ) ( )

( )

∑

( )

∑ ( )

( )

( )

[∑ ( )( )

∑

( )

]

0∑

∑

1 , ( ) ( ) ( ) ( )- ( ) ( ) ( ) (Answer)

Solution 2 by Ovwie Edafe-Nigeria

∫ ( )

6 ( )

7

∫ ( )

∑

∫

∫

( )

( )

∑

( )

∑

∑

∑

, ( ) ( ) ( )- ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

∑

.

/ ( )

∑ ( )

( )

www.ssmrmh.ro

48

831. Prove that:

∫ ( ) ( )

(

* ( ) ( ) ( )

( )



∫ ( ) ( )

∫ ( ) ( )

( ) ∫

( )

⏟

∫

( ) .

/

⏟

(1)

( ) ∫

( )

( ) (2)

∫ ( ) .

/

∑

∫ ( )

∑

< ( )

∫ ( )

=

( )∑

∑

∫ ( )

( ) ( ) ∑

< ( )

∫

=

( ) ( ) ( )∑

∑

[

]

( ) ( ) ( ) ( ) ∑

∑

( ) ( ) ( ) ( ) ( ) ∑

.

/

( ) ( ) ( ) ( ) ( ) .

/ (3)

Rewriting (1) with (2) and (3), we get:

www.ssmrmh.ro

49

( )∫

( )

∫ ( ) .

/

( )

( ) ( ) ( ) ( ) ( ) (

*

Note: ( ) ∑

( ) ∑

832. ( )

∏ ( √

√ *

Prove that:

∫

( )(

*


Solution by Samir HajAli-Damascus-Syria

( )

∏ 4 √

√

5

∏ : ( )

( )

;

∏ 4 ( )

5

( ( ))

( )

( )

∫

( )

(

* ∫

( )

(

*

∫ ( )

∫ ( )

∫ ( )

∑∫ ( )

∑

( )

∑

( )

∫ ( )

∑∫ ( )

∑

( )

𝛀 𝝅𝟐

𝟐𝟒 𝟏

𝟑 𝝅𝟐

𝟖 𝝅𝟐

𝟏𝟐

www.ssmrmh.ro

50

833. For any complex numbers with Re( ), Re( )

If

( ) ∫( )

( )

then show that

:∫ ( )

;

:∫ ( )

;

4

( )

( )5


Solution by Nassim Nicholas Taleb-USA

∫ (

*

, -

We substitute , so ∫4

5 , -

We have

∫

, -

, -

, ( )- .

/, where is the

Laplace transform.

From tables, , ( )-( )

, so

, and reversing the order of

integration – deriv: ∫

[ ] |

[ ] |

[ ]

[ ]

6

7

Repeating for and summing gets the required result.

834. Find:

∫4

( ) ( )5

Proposed by Radu Diaconu-Romania

www.ssmrmh.ro

51

Solution by Zaharia Burghelea-Romania

Let

( )

∫

( ) ( )

∫ .

/

( ) . /

. /

( )

∫( ) .

/

, -,( ) ( )-

∫

( ) ( )

∫

( ) ( )

∫

( ) ( )

√ ( ) 4

( )

√ ( )5|

√ ( ) (

√ ( )

+

835. Prove that:

∫

4

5

( )

( )


Solution by Zaharia Burghelea-Romania

∫

4

5

⏞

∫ (

*

∫ 4

5

∫ (

*

∫ 4

5

⏞

∫

∫

∫

( )

∫ (

*

⏞

∫ ( )

√

⏞

∫ ( )

( )

www.ssmrmh.ro

52

( )

( )

836. Find:

∫ ( ) ( ) , n



( )

∫

∫

∫

∫

( )

837.

∫

Proposed by Igor Soposki-Skopje-Macedonia

Solution by Tobi Joshua-Nigeria

∫

∫

∫

∫

∫

∫

( )

( ) ( )

∫ ( )

( ) ( )

6

∫

( )

∫

∫( )

( )7

www.ssmrmh.ro

53

[

∫

( )

. /

4√ 5

∫

. /

4√ 5

∫(

*

( *

]

[

√ (

√ *

√ (

√ *

∫

. /

. /

(√ )

]

6

√ (

√ *

√ (

√ *

√ 4

√

√ 5 7

6

√ (

√ *

√ (

√ *

√ 4

√

√ 5 7

6

√ (

√ *

√ (

√ *

√ 4

√

√ 5 7

838. Find:

( ) ∫ (

( )( )*


Solution by Kamel Benaicha-Algeirs-Algerie

( ) ∫ (

( )( )*

∫ (

( )( )*

( ):∫

∫ 4

5

;

( )4

4

√ 5 | 5

| |

( )

( ) ∫

∫

( ) {

| |

( )

www.ssmrmh.ro

54

839. Let be positive and continuous when . Prove:

:∫ ( )

;

∫( ( ))

Proposed by Jalil Hajimir-Canada

Solution by Soumitra Mandal-Chandar Nagore-India

is a positive continuous function on , -

( ) for all , - hence ∫ ( )

Now ∫ ( )

.∫ ( )

/

. Let ∫ ( )

We have to prove, ( ) ( )

( )( ) ( ) ( )

Hence .∫ ( )

/

.∫ ( )

/

∫ ( )

:∫ ( )

;

:∫ ( )

;

∫ ( )

:∫ ( )

;

840. If

then:

∫∫( ( )

( )*

√

( )



Denote

( )

( ) ( )

www.ssmrmh.ro

55

⏞

.

/ ⏞

.

/

. /

.

/

√

∫∫4 ( )

( )5

∫∫4 √

5

√

:∫

;

√

( )

Equality holds for .

841. If

then:

∫∫4 ( )

( )5

√ ( )


Solution 1 by Jalil Hajimir-Canada

( )

√

√

if

∫∫

( )

√

∫∫

√

( )

( ) ( )

Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand

For and

, we have as following:

1. since ( )

Hence ( ( )) √

and since

( ( ))

( )

Hence

( ( ))

( ) ( √ )

( )

www.ssmrmh.ro

56

Hence √

( ( ))

( )

( √ )

Hence ( )

( ) √

Hence ∫ ∫

( )

( ) ∫ ∫

√

√

|

√

( )( ). Therefore, it is true.

Solution 3 by Ali Jaffal-Lebanon

Let ( ) 8

1

1

we have ( ) 0

1

( )

Let ( ) 0

1 ( )

then ( )

and

( )

( )

So, ( ) 0

1 then ( ) 0

1 so, is concave on 0

1

Let 0

1 where we have .

/

( ) ( ) ( )

so,

.

/

√

by GM-AM inequality we have:

(

*

(

*

4 √

5

√

So, ∫ ∫

( )

∫ ∫

√

( ) √

www.ssmrmh.ro

57

842. If then:

∫:∫:∫(√( )( )( )

( )( )( )+

;

;

(

*


Solution 1 by Lazaros Zachariadis-Thessaloniki-Greece

√∏( )

∏( )

∏( )

( )

so, ∫ ∫ ∫√∏( )

∏( )

∫ ∫ ∫

∫∫

( )

( )

( ) ( )

( )

. . //

Solution 2 by Sanong Hauyrerai-Nakon Pathom-Thailand

For , we have ( ) ( ) ( )

and ( )

and since ( )

( )

( )

Hence ( ) ( ) ( ) ( )

( )

( )

( )

( )

( )

√( )

( )

√ ( )

. Hence for

√( )( )( )

( )( )( )

√

(( ) )

Hence for

∫:∫:∫√( )( )( )

( ) ( )( )

;

;

www.ssmrmh.ro

58

∫:∫:∫

;

;

|

( )( )( )

.

/

. Therefore, it is true.


Let √( )( )( )

( )( )( ) since √

√

√

,( )( )( )-

∫∫∫

∫∫∫

( )

843. If (, -) | ( )| , - then:

∫ ( )

:∫ ( )

;

Proposed by Jalil Hajimir-Canada, Dinu Șerbănescu-Romania


( ) ( ) ⏞

( )( )

( ( )) ( ) ( ( )) ( ) ( )

( ) ( ( ) ( )) ( )

∫∫( )

∫∫( ( ) ( ))

∫∫ ( ) ∫∫ ( ) ( ) ∫∫ ( )

∫ ( ) ∫ ( ) ∫ ( ) ∫ ( )

www.ssmrmh.ro

59

∫ ( ) :∫ ( )

;

∫ ( ) :∫ ( )

;

844. If then:

∫∫ (

*

( )∫

∫∫ (

*

( )∫



Let ( ) , where - ,

( )

( )

( )

Let ( )

( ) ( ) has the same sign as since for all

. We have 0

1 [.

/

] . So,

( )

( )

then ( ) for all then ( ) and is convex.

By Jensen’s inequality: .

/

( ) ( )

∫∫ (

*

( )

∫( )

( )

∫ ( )

www.ssmrmh.ro

60

4 ( )

( )

5:∫( )

;

( )∫

( )∫

Therefore: ∫ ∫ .

/

( ) ∫

( )∫

∫∫ (

*

Solution 2 by Soumitra Mandal-Chandar Nagore-India

Let ( ) for all

( )

( )

( )

Let ( ) for all .

Then ( ) {.

/

} for all . Hence is an increasing function

( ) ( ) then ( ) for all .

Hence is a convex function, so by definition:

( )

( ) .

/

( )

( ) (

* (

*

.

/ .

/ [generalizing by ]

(

* (

*

( )∫

∫∫ (

*

( ) ∫

∫ ∫ .

/

. Proved.

845. If then:

www.ssmrmh.ro

61

∫

∫

∫

√ ( )


Solution 1 by Lazaros Zachariadis-Thessaloniki-Greece

( ) ( )

( ) convex, so:

∫ ( )

(

*

( )

Thus ∑ ( )

√ ( )

( )

( )

√ ( )

( )

( )

√ ( )

√ ( )


Let ( ) , , ( )

( )

( )

then is convex on , , so,

∫ ( )

( ) .

/ for all then

∫ ( )

∫ ( )

∫ ( )

(

* (

* (

*

√ .

/ .

/ .

/

by GM-AM inequality

√ ( ) ( ) ( )

√ ( ) ( ) ( )

but ( ) ( ) ( )

(( ) ( ) ( ))

since ( ) is convex on , , then

√ ( ) ( ) ( )

√ ( )

Therefore:

∫

∫

∫

√ ( )

www.ssmrmh.ro

62

846. If then:

∫ ∫ ∫ :√

√

√

;



√

√

√

√

( )

√

( )

√

⏞

( ) √

(

√

)

(

√

)

(

√

)

( ) √

( ) ( )

( ) √

( ) ( )

⏞

( ) ( ) ( ) ( )

( )

( )

√

√

√

∫ ∫ ∫ :√

√

√

;

∫ ∫ ∫

( )( )( )

847. If then:

∫ ∫ ∫ :√ √

√ √ √ ; √


www.ssmrmh.ro

63


( ) ( ) ( ) . By Popoviciu’s inequality:

∑ ( )

(

*

∑ (

*

∑

∑

√

(√ √ √ )

√

√ √ √ √

√

√ √ √ √

∫ ∫ ∫ :√ √

√ √ √ ; ∫ ∫ ∫ √

√


848. If

then:

( ) ( ) :∫

; ∫

. /

:∫

;:∫

;



We have ( ) is continuous and convex function on . Since ( ) for

all then

:

∫ (

*

;

∫ ( (

**

But . .

//

. So, .

∫ .

/

/

∫

We have ( ) ( ) .∫

/

∫ .

/

( ) ( ) :∫

;(

*∫

www.ssmrmh.ro

64

( ) :∫

;:∫

;

But ( ) .∫

/

4∫ .

/

.

/

5

By Cauchy-Schwarz .∫

/.∫

/

Therefore: ( ) .∫

/ .∫

/ .∫

/ .∫

/

Therefore ( ) ( ) .∫

/

∫ .

/

∫

∫

Solution 2 by Soumava Chakraborty-Kolkata-India

( ) ( ):∫

; ∫

. /

( )

:∫

;:∫

;

The continuous analogue of AM-GM inequality:

∫

( )

( ) ∫

. /

and

.

/ ∫

∫

( )

(1).(2) .∫

/ .∫

/ ( )

( ) .∫

/

∫ .

/

Again, continuous analogue of reverse CBS inequality

:∫

;:∫

; ( )

:∫√

√

;

( )

(3) (4) RHS of (a) LHS of (a) (a) is true (Proved)

849. If , - ( ), continuous, then:

∫∫:√ ( ) ( )

√ ( ) ( );

( )∫ ( )


www.ssmrmh.ro

65

Solution 1 by Yuval Peres-USA

It suffices to verify that , -

√ ( ) ( )

√ ( ) ( )

( ) ( ) (*)

and then integrate both sides. Squaring, (*) is equivalent to:

( ) ( )

( ) ( ) √

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

that is: √, ( ) ( )- ( ) ( ) [ ( ) ( )] ( ) ( )

This is just the AM-GM inequality √

Solution 2 by Chris Kyriazis-Athens-Greece

It holds that √ ( ) ( )

( ) ( )

and √ ( ) ( )

( ) ( )

So, adding those inequalities, we have:

√ ( ) ( )

√ ( ) ( ) ( ) ( )

Integrating from to , we have:

∫∫<√ ( ) ( )

√ ( ) ( )=

∫∫, ( ) ( )-

( )∫ ( )

(*)

(*) ∫ 2∫ , ( ) ( )-

3

∫ 0∫ ( )

( )( )1

∫ ( )

( ) ∫ ( )

( ) ( )∫ ( )


We know for any two positive real numbers and

( ) ( )

www.ssmrmh.ro

66

∫∫[ ( ( ) ( )) ( ( ) ( )) ]

∫∫( ( ) ( ))

∫∫√ ( ) ( )

√ ( ) ( ) ( )∫ ( )

850. If then:

∫∫∫ √ (

*

( ) ∫ √


Solution by Ali Jaffal-Lebanon

Let ( ) ( )

( )

( )

( )

( )

(

* ( )

we have ( ) for all , - then is concave on , -

Let , -, so, by Jensen’s inequality we have:

(

*

( ) ( ) ( )

∫∫∫ (

*

∫∫∫ ( ) ( ) ( )

∫∫∫ ( ) ( ) ( )

∫

∫

∫ ( )

∫

∫

∫ ( )

∫

∫

∫ ( )

www.ssmrmh.ro

67

( ) ∫ ( )

( ) ∫ ( )

( )

∫ ( )

( )( )

∫ ( )

( ) ∫ ( )

. So,

∫∫∫ (

*

( ) ∫ ( )

Therefore: ∫ ∫ ∫ √ .

/

( ) ∫ √

851. If – continuous, ( ) ( ) ( ) then:

∫∫∫ ( )

∫ ( )


Solution by Adrian Popa – Romania

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

Similarly: ( ) ( ) ( ) ( )

( )

( ) ( ) ( )

So, ( ) ( ( ) ( ) ( ))

∫∫∫ ( )

∫∫∫( ( ) ( ) ( ))

∫ ( )

∫ ( )

∫ ( )

852. ( ) ∫ (

*

( ) ∫ (

*

www.ssmrmh.ro

68

Prove that:

( ) ( )

0

1


Solution by Șerban George Florin – Romania

√

( )(

)

√

( )(

) √

( )( ) √

√

because: ( ) , - and

( ) , -

( ) ( ) ∫(

+

∫√

∫

|

|

0

1 , - and

because , - ( ) ( )

, true, because

853. Prove without softs:

∫∫∫ √

( )( )


Solution by Soumava Chakraborty-Kolkata-India

√ √

www.ssmrmh.ro

69

∫∫∫ √

( )( )

⏞( )

∫∫∫

( )( )

∫

.

/

√

∫

√

∫√

⏞( )

∫√

∫√ ∫√

∫(√ √ ) √ ∫( )

√

√ ∫ ( )

√ ( ) √ ∫

√ ( )

⏞( )

√ ( )

∫(√ √ ) √ ∫( )

√

√ ∫( )

√( ) √ ∫

√ ( )

⏞( )

√ | √ |

( ) ( ) ∫√

√ ( ) √ | √ |

∫√

√ .

/

√ |

√

| √ ( ) √ | √ |

√ (√ ) √ .

/

∫√

⏞( )√

4

(√ )5

www.ssmrmh.ro

70

( ) ( ) ∫

⏞( )√

4

(√ )5

∫

∫( )

∫

∫

⏞( )

( ) ( ) ∫∫∫ √

( )( )

4√

54

(√ )5∫∫

√ 4

(√ )5 (

*∫

√

4

(√ )5 ( )

( )

854. Let be continuous on , -. If ( ) ( ) , -, prove:

∫ ( )


Solution by Khaled Abd Imouti-Damascus-Syria

Suppose: √ . So: (

√ *

(

√ *

So:

√

√ ( ) ( )

√

For: , -. Then: ( ) ( )

∫ ( )

∫ ( )

2

8

∫ ( )

∫ ( )

{

{

www.ssmrmh.ro

71

∫ ( )

∫ ( )

∫ ( )

∫

, -

∫ ( )

855. If ( ) then:

∫∫(( ( ) ( )) 4 ( ) ( )

5+

( )∫( ( ) ( ( )))


Solution by Florentin Vișescu-Romania

( ) ( )

( )

( )

( )

( )

( ) convexe .

/

( ) ( )

( )

( )

( ) ( )

∫∫, ( ) ( )-

( ) ( )

∫∫ ( )

( ) ( ) ( )

∫∫( ( ) ( ))

( ) ( )

( )∫ ( )

( )


∫ ( )

www.ssmrmh.ro

72



, - ( ) ⏞

(

*

( ( )) (

*

∫ ( ) ∫(

*

4(

*

(

*

5

∫ ( )

(

*

857. If .

/ continuous then:

∫ ( ( ))

∫ 4

( )5

( )


Solution by Khaled Abd Imouti-Damascus-Syria

∫( ( ( )) 4

( )5

( )

Let be the function: ( ) ( ) .

/ 1

0

( )

.

/ 1

0

, ( )-

, ( )-

( )

.

/ ( )

.

/

( )

( )

So: ∫ ( ( ( )))

(

( )* ∫

( )

www.ssmrmh.ro

73


∫√ ( )

( )

√



∫4 ( )

( ) 5

⏞

∫4( ) ( )

( ) 5 ( )

∫4( ) ( )

( ) 5

∫4( ) ( )

( ) 5

∫4( ( ) ) ( )

( ) 5

∫( ( ))

∫:√ ( )

( ) ;

⏞

√ ∫

√


∫ (( ) ( )

)



[

] [

]

√

∫ (( ) ( )

)

⏞

www.ssmrmh.ro

74

∫ (( √ )

( √ )

* ⏞

∫ 4√( √ )

5

⏞

∫ 4√( √ )

5

∫ 4√(√ ) 5

∫ (√ )

√ √

(

*

√

√

860. If then:

:∫

;:∫

; :∫

;:∫

;

Proposed by Daniel Sitaru-Romania


:∫

;:∫

; :∫( )

;:∫(

)

;

⏞

:∫( )

;:∫( )

; :∫

;:∫

;


Solution 2 by Sergio Esteban-Argentina

Let ( ) , - since is positive and nonincreasing, it follows that

∫<∫ ( )

( )( (

) ( )

) =

we would just have to prove that:

(

) ( )

www.ssmrmh.ro

75

The function ( ) is a convex on , so according to Karamata’s inequality it

suffices to prove that without loss of generality such that ( )

majorizes ( )

i) and ( ) ( )

ii) ( ) ( ) true! Expanding , we obtain

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

∫ ( )

the proof is now complete.

861. Let , - be continuos and increasing. Prove that:

∫∫∫4 ( ) ( ) ( )

5 ( ) ∫ ( )



Let , -. WLOG: .

increasing ( ) ( ) ( ) ( ) ( ) ( )

By Cebyshev’s inequality: ( ) ( ) ( )

( )( ( ) ( ) ( ))

( ) ( ) ( )

( ( ) ( ) ( ))

∫∫∫4 ( ) ( ) ( )

5

∫∫∫4

( ( ) ( ) ( ))5

www.ssmrmh.ro

76

∫∫∫ ( ) ( ) ∫ ( )

862. If ( ) – continuous, then:

∫∫

(

(

( ( ))( ( ))

( ( ) ( )

*

)

)

( )∫ ( )

:∫ ( )

;


Solution 1 by Adrian Popa-Romania

( ( ))( ( ))

4 ( ) ( )

5

4 ( ) ( )

5

( ( ))( ( ))

( ( ) ( )

*

( ( ))( ( ))

( ( ) ( )

*

∫ ∫ ( ( ))( ( ))

. ( ) ( )

/

(1)

:∫( ( ) )

;

∫ ( )

∫

|

∫ ( )

( )∫ ( )

( )∫ ( )

.∫ ( )

/

(2)

From (1) and (2) ∫ ∫ ( ( ))( ( ))

. ( ) ( )

/

( )∫ ( )

.∫ ( )

/


Let ( ) ( ) for all then ( )

( )

( )

hence is convex function ( ) ( ( )) ( ) ( ( ))

4 ( ) ( )

5

4 ( ) ( )

5

( ) ( )

4 ( ) ( )

5

( ( ))( ( ))

4

( ) ( )

5

www.ssmrmh.ro

77

( ) ( ) ( ) ( )

( ( ))( ( ))

( ( ) ( )

*

∫∫ ( )

∫∫ ( )

:∫ ( )

;:∫ ( )

;

∫∫ ( ( ))( ( ))

( ( ) ( )

*

( )∫ ( )

:∫ ( )

;

∫∫ ( ( ))( ( ))

( ( ) ( )

*

863. Prove:

∫√

( )


Solution by Soumitra Mandal-Chandar Nagore-India

∫√ ( )

∫√

(

*

∫√

(

*

<

√ ( )=

<

√ ( )=

. Again,

√ ( ) for all

Let ( ) for all

( ) ( ) ( ) ( )

for ( )

( )

www.ssmrmh.ro

78

.

/ √ hence has local maximum at

So,

( )

√

√

√ √

√ √

( )

√ ( )

√ √

√ ∫

∫√ ( )

√ √ ∫√ ( )

√ ∫ √

( )

√ √

. Hence proved

864. , - ( ) – continuous. Find: such that:

∫∫∫.( ( ) ( )) ( )( ( ) ( ))

( )( ( ) ( ))

( )/

∫ ( )


Solution by Tran Hong-Dong Thap-Vietnam

Let ( ) ( ) ( )

We choose such that . Using Jensen’s Inequality:

( ) ( ) ( ) ( , -)

( ) ( ) ( ) ( , -) ( )

( )

( )

∫∫∫( ( ) ( )) ( )( ( ) ( ))

( )( ( ) ( ))

( )

∫∫∫, ( ) ( ) ( )-

<>∫ ( )

∫

∫

? >∫ ( )

∫

∫

? >∫ ( )

∫

∫

?=

www.ssmrmh.ro

79

<∫ ( )

∫ ( )

∫ ( )

=

∫ ( )

∫ ( )

865. If * + then:

( ) ∫(∏ ( )

+

∏∫ ( )



( )

√ ∫

( )

√

[for all ]

( ) in , - * +

Also, we know for ( ) ( ) for any in the interval

, - * + if ( ) ( ) , -

( ) ( ) is monotonically increasing in , - * +

also ∫ ( )

∫ ( )

[ ( )

√ ]

[ ( ) ( )

√ ]

( ) ( ) is the integrable between , - * +

( ) ∫ ( )

( ) ( )

∫ ( )

∫ ( )

∫ ( )

[Let us denote ( ) ( ) ( ) ( ) ( ) ( )]

( ) ∫ ( )

( ) ( ) ∫ ( )

∫ ( )

∫ ( )

www.ssmrmh.ro

80

( ) ∫∏ ( )

∏∫ ( )

866. If , - ( ) – continuous then:

:∫√ ( )

;

∫ ( )

:∫ √ ( )

;



Using Cauchy – Schwarz inequality:

:∫√ ( )

;

:∫ √ ( )

;

∫

∫ ( )

∫ ( )

:∫√ ( )

;

:∫ √ ( )

√ ( )

√ ( )

;

∫ 0√ ( )

1

∫ 0√ ( )

1

∫ 0√ ( )

1

:∫ √ ( )

;

:∫√ ( )

;

∫ ( )

:∫ √ ( )

;

867. If then:

∫ (

)

∫

:∫

;



We can notice that: ( ) ( ) and ( )

www.ssmrmh.ro

81

as ( )

( ) is strictly convex. By integral form of Jensen’s inequality

∫

:∫ ( )

; ∫

:∫

;

868. If

then:

( ) ∫

( )


Solution by Florentin Vișescu-Romania

.

/ ( )

( )

( )

( )

( ) .

/

( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

. So, has a root in .

/

( )

( )

. So ( ) and has a root in

.

/

www.ssmrmh.ro

82

( )

( )

So, ( ) and has a root in ( ). It doesn’t matter where is towards

( )

( )

So, ( ) .

/. Then:

∫

( )

( )

( ) ∫

( )

869. If , - has a continuous second derivative and ( ) ( )

then:

( ( )) ∫( ( ))



:∫( ( ))

;:∫

; ⏞

:∫ ( )

;

:∫( ( ))

;

: ( ) ( ) ∫ ( )

;

( ( ) : ( ) ( ) ∫ ( )

;,

: ( ) ( ) ∫ ( )

;

( ( ) ( ) ( ) ( ( ) ( ))) ( ( ))

www.ssmrmh.ro

83

:∫( ( ))

; ( ( ))

870. If then:

∫∫

√ ∫∫

√

(

*



√ √

( ) ( )

√ ( )

( )

( )

( ) ( )

√ √

∫∫

( )

√ ∫∫

√

( )

∫∫(

*

∫∫

∫∫

6

7 , -

6

7 , -

( )( ) (

*( )

(

*( )

( )

871. If

then:

www.ssmrmh.ro

84

∫∫( ( )

( )*

√ ( )



( ) √ ( )

( ) √ ( ) √

4

√

5 ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) √

Hence f .

/

√

√

( )

( ) , ( )-

( ) √

. .

/ ( ) /. Hence

∫∫( ( )

( )*

∫∫4 √

5

√ ( )

Proved. Equality for a=b.


∫( ( ))



( ) ( ) ( )

( )

( ) ⏞

We compute: .

/ .

/ ( )

www.ssmrmh.ro

85

( ) ( ) ( )

( ) ( ) .

/

( )

∫

∫ ( )

∫

∫ ( )

873. If then:

∫

( )

( )( )


Solution 1 by Sujit Bhowmick-North Bengal-India

( )

∫

( )

( )( )

∫

( ) ( )

⁄

( )

:

( ) ( ) ( ) ( ( ))

;

( ) ( ( ))

∫

||

( )

( )

( )

( )

www.ssmrmh.ro

86

Solution 2 by Ravi Prakash-New Delhi-India

( )

( )( )

( ( ))

( )

4 ( ( ))( ) ( ( ))

( )5

( )

4 ( )

5

∫ ( )

( )( )

∫

4 ( )

5

( )

| ( )

⁄

∫

( )

( )( )

( )

( )

874. If , - continuous then:

∫∫∫ √( ( ) ( ) ( ) ( ) ( ) ( ))

∫ ( )


Solution by Florentin Vişescu-Romania

( )( )

( ) ( ) ( )

We must show: ( ) ( ) ( )

( ) ( ( ) ( ) ) (( ) ( ))

( ) ( )

( ) ( )

( ) ( ) ( )

www.ssmrmh.ro

87

( ) ( ) true from:

⇒ {

( ) ( ) ( ). So,

( ) ( )

∫∫∫ √( ( ) ( ) ( ) ( ) ( ) ( ))

∫∫∫( ( ) ( ) ( ))

∫ ( )

875. Evaluate the following sum:

∑(

( )

( )

( )

( ) *

Proposed by Prem Kumar-India

Solution 1 by Avishek Mitra-West Bengal-India

∑

( )

∑

. /

(

*

( )

∑

( )

∑

( )

( )

( ) ( )

∑

( )

∑

. /

(

*

( )

∑

( )

∑

( )

www.ssmrmh.ro

88

Solution 2 by Nelson Javier Villaherrera Lopez-El Salvador

∑[

( )

( )

( )

( ) ]

∑[

( )

( )

( )

( ) ]

∑{

, ( ) -

, ( ) -

, ( ) -

( ) }

∑[

( )

( )

( )

]

∑[

( )

( ) ]

[∑

( )

∑

]

∑( )

( )

[∑

∑

( )

∑

]

(∑

∑

+

∑

( )

( ) ∑

876. Find:

:∑∑∑∑

( )( )( )( )

;



Let ∑ ∑ ∑ ∑

( )( )( )( )

∑∑∑

( )( )( )

www.ssmrmh.ro

89

∑∑

( ) ( )

∑∑

( )( )

∑

( )

We will prove that:

In first note that there are of permutations of

( )

( )

( )

( ) let’s

discuss the 04 cases:

Case 01: if all 4! combinations are then contained with in

Case 02: if and are distinct we need 6 of because some possibilities are

contained in

Case 03: if we need 3 of because some possibilities are contained in

and

Case 03: if we need 8 of because some possibilities are contained in

and

Case 03: this appears exactly once in every sum, which thus us 12

copies of

Since: ∑

( )

∑

( )

∑

∑

( )

( ) ( )

( ( ) )

∑

( )

∑(

( ) *

∑(

( ) *

∑

( )

∑(

( ) *

∑(

( ) *

∑(

( ) *

( ) ( ) ( ) ( )


∑∑∑∑

( )( )( )( )

www.ssmrmh.ro

90

∑∑∑∑

( )( )( )

{

}

∑∑∑

( )( )( )

(

*

∑∑

( )( )

. ( ) /

∑∑

( )

{ ( )

( )

} ∑∑

( )

{

}

∑

( )

∑{( ( )

( )

+

( ) }

∑

( )

(

*

∑

( )

< ( )

∑{

( )}

2 ( ) 3= ∑

( )

∑

( )

[ ( )

( ) ] ( )

∑

( )

[

( )

( ) ]

∑

( )

∑ ( )

( )

∑ ( )

( )

, * ( ) +- ∑{

( ) }

( ) ∑{

( )

( )

}

∑{

( )

( )

}

∑ ( )

( )

∑[{ ( )

( )

}

( ) ]

∑ ( )

www.ssmrmh.ro

91

{ ∑(

*

∑

( )

} ∑ ( )

∑ ( )

* ( ) + ∑ ( )

877. Find:

:∫4

5

;


Solution 1 by Naren Bhandari-Bajura-Nepal

Here

( ) ∫

∫(∑

4

5

+

∑

( )

and hence we have the limit: ( ( )) . .∑

( ) //

(

( ∑

( )

++

:

(( ) ∑

( )

+

;

(

(

** ( ) √

Solution 2 by Abdul Hafeez Ayinde-Nigeria

Let ∫

and ( )

∫

∑

∫

www.ssmrmh.ro

92

(∑

( )

+

(∑

( )

+

( ∑

( )

+

∑( )

(∑

( )

+

{∑( )

(

*

}

8

(

*

9

{

( )}

{

( )}

878. Find:

(

:∑√

∑

√

∑(√ √ )

√

;,


Solution by Remus Florin Stanca – Romania

∑√

∑

√

∑(

√ √ *

√ (√ √ )

√ (√ √ )

√ √ √ (

√

√ * √ (

√

√ * √

√

∑ √

∑ √

www.ssmrmh.ro

93

∑ :√

√

;

∑ 4 √

√ 5

∑(√ √ )

√

( )

∑

(√ √ )

√

∑(√ √ )

√

∑√

∑

√

∑(√ √ )

√

∑√

∑

√

∑(√ √ )

√

∑(√ √ )

√

∑(√ √ )

√

:∑√

∑

√

∑(√ √ )

√

;

879. Find:

(

∫ 4

5

√ ∑ .

/

∫ 4

5

√ ∑ .

/

)



Let’s compute ∫

and ∫

∫

∫

∫( )( )

∫( )( )

∫( )

So,

www.ssmrmh.ro

94

“ ”

√ ∑

(

√ ∑

* (

√ ∑

*

√ ∑

.

√ ∑

/ .

√ ∑

/ (1)

Let’s compute

√ ∑

∑

(√ )

∑

(√ )

(√ )

( ) ( ) ( )

( )

(√ )

4

√ 5

(

.

/

√ )

( )

.

/ .

/


We have: ∑

∑

∑

. So,

∑

( )

( )

( )

Where ( )

So,

√ ∑

√

√

( )

( )

Then,

√ ∑

tends to

when

and

√ ∑

√

( )

Let

then

www.ssmrmh.ro

95

∫

∫

∫

∫

∫

∫

∫( )( )

∫

∫( )

Let ∫

and ∫ ( )

∫

We have

∫

∫( )( )

∫( )

So,

then

Therefore

880. Find:

(. /

:

∑:

. /

.

/;

;,


www.ssmrmh.ro

96


Let ∑. /

.

/

∑ .

/

( ) ( )

( )

( )∑. /

( ) ( )

( )

( )∑. /

∫

( )

( )∫ ( )

∑. /

( )

( )∫ ( )

( ) ( ) ( )

( )

( )

( ) ( )( )

( )

( ) ( )( )

( ) ( )( )

Now: (. /

*

( )

( ) ( )( )


Let ( )

∑

∑

( ) ( )

( )

∑

( ) ( )

( ) ( )

∑

∫

( )

∫∑ ( )

( )

∫( )

( )

∫ ( )

www.ssmrmh.ro

97

( ) ∫ ( )

Let ∫ ( )

( ) ( )

( )

( ) ( )

( )

( )

So, . Then

881.

(

(

∑

,

(

∑

+

)

(

(

∑

+

(

∑

,

)

Find:

(∏

+



Let

∑

∑

and √∏

By GM-AM inequality: then: ( ) ( )

( )

We have ( ) ( )

. So,

( ) ( ) (*)

( ) ( ) (**)

where ( ) ( )

by (*) and (**) we have: ( ) ( )

So: ( ) then ( )

And also, by (*) and (**) we have:

www.ssmrmh.ro

98

( ) ( )( ) ( ) ( ) ( )

then ( ) ( ) ( ) ( ) ( )

( )

( )

So, then ( ) but ( )

then

( ) . we have ( )

( ) ( )

then by Sandwich theorem: ( ) ( )

( ) then

( ) and ∏

882. If , different in pairs, then:

( ) ( ) ( ) :∏:√

√

;

;

( ) ∑(

4

5

+



( ) ∑

4

5

∑

4

5

4

5

(

*

4

√ 5 :√

√

;

∑ ( )

∑ :√

√

;

.

/

∑ :√

√

;

( )

www.ssmrmh.ro

99

∑ ( )

∑ :√

√

;

Solution 2 by Adrian Popa-Romania

( )

?

|∫

( ) ∑

( )

( )

∑

4(

*

5

4

( )

( ) 5

( )

( )

4

√ 5

√ 4

√

√ √

√ 5

We must prove that: ∑ (√

√

√

√ * ∑ (√

√

*

⏟

(

:√

√

;, (

:√

√

;,

:√

√

;

(√

√

* (√

√

*

(True) because (1)

Similarly: (

(√

√

*+ (√

√

* (2) and

(

4√

√

5+ 4√

√

5 (3)

(1) (2) (3) ( ) ( ) ( ) (∏(√

√

**


( ) ∑ (

.

/

* , let

, we also know that

www.ssmrmh.ro

100

because , so (1)

( ) ∑4

5

∑4∫4

5

5

∑(∫ *

∫(∑

+ ∫

4 ( )

5 ∫

∫

(| |)

( )

4

5

.

( ) / (

√

* because

( ) 4 √

5 ( ) ( ) ( )

∑ 4 √

5

(

∏ 4√

√ 5

)

( ) (∏ (√

√

* * (2)

we know that ( ) ( )

:∏:√

√

;

; ( ) :∏:√

√

;

; ( )

( ) ( ) ( ) :∏:√

√

;

;

883. Find:

(

(( )

( ) ( )

*

( ) ( ) ( ),


www.ssmrmh.ro

101


We have by GM-AM inequality:

4( )

( ) ( )

5

.√( ) ( ) ( )

/

(

*

, ( )( ) ( )-

.

/ , ( ) ( ) ( )-

(*)

And ( ) ( ) .

/ , ( ) ( )-

(**)

Let

(( ) ( )

( ) )

( ) ( ) ( )

( ) 4

( ) ( )

5

( ) , ( ) ( ) ( )-

( ) ( )

.

/

.

/ . So,

but then so,


(

( (

( ) ( )

**

( ( ))

( ( ))

( ) ( )

)

4 4( )

( )

55

( ) ( ), let

4 4( )

( )

55

( ) ( )

:

4( )

( )

5

4( )

( )

5

;

4

( ) ( )

5

( ) (1)

(( )

( )

*

( )

(( )

( )

*

( )

www.ssmrmh.ro

102

(( )

( )

( ) ( )

*

.

/

(( )

( )

( ) ( )

*

.

/

(( )

( )

( ) ( )

*

( ) ( )

( ) ( )

(( )

( ) ) ( )(( )

( ) )

( )(( ) ( ) )

( )( )

(

( )( ) *

( )( )

( ) ( )

( )

(( )

( ) ) ( )(( )

( ) )

( )(( ) ( ) )

( ) ( )

( ) ( )

( ) ( )

( )

( ) ( )

( ) ( ( )

( ) ( )

)

( ( ) (

( ( )

( )* *

( ) ,

4 ( )

( ) ( )

( )5

(4 ( )

5

+

( (

4 ( )

( )5

++

4 ( )

( ) ( )

( )5

www.ssmrmh.ro

103

( )

4 ( )

( ) ( )

( )5

( )( ) ( )

( )

( ) ( ) ( )

( ) .

/

( ) .

( )

( )/

( )

(2)

4( )

( )

( ) ( )

5

(( )

( )

*

4( )

( )

( ) ( )

5 ( )

(( )

( )

*

( )

( )

4( )

( )

( ) ( )

5

.

/ .

/

( ) .

/

( )

(

( )( )

*

( )( )

( )( )

(3) ( ) ( ) ( )

Because : 4

( ) ( )

5

4( )

( )

5

;

:( )

( )

( ) ( )

;

:( )

( )

;

884. Find:

( (∑4 . /5

+ ∑

+

Proposed by Florică Anastase – Romania

Solution 1 by Marian Ursărescu – Romania

Let ∑

. We have:

www.ssmrmh.ro

104

∑

∑

∑

∑

∑

(1)

But ∑

(

*

(2)

From (1) (2)

and (geometric progress). We must find:

( ∑

+

( ∑

+

∑

(3)

Now, using Cesaro-Stolz for

:

∑

a) strictly decreasing

b)

∑

∫

c)

( )

( )( )

( )

( )

( )( )

(4)

From (3) (4)


Let:

www.ssmrmh.ro

105

∑. /

∑.

/

∑.

/

.

/

∑.

/

⏞

(

.

/ ∑.

/

⏞

)

.

/ . /

.

/ 4.

/ .

/5

.

/ .

/

. Also:

∑

∑

(∑

∑

+ ( )

Now: ( ( ) ) ( )

.

/

( ) (

*

( )

(answer)

885. Dedicated to teacher Mehmet Șahin

( ) ( ) ( ) ( )

Find:

:

∑ 4 (

* (

*5

;


www.ssmrmh.ro

106

Solution by Florentin Vișescu – Romania

( ) ( ) ( ) ( )

( ( ) ( ) ) , ( ) ( )-

( ) ( ) ( ) ( )

( ) not verify; ( )

( ) ( )

( )

( )

( ( )

( )* ( ( )

( )*

( )

( ) ( ) ( ( ) ) ( ) ( ) ( )

( )

⟨

( ) [

]

( )

( ) [

]

( )

( ) ( )

:∫

;

. | /

∑ 4 (

* (

*5

.∑ .

/

/

∑ . /

: ∑ 4 (

* (

*5

;

[

∑ (

*

]

∑ (

*

:∫ ( )

;

∫ ( )

:∫ ( )

;

886.

www.ssmrmh.ro

107

∫ .

/

( )



Let ∫ .

/

( )

∫ .

/

( )

∫ .

/

( )

We have | .

/

( )|

( )

Since | | | |

|∫ .

/

( )

| ∫

∫

then ∫ .

/

( )

we have:

|∫ .

/

( )

| ∫

∫

∫

(

√ *

√ [

(

√ *]

√

then ∫ .

/

( )

therefore


Here ∫ .

/

( )

Consider ( ) .

/

( ). We note that at

( ) and we have ( )

( )

www.ssmrmh.ro

108

( )

( .

/

*

( )

( )

| ( )| ( ) . Then, by dominating convergence, we have:

∫ ( )

∫ ( )

∫

887. Find:

∑4

(∑ )(∑

)

5


Solution 1 by Adrian Popa – Romania

∑4

(∑ )(∑

)

5

∑(

*

We know that


We write the sum without notation

( )

( )( )

( )( )

Now, by partial fraction decomposition we have:

www.ssmrmh.ro

109

(

* (

*

Since is a telescoping series with nth partial sum as

6

∑

7


Here ∑ (

∑ ∑

*

∑ ( ( )

∑ ( ( ) ) ∑ ( ( ) )

*

( )

( ( ) )( ( ) ( ) )

( )

( ( ) ( ) )( ( ) ( ) ( ) )

( )

( ( ) ( ) ( ) )( ( ) ( ) ( ) )

( )

( ) ( )

( ) ( )

( ) ( ) ( )

We have telescoping sum give us

( ) (∑(

*

+

888. Find:

(

( )∑4 4

55

+



Let ( )

, ,

www.ssmrmh.ro

110

( )

( ) then is increasing on , ,

Let and , so, ( ) ( ), then

then

then

( )

( ) ( )

( ) (*) but

( )

So,

(**). Then by (*); (**) we have:

( )

( ) ( )

( )

but is increasing, so: .

/ .

( )

( )/ (.

/

*

So,

( )∑ .

/

( )∑ ( ) (.

/

*

Where

( )∑ .

( )

( )/

We have

∑

(.

/

* ∫

( )

Let ( )

then ∫

( )

∫

, ( )-

We have

( )∑ (.

/

*

and

( )∑ ( ) (.

/

*

( )∑

4(

*

5

Then


We know that ∑ ( ) ( ) ∫ ( )

continuous and

, - and || || where || ||

( ), and and

, let ( )

( )

( )( )

( ) ( )

( )

( )

( ) ( )

( )

and ( )

|| || and ( )

( ) and ( ) ( ) continuous

www.ssmrmh.ro

111

∑ ( )

( )

4

5 ∫ ( )

∑

( )

.

/

∫ ( )

∑

( )

.

/

(1)

( ) increasing .

/ .

/

( )∑ .

/

( )∑

and

( )∑ 4

5

( )

( )

.

( )∑ . .

//

/

∫ ( )

(2)

∫ ( ) ∫ ( ) ( ) ∫

( )

∫

( )

( )

( )

( )

(

( )∑4 4

55

+

4

( )

5

( )

( )

( )

Solution 3 by Marian Ursărescu-Romania

(another approach) Let , - ( ) Riemann integrability

Let .

( )

( )

( )

( )

( ) /

|| ||

( )

( ( ) ( )( ))

( )

( )

|| || . Let ,

-, so that:

( )

( ) (

) ∑ 4 ( )

( )5

( )

( )∑

www.ssmrmh.ro

112

( )∑

∫

∫

|

∫

( ) |

889. Find:

:

∫. .

//

;



Let

∫ .

/

∫

∫ .

/

∫ .

/

∫ 4

5

then| |

∫ .

/

We have ( ) and then

so, .

/

then | |

∫

so | | then


We consider ( ) (

*

. We observe that at , the limit of ( )

respectively when . Limit of ( )

when . Hence by Dominated

Convergence Theorem, we have:

∫ ( )

∫

890. Find the closed form:

www.ssmrmh.ro

113

∑ ( )( ) ( )

( )( )( ) ( )

Proposed by Florică Anastase-Romania


∑ ( )( ) ( )

( )( ) ( )

Consider the series ( ∑ ) ⁄ ( )( ) ( )

( )( ) ( ), so:

( ) ∑( )( ) ( )

( )( ) ( )

⁄

( )

( ) ( )∑ ( ) ( )

( )

( )

( ) ( )∑∫ ( )

( )

( ) ( )∫∑

( )

( )

( ) ( )∫ ( )

( )

( ) ( ) ( ) ( )

( )

( )

( ) (1)

Consider the function ( ) ( )

(- ,) and ( )

( )

So, for ( ) ( ) ( )

( )

Let

So, , using the result (1):

www.ssmrmh.ro

114

∑ ( )( ) ( )

( )( ) ( )

Solution 2 by Samir HajAli-Damascus-Syria

Let put: ( ) ∑ ∏

We have: ∏

( )

( ) ( )

( ) ( ) ( )

( )

( )

( ) ( )

( ) ( )

Where ( ) is Beta function. Now: ( )

( ) ∑ ( )

But: ( ) ∫

( ) .

Since: ∑ uniformly converge

( ) ( ) ∫∑

( ) ∫ ( )

∫ ( )

( )( ) ( )

So, ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

Now, for:

Where:

We find: ∑ ( ) ( )

( )( ) ( )


∑ ( )( ) ( )

( )( ) ( )

( )

( ) ( )∑

( )

( )

( ) ( )∫

( )

∑

www.ssmrmh.ro

115

( )

( ) ( )∫

( )

( )

( ) ( ) ( ) ( )

( )


∑ ( )( ) ( )

( )( ) ( )

∑ ( )( ) ( )( ) ( )( ) ( )( )

( )( ) ( )

∑4

( )( ) ( )

( )( ) ( ) ( )( ) ( )

( )( ) ( )5

4

( )( ) ( )

( )( ) ( )5

891. ( ) .∑ .

/

/

Find:

:

:

;

;



Since: ∏ .

/

( .

/*

∑ . .

//

( ( .

/*

+

Now, by differentiable both side with respect to we get:

∑

( (

**

(

( . /*

,

www.ssmrmh.ro

116

∑ (

*

4 .

/5

∑ .

/

( ) (( ). /*

(( ) )

Let:

4

.

/5

4

(( ) ) 4

.

/55

(

:

.

/;

(( ) )

)

(

( ) ( )

( ) (( ) )* 4

.

/

5. Now:

: .

/

; (

(

**

√

892. Find:

4√( ) √( )

√( ) √

5


Solution by Naren Bhandari-Bajura-Nepal

4√( ) √( )

√( ) √

5

:( ) 4√ ( ) (

*

5

( )

( ) (√ .

/

*

;

√

:(

*

√ ( )

( ) (

*

√

;

www.ssmrmh.ro

117

√

:( ) (

( )*

( )

( ) (

*

;

√

( 4

( ) (

( ) *5 4

(

*5+

√

(

* √

893. If: ( )

( )

( ) ( )

Find: ( √

*



( ) .

/ .

/ .

/ ( ) .

/

∫( ) ( ) |

∫( )

∫( )( ) ∫( ) ∫( )

( )

( )

√

√ ( )

( )

⏞

( )(( ) )

( ) ( )

( )

4 √

5

√


www.ssmrmh.ro

118

We write: ∑ ( ) .

/

∑ ( )

. /∫

∫(∑( )

. / +

∫( )

∫( )

< ( )

. / (

*=

[

√

( ) √

]

( ) ( )

( )

Now, we solve for

4 √

5

4

√

5 :

√( )

√( )

( )

;

4

√

√

5 4

√

5

Note for all inductively we can show that:

Set and hence by Squeeze theorem

894. Find:

(

√ ∑ (

√ *

+


Solution 1 by Florentin Vișescu-Romania

We know that:

√

(

√ *

√

√

(

√ *

(

√ *

www.ssmrmh.ro

119

∑

√

∑

∑

√

√ ∑

∑

√ ∑

( )

√

( )

∑

√

( )

√

( )

√ ( )

( )

⏟

√ ( )

⏟

√ ∑

√

( )

⏟

( ) grade 6 polynomial in .


(

√ ∑ (

√ *

+

∑(

√ (

√ **

∑4

√ 4

√

√ 5 5

∑

∑ ( )

( )

( )

( )

We note that:

( )

∑(∑ ( )

( ) √

+

∑

( )

4

( )

( ) √( )

√

5

∑ ( )

( )

( )

895. Find:

www.ssmrmh.ro

120

( ) ∏4

( ) 5



( ) ∏ . ( )

( ( )) /

. Let ( ) ( ( ))

( ) ∏

( )

( ) ( )

( )

If then: ( ) . If then ( )

If then ( ) ( )

( ). Let ( )

( )

then ( )

( )

So, for: and for: and

so: ( ) ( ) ( ) and ( ) ( ( ))

we know: ( ) . For

( ( ) ( ) ( )

( ) ( ) ( ) *

for: ( ) ( ) so: ( ) , then ( )

( ( ))

( )

( ) ( ) ( ( )) ( ) ( ( )) . So: ( )

∏4 ( )

( ( )) 5

896. Evaluate the limit:

:∫( ( ) ( ))

;



(∫ ( ( ) ( ))

)

. Let ( ) ∫ ( ( ) ( ))

www.ssmrmh.ro

121

∫

( ( ) ( ))

. /∫

. /∫ . /

(

*

- ,

. /

(

4 .

/5

. /

)

. /:

. /

. /;

. .

/ .

//

.

/ .

/

. So: . .

/ .

// .

/

[ (

* . .

/ .

//]

4 4 .

/5 . .

/ .

//5

. ( ( )) . .

/ .

///

( ( ( ))

( ( ))*

Then, we have: ( ( ))

( ) and,

( ( ))

So:

(∫ ( ( ) ( ))

)

( denote Euler – Mascheroni constant)


Since: ∫ ( )

.

/ .

/

√ and ∫ ( )

.

/ .

/

√ where: and

we obtain:

:∫( ( ) ( ))

;

( . /

. .

/ .

//,

(

. . .

/ .

// ( ( ))/*

www.ssmrmh.ro

122

(

. / .

/

. / .

/

( ), .

/

897. Find:

: ( ∑( )

( ) ( )

+;


Solution by Remus Florin Stanca – Romania

(∑( )

( ) ( )

∑

+

(∑( ) ( ) ( )

( ( ) ( ))( )

+

(∑ ( )

( ( ) ( ))( )

+

We know that ( )

( ( ) ( ))( )

( ( ) )( )

( ( ) )( )

∑ ( )

( ( ) ( ))( )

( )

( )( ( ) )

( )

( )( ( ) ) .∑

( )

( ) ( ) ∑

/ (*)

∑

∑

(1)

( )

continuous, let

|| ||

∑ ( ) ( ) ∫ ( )

where , - and

and and let

∑

∫

( ) ( )

www.ssmrmh.ro

123

( )

( )

∑( )

( ) ( )

( )

and and let

∑

∫

( ) ( )

∑

( )

( )

∑( )

( ) ( )

( )

( ( ) ∑( )

( ) ( )

+

( ) ∑( )

( ) ( )

(

( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) )

( )

(

( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) )

4 ( )

( ) ( )

5

4 ( )

( ) ( )5

4 ( )

( ) ( )5

898. Find ( – golden ratio):

www.ssmrmh.ro

124

:

√

√

√

;



We have by Bernoulli’s inequality:

( ) for all and then .

/

Let ∑ .

/

. So,

∑

but

then

∑

we know that ∑ .

/

.

/

and .

/

since

then therefore

899. Find:

(

√∑( ) .

/

)



∑( )

. / .

/ ( ) .

/ ( ) .

/

( ) . / . ( )/

.

/

. / ( ) .

/ ( ) .

/ ( ) .

/

( ( )) .

/

www.ssmrmh.ro

125

0. / ( ) .

/ ( ) .

/ ( ) .

/ ( ) .

/1

6.

/ ,( ) ( ) ( ) -7

√∑( ) .

/

√.

/ ( )( )

Degree of .

/ ( )( )

is equal to

So: degree of

√.

/ ( )( )

is

4

√∑ ( ) .

/

5 4

5 (*)

But 4

5 4

( )

5 4

( )

5

So: ∑ ( ) .

/

Solution 2 by Khaled Abd Imouti-Damascus-Syria

We know that ∑

then for

Let

√∑ ( ) .

/

we have

√∑ ( )

√∑ ( )

. We know that:

∑ ( )

∑ ( )

∑

( )( )

Then

. ( )( )

/

But . ( )( )

/

(

( )( )

*

then

Solution 3 by Artan Ajredini-Presheva-Serbie

www.ssmrmh.ro

126

Let ∑ ( ) .

/ and ( ) (

) . /

Since ( ) we have that: . Now,

( )

. Hence, √ . Definitely,

√


Let

√∑ ( )

we know that ∑

then ∑

So, for all then

: ∑ ( )

;

: ∑ ( )

;

: ∑ ( )

;

:∑

;

:∑

;

but


Let be the limit and we write ∑ ( ) . /

. Now from

Bernoulli inequality we have: √

√

and hence we have

.

/ and by Stolz Cesaro theorem we have

( ) (∑( ) .

/

+

( ) (∑( ) .

/

+

( ) . /

( )

( ) . /

( )

and hence by Squeeze theorem we have .

www.ssmrmh.ro

127

900. ( ) .∑ ( )( ) ( )

( ) / * +

Find:

∑

( )


Solution by Samir HajAli-Damascus-Syria

We have: ∑

( )

∑

( ) because the series converge uniformly when | |

∑ ( )

( ) (Similarly)

We find: ∑ ( )

( )

( )

( )

( )

Then: ∑ ( )( )

( ) | |

Therefore: ∑ ( )( )

( )

( )

Similarly step by step we find:

∑ ( )( ) ( )

( ) | |

Hence: ( ) ∑ ∏ ( )

( )

( )

(

*

. /

and: ∑

( ) ∑

∑

www.ssmrmh.ro

128

It’s nice to be important but more important it’s to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

Documents

ROMANIAN MATHEMATICAL MAGAZINE · R M M ROMANIAN MATHEMATICAL MAGAZINE Founding Editor DANIEL SITARU Available online ISSN-L 2501-0099 RMM - Calculus Marathon 801 - 900