(Everything a physicist needs to know about)Bessel functions Jn(x) of integer order

(and also Hankel functions H(1,2)n )

Nikolai G. Lehtinen

August 26, 2019

Abstract

Some properties of integer-order Bessel functions Jn(x) are derivedfrom their definition using the generating function. The results may beof use in such areas as plasma physics. Now with a Section on Hankel

functions H(1,2)n (x)! We assume that the reader knows some complex

analysis (e.g., can integrate in the complex plane using residues).

1 Basic properties

1.1 Generating function

We derive everything else from here, which will serve us the definition of theinteger-order Bessel functions (of the first kind):

g(x, t) = ex2 (t− 1

t ) =+∞∑

n=−∞

Jn(x)tn (1)

We immediately make our first observations that since g(x, t) is real for realx, t then Jn(x) must be real for real x. From now on,∑

n

or just∑

will mean+∞∑

n=−∞∫dθ

2πwill mean

∫ 2π

0

dθ

2π

1

1.2 Some useful formulas

1. t→ −t in (1) =⇒ e−x2 (t− 1

t ) =∑

(−1)nJn(x)tn =∑Jn(−x)tn =⇒

(equate powers of t) =⇒

Jn(−x) = (−1)nJn(x)

2. t → 1t

in (1) =⇒ e−x2 (t− 1

t ) =∑Jn(−x)tn =

∑Jn(x)t−n =⇒

(equate powers of t and use previous expression) =⇒

J−n(x) = (−1)nJn(x) = Jn(−x) (2)

3. t = eiθ =⇒ 12

(t− 1

t

)= i sin θ =⇒

eix sin θ =∑

Jn(x)einθ (3)

4.∫

dθ2πe−imθ × (3), switch m→ n, =⇒

Jn(x) =

∫dθ

2πeix sin θ−inθ (4)

1.3 Recursion formulas

1. ∂∂x

(1) =⇒ 12

(t− 1

t

)g(x, t) =

∑J ′n(x)tn =⇒∑

12Jn(x)(tn+1 − tn−1) =

∑J ′n(x)tn =⇒∑

12(Jn−1(x)− Jn+1(x))tn =

∑J ′n(x)tn =⇒

J ′n(x) =1

2(Jn−1(x)− Jn+1(x)) (5)

2. ∂∂t

(1) =⇒ x2

(1 + 1

t2

)g(x, t) =

∑nJn(x)tn−1, multiply by t, =⇒

x2

(t+ 1

t

)g(x, t) = x

2

∑Jn(x)(tn−1 + tn+1) =

∑nJn(x)tn =⇒

x2

∑(Jn+1(x) + Jn−1(x))tn =

∑nJn(x)tn =⇒

nJn(x) =x

2(Jn−1(x) + Jn+1(x)) (6)

3. Combining the two we can get

J ′n =n

xJn − Jn+1 (7)

J ′n = Jn−1 −n

xJn (8)

2

1.4 Differential equation

Bessel equation arises when we solve Helmholz equation ∇2φ + p2φ = 0 in2D, in cylindrical coordinates. It is more naturally understood when we goto Fourier coordinates k (see Section 3.1 below), in which operator ∇ = ik,∇2 = −k2. Here we brutally derive the Bessel equation from the recursionformulas.

Let us use (7) to find J ′n, differentiate it again and substitute expressionJ ′n+1 = Jn − n+1

xJn+1 derived from (8). We have

J ′n = nxJn − Jn+1

J ′′n =(nxJn)′ − Jn + n+1

xJn+1

= − nx2Jn + n2

x2Jn − n

xJn+1 − Jn + n+1

xJn+1

= 1x(−n

xJn + n2

xJn − xJn + Jn+1)

By eliminating Jn+1 from these equations we obtain

x2J ′′n + xJ ′n + (x2 − n2)Jn = 0 (9)

2 Sums

2.1 Sums of the first power of Jn(x) or J ′n(x)

The basic idea is to substitute t = 1 in various derivatives of g(x, t) in respectto both x and t.

1. t = 1 in (1) =⇒ ∑Jn(x) = 1 (10)

2.∑

(6) =⇒ ∑nJn(x) = x (11)

3.∑nkJn(x) may be found by differentiating g(x, t) over t and multiply-

ing by t, k times. Example:∑n2Jn(x)tn = t ∂

∂tx2

(t+ 1

t

)g(x, t) = . . .

3

Figure 1: Graf’s addition theorem

2.2 Sums of the second power of Jn(x) or J ′n(x)

The basic idea is to multiply two g(x, t) or its derivatives in respect to bothx and t. One can take the second g to be a function of u = 1/t instead. Notethat for equal first arguments g(x, t)g(x, u) = 1 and we have a polynomial int. By equating powers, we get various sums. Here are some examples.

1. g(x, t)g(x, 1/t) = 1 =∑

n

∑m Jn(x)Jm(x)tn−m =⇒ all coefficients

are zero except at t0 =⇒∑n

JnJn−k = δk,0

2. g(x, t)g(y, t) = exp(x+y2

[t− 1/t])

=⇒∑k Jk(x+ y)tk =

∑n

∑m Jn(x)Jm(y)tn+m =⇒

Jk(x+ y) =∑n

Jn(x)Jk−n(y) =∑n

Jn(x)Jn−k(−y)

where we used (2).

2.3 Graf’s addition theorem

See Abramowitz and Stegun (1965, eq 8.1.79).From Figure 1, OB = z sin(θ + ψ) = OA + AB = x sin θ + y sin(φ − θ).

From (4), Jn(z) =∫

dθ2πeiz sin(θ+ψ)−in(θ+ψ) =⇒ einψJn(z) =

∫dθ2πeix sin θ+iy sin(φ−θ)−inθ

Substitute eix sin θ =∑

k Jk(x)eikθ, eiy sin(φ−θ) =∑

m Jm(y)eim(φ−θ) =⇒

4

einψJn(z) =∑

k

∑m Jk(x)Jm(y)eimφ

∫dθ2πe−inθ+ikθ−imθ

Use∫

dθ2πeilθ = δl,0 for integer l =⇒

einψJn(z) =∑m

Jm+n(x)Jm(y)eimφ (12)

In particular, for n = 0

J0(z) = J0(√x2 + y2 − 2xy cosφ) =

∑m

Jm(x)Jm(y)eimφ (13)

3 Integrals

3.1 2D Fourier transforms

Define direct and inverse Fourier transform of a function F (r) in 2D positionspace r = {x, y} into function F (k) in 2D wavevector space k = {kx, ky} as

F (k) =

∫d2r e−ik·rF (r)

F (r) =

∫d2k

(2π)2eik·rF (k)

where d2r = dx dy and d2k = dkx dky and the integration is over the wholeplane. Below, we also use polar coordinates r = |r|, θ = atan2(y, x) inposition space and k = |k|, χ = atan2(ky, kx) in wavevector space.

Let F (r) = δ(r − a)e−inθ. The Fourier image is

F (k) =

∫d2r e−ik·rF (r) =

∫ ∞0

r dr

∫ 2π

0

dθ δ(r − a)e−inθ−ikr cos(θ−χ)

Using φ = θ−ξ−π/2, we get cos(θ−χ) = − sinφ and θ = φ+χ+π/2 =⇒

F (k) = 2πae−inχ−inπ/2∫dφ

2πe−inφ+ika sinφ = 2πai−ne−inχJn(ka)

We can use the same formula by substituting r ↔ k and i ↔ −i andtaking care of coefficient (2π)2. It summary,

F (r) ⇐⇒ F (k)

δ(r − a)

aeinθ ⇐⇒ 2πi−nJn(ka)einχ (14)

inJn(pr)einθ ⇐⇒ 2πδ(k − p)p

einχ (15)

5

3.2 Weber’s First Integral

See Abramowitz and Stegun (1965, eq 11.4.28 with µ = 2, ν = 0).The integral I1 =

∫∞0r dr J0(pr)e

−q2r2 is found by representing this inte-gral in a 2D plane (2πr dr = d2r) and going to Fourier space r→ k.

I1 =∫d2rF1(r)F2(r), where F1(r) = e−q

2r2 , F2(r) = 12πJ0(pr), where

r = |r|. We use the fact that∫d2rF1(r)F2(r) =

∫d2k(2π)2

F ∗1 (k)F2(k), where

F1,2(k) =∫d2r e−ik·rF1,2(r) are the Fourier images. We have F1(k) = π

q2e− k2

4q2

and using the Fourier transform result from above, we obtain F2(k) = δ(k−p)p

.Thus, ∫ ∞

0

x dx J0(px)e−q2x2 =

1

2q2e− p2

4q2 (16)

3.3 Weber’s Second Integral

I2 =

∫ ∞0

x dx Jn(ax)Jn(bx)e−q2x2

From Graf’s addition theorem (13) =⇒ Jn(x)Jn(y) =∫

dφ2πJ0(z)e−inφ =⇒

Jn(ax)Jn(bx) =∫

dθ2πJ0(x√a2 + b2 − 2ab cos θ)e−inθ =⇒

I2 =∫

dθ2πe−inθ

∫∞0x dx J0(x

√a2 + b2 − 2ab cos θ)e−q

2x2 =⇒ [use (16)] =⇒

I2 =∫

dθ2π

12q2

exp[−a2+b2−2ab cos θ

4q2− inθ

]= 1

2q2e−a

2+b2

4q2∫

dφ2π

exp[i(−i ab

2q2) sinφ− inφ+ inπ

2

](we substituted θ = φ− π/2)

I2 = 12q2e−a

2+b2

4q2 Jn(−i ab2q2

)in

We use the definition of the modified Bessel function (of the first kind)

In(x) = inJn(−ix) = i−nJn(ix)

=⇒ ∫ ∞0

x dx Jn(ax)Jn(bx)e−q2x2 =

1

2q2e−a

2+b2

4q2 In

(ab

2q2

)(17)

In particular, for a = b we have∫ ∞0

x dx J2n(ax)e−q

2x2 =1

2q2e− a2

2q2 In

(a2

2q2

)=

1

2q2Λn

(a2

2q2

)where we introduced another function Λn whose properties are explored inmore detail below in Section 5.

6

3.4 Lipschitz Integral

I =∫∞0e−axJ0(bx) dx =

∫dθ2π

∫∞0e−ax+ibx sin θ dx =

∫dθ2π

1a−ib sin θ

Substitute z = eiθ, integration is now around the unit circle in the complexz-plane:I = 1

2π

∮ dz/(iz)

a− b2(z−1/z) = − 1

2πi2b

∮dz

z2− 2abz−1 = − 1

2πi2b

∮dz

(z−z1)(z−z2) , where

z1,2 = 1b(a±

√a2 + b2).

Note that z1 is outside the unit circle, z2 is inside. We use the residuecalculus and find I = −2

b1

(z2−z1) = 1√a2+b2

.∫ ∞0

e−axJ0(bx) dx =1√

a2 + b2(18)

3.5 1D Fourier Integral

Calculation is similar to the Lipschitz integral.∫ +∞−∞ e−iaxJn(bx) dx =

∫dθ2πe−inθ

∫ +∞−∞ e−iax+ibx sin θ dx =

∫ 2π

0e−inθδ(a −

b sin θ) dθwhere we used the integral representation property of the delta function.

Its argument may be transformed using:

δ(f(x)) =∑k

1

|f ′(xk)|δ(x− xk)

where xk are the zeros of f(x). If |a| ≥ |b|, there are no zeros. Otherwise,there are two: θ1 = arcsin(a/b), θ2 = π − arcsin(a/b).

We have δ(a− b sin θ) = 1|b cos arcsin(a/b)| [δ(θ − θ1) + δ(θ − θ2)] and finally∫ +∞

−∞e−iaxJn(bx) dx =

H(|b| − |a|)√b2 − a2

[u−n + (−u)n

]where

u = ei arcsin(a/b) =

√1−

(ab

)2+ i(ab

)and H is the Heaviside step function.

7

4 Asymptotics

4.1 Taylor series expansion

We will derive the expansion of Jn(x) in the vicinity of x = 0 using (4).Expand

eix sin θ =∞∑m=0

xm(i sin θ)m

m!=

∞∑m=0

(eiθ − e−iθ)m

m!

(x2

)mWe use the binomial expansion

(eiθ − e−iθ)m

m!=

m∑k=0

(−1)kei[(m−k)−k]θ

(m− k)!k!

The integral over θ “cuts out” only the exponents with the correct argument:

Jn(x) =

∫dθ

2πeix sin θ−inθ =

∞∑m=0

m∑k=0

δn,m−2k(−1)k

(m− k)!k!

(x2

)mLet us assume n ≥ 0 (the formulas for n < 0 are obtained trivially usingequation 2). After substitution m = n+ 2k instead of double summation wehave summation in k only:

Jn(x) =∞∑k=0

(−1)k

(n+ k)!k!

(x2

)n+2k

(19)

4.2 Asymptotics for large x

Again, use (4). The phase φ = x sin θ−nθ changes fast for large x, thereforewe should use the stationary phase integration method. The stationary phaseis at θ = ±θ0 found from φ′(θ) = x cos θ − n = 0 =⇒ cos θ0 = n/x� 1,we can use sin θ0 ≈ 1, θ0 ≈ π/2 and φ(θ0) ≈ x−nπ/2. The second derivativeat θ0 is φ′′(θ0) = −x sin θ0 ≈ −x. The contribution at −θ0 is a complexconjugate. Thus, we extend integration over ζ = θ− θ0 to infinite limits andget

Jn(x) ≈ exp[iφ(θ0)]

∫ +∞

−∞

dζ

2πexp[iφ′′(θ0)ζ

2/2] + c.c.

= exp[i(x− nπ/2)]

∫ +∞

−∞

dζ

2πexp[−ixζ2/2] + c.c.

8

Using gaussian integral∫e−ζ

2/(2σ2) dζ =√

2πσ with σ =√

1/(ix), we find

Jn(x) ≈√

2

πxcos(x− nπ

2− π

4

), x→∞ (20)

and for modified functions,

In(x) ≈ ex√2πx

, x→∞

5 Properties of Λn(x)

Introduce Λn(x) = e−xIn(x). First, some properties of In(x) = inJn(−ix) =i−nJn(ix) which are derived elementarily from the corresponding propertiesof Jn:

x2I ′′n = −xI ′n + (x2 + n2)In

I ′n =1

2(In−1 + In+1)

nIn =x

2(In−1 − In+1)

I−n = In∑n

In(x)einθ = ex cos θ

From here, it follows that Λ′n = e−x(I ′n − In). Some important properties ofΛn: ∑

n

Λn(x) = 1 (follows from the sum of In above with θ = 0)∑n

nΛn(x) =x

2

∑n

(Λn−1 − Λn+1) = 0

5.1 Taylor series

Let us derive Taylor series expansion of Λn. We can assume n ≥ 0 becauseotherwise Λn = Λ−n. We start with (see equation 19)

e−x =∞∑k=0

(−x)k

k!, In(x) =

∞∑l=0

1

l! (n+ l)!

(x2

)2l+n9

Multiplying, we get

e−xIn(x) =∞∑k=0

∞∑l=0

(−2)k

k! l! (n+ l)!

(x2

)2l+n+k=∞∑p=n

Ap

(x2

)pThe goal is to find a tractable expression for

Ap =∞∑k=0

∞∑l=0

∞∑m=0

∣∣∣∣∣ k+l+m=p

m−l=n

(−2)k

k! l!m!

We use the polynomial expansion

(a+ b+ c)p =∞∑

k,l,m=0

k+l+m=p

p!

k! l!m!akblcm

Take a = −2, need to single out term blcm such that m− l = n. Take b = eiθ,c = e−iθ, we get blcm = e−inθ which may be singled out by integration over θof this expression multiplied by einθ:

Ap =1

p!

∫ 2π

0

(−2 + eiθ + e−iθ)peinθdθ

2π

Let us simplify using −2+eiθ+e−iθ = −(2 sin θ′)2, with θ′ = θ/2. Integrationover θ′ is from 0 to π but may be extended to [0, 2π] because of the symmetryof the expression under the integral. After dropping the dash from θ′ andusing 2 sin θ = −i(eiθ − e−iθ):

Ap =1

p!

∫ 2π

0

(eiθ − e−iθ)2pe2inθ dθ2π

Use the binomial expansion:

(eiθ − e−iθ)2p =

2p∑j=0

(2p)! (−1)2p−j

j! (2p− j)!eiθ(j−2p+j)

From this expansion, the integration over θ only extracts the term ∝ e−2inθ,i.e., j = p− n. Finally,

Ap =(2p)! (−1)p+n

p! (p− n)! (p+ n)!

10

and we have our final answer

Λn(x) =∞∑k=n

(2k)! (−1)n+k

k! (k − n)! (k + n)!

(x2

)k(21)

5.2 Large x

We use (20) to find

Λn(x) = inJn(−ix)e−x ≈√

1

2πx, x→∞

6 Hankel functions

This is extra material which is used in completely different situations. Inplasma physics, Jn is usually used to describe the field of a plane wave seen byan oscillating particle, using the expansion (3). On the other hand, Hankel

functions H(1,2)n (x) (also called first and second kind, respectively) are

usually used to describe waves in a cylindrical system (to which Jn may alsohave a limited applicability).

6.1 Cylindrical waves

We will heavily use notations from Subsection 3.1. Let us consider theHelmholtz (wave) equation in 2D. From equation (15) we see that the func-tion Φ(r) = inJn(pr)einθ in configuration r-space becomes Φ(k) = 2π

pδ(k −

p)einχ in the wave vector k-space which satisfies the uniform Helmholtz equa-tion (−k2 + p2)Φ(k) = 0 or (∇2 + p2)Φ(r) = 0. Notice that another way toshow this is that by separating variables θ and r we get the differential equa-tion (9). Let us consider a Helmholtz equation with sources:

(∇2 + p2)Φ(r) = −S(r)

or, in k-domain,(−k2 + p2)Φ(k) = −S(k)

(the minus sign is for convenience).The simplest source would be a δ-function in the origin, however, it is

cylindrically symmetric and can only generate the azimuthal harmonic with

11

n = 0. As a more general source, let us take, for example, a thin (δ-function)source at distance a from origin, such as given by equation (14) (we switchn→ −n for later convenience), with Fourier image

S(k) = 2πi−nJn(ka)e−inχ = s(k)i−ne−inχ

The solution in r-domain is

Φ(r) = i−n∫

d2k

(2π)2s(k)eik·r−inχ

k2 − p2

The azimuthal dependence of the answer will be Φ(r) = Φ(r)e−inθ, so itis sufficient to find Φ at, say, r = ry with r > 0, i.e. Φ(r) = Φ(ry)ein(π/2) =inΦ(ry). Let us evaluate this integral in Cartesian coordinates kx, ky:

Φ(r) =

∫d2k

(2π)2s(k)eik·r−inχ

k2 − p2=

∫ +∞

−∞

dkx2π

F (kx)

where F (kx) is the result of integration over ky:

F (kx) =

∫ +∞

−∞

dky2π

s(k)eikyr−inχ

k2y − (p2 − k2x)

where k =√k2x + k2y and χ = atan2(ky, kx).

For the ease of integration in complex ky plane, we assume s(k) is anarbitrary function that is analytic in the upper half-plane (=ky > 0) and alsonot very big so that s(k)eikyr → 0 at =ky → +∞. For example, we may takethe above expression for a thin source with a → 0 (a small source) so using(19) s(k) ≈ 2π(n!2n)−1(ak)n [1 +O(ak)] ∝ kn. Note that for a finite-sizesource, s(k) = 2πJn(ka), at ky = iX, X → +∞ asymptotically becomes,using k ≈ ky = iX:

s(k) ≈√

8π

iXacos(iXa− nπ

2− π

4

)≈ in

√2π

XaeXa

so the product s(k)eikyr ∼ eX(a−r) → 0 only when r > a (makes sense becausethe free waves are outside the source).

12

6.2 Outward waves: H(1)n (x)

To get an outward propagating wave solution, assume that p has a smallpositive imaginary part, p→ p+ i∆ (i.e., the medium has a small absorptionwhich makes the inward wave infinite at r → ∞ and therefore the outwardwave is the only one that we get). This gives us the rule for going aroundthe pole at k ≈ p. Since r > 0, we must close the integration contour in theupper plane, i.e., go counterclockwise around the pole at ky =

√p2 − k2x if

p > |kx| and the pole at ky = i√k2x − p2 when p < |kx|. In both cases, we

can write

F (kx) =i

2

s(p)eipr sinχ−inχ

p sinχ

where

p cosχ = kx

p sinχ =

{ √p2 − k2x when p > |kx| or |cosχ| < 1

i√k2x − p2 when p < |kx| or |cosχ| > 1

i.e., p sinχ is the ky value at the pole. Changing the integration variabledkx = −p sinχdχ we finally get

Φ(r) = −s(p) i2

∫C

dχ

2πeipr sinχ−inχ

where the contour C is such that cosχ goes through the values {−∞,−1, 0, 1,+∞}and sinχ goes through the values {+i∞, 0, 1, 0,+i∞}. This corresponds toa contour in χ going through points {π − i∞, π, π/2, 0,+i∞}. For conve-nience, usually a contour C1 is chosen which goes in the opposite direction.We define (compare to (4)!)

H(1)n (x) = 2

∫C1

dχ

2πeix sinχ−inχ, C1 = {+i∞, 0, π/2, π, π − i∞} (22)

Thus

Φ(r) = s(p)i

4H(1)n (pr)e−inθ

In particular, for a point source S(r) = δ(r) we have n = 0 and s(p) = 1 sothe “outward” Green’s function of Helmholtz equation is

G(r) =i

4H

(1)0 (pr)

13

6.3 Inward waves: H(2)n (x)

If we look for inward waves, we set p→ p− i∆ and the pole in the upper halfof ky-plane is at ky = −

√p2 − k2x if p > |kx| and at ky = i

√k2x − p2 when

p < |kx|. We get the same expression for F (kx) but now

p cosχ = kx

p sinχ =

{−√p2 − k2x when p > |kx| or |cosχ| < 1

i√k2x − p2 when p < |kx| or |cosχ| > 1

Using dkx = −p sinχdχ we get

Φ(r) = −s(p) i2

∫C2

dχ

2πeipr sinχ−inχ

where the contour C2 is such that cosχ goes through the values {−∞,−1, 0, 1,+∞}and sinχ goes through the values {+i∞, 0,−1, 0,+i∞}. This correspondsto a contour in χ going through points {π − i∞, π, 3π/2, 2π, 2π + i∞}. Wedefine

H(2)n (x) = 2

∫C2

dχ

2πeix sinχ−inχ, C2 = {π − i∞, π, 3π/2, 2π, 2π + i∞} (23)

and

Φ(r) = −s(p) i4H(2)n (pr)e−inθ

The “inward” Green’s function is

G(r) = − i4H

(2)0 (pr)

Because of periodicity in χ with period 2π, we can use C2 = {−π−i∞,−π, 0,+i∞}(for integer n, of course).

6.4 General properties

H(1,2)n (x) satisfy the same recursion relations (5–6) and the differential equa-

tion (9) as Jn(x) because they are defined through an integral similar to (4),although the contours of integration are different. Let us consider the sym-metry relations analogous to equation (2) in more detail. In (22) for −n,

14

change variable to α = π − θ (θ = π − α). The contour C1 changes to −C1

(by minus we mean oppositely directed):

H(1)−n(x) = 2

∫α∈−C1

d(−α)

2πeix sin(π−α)+inπ−inα = einπH(1)

n (x)

which is also valid for non-integer n if we postulate the definition (22). Forinteger n, einπ = (−1)n.

Changing the variable to α = −θ∗, and using sinα∗ = (sinα)∗, we get(the contour C1 for θ becomes −C2 for α)

H(1)n (x) = 2

∫α∈−C2

d(−α∗)2π

e−ix sinα∗+inα∗

=(H(2)n (x∗)

)∗In summary,

H(1,2)n (x) = (−1)nH

(1,2)−n (x) =

(H(2,1)n (x∗)

)∗(24)

There is no relation of symmetry when changing x→ −x because for <x < 0the defining integral (22,23) becomes infinite.

We notice that C1 + C2 = [0, 2π] (the parts going over complex θ cancel,also due to 2π periodicity). Thus,

Jn(x) =

∫ 2π

0

dθ

2πeix sin θ−inθ =

1

2

(H(1)n (x) +H(2)

n (x))

so Jn(x) = <H(1,2)n (x) for real x. We may introduce Bessel functions of the

second kind (Weber or Neumann functions) Yn(x) which are real for real x:

H(1)n (x) = Jn(x) + iYn(x)

H(2)n (x) = Jn(x)− iYn(x)

But they are only useful as linearly independent from Jn(x) solutions ofHelmholtz equation (or Bessel equation) and don’t have extra physical mean-

ing compared to H(1,2)n (x). Moreover, Yn(x) are infinite at x→ 0. This means

there is no Taylor expansion at x→ 0.One of the recursion formulas (5) is also valid for H(1,2). The reason is

that it may be derived from (4):

d

dxZn(x) =

d

dx

∫C

dθ

2πeix sin θ−inθ =

∫C

dθ

2πi sin θ eix sin θ−inθ

=1

2

∫C

dθ

2π

(eix sin θ−inθ+iθ − eix sin θ−inθ+iθ

)=

1

2(Zn−1(x)− Zn+1(x))

15

where the contour of integration C is different for different Bessel functionsZ: (i) for Z ≡ J , C = {χ, χ+ 2π} with arbitrary shift χ, due to periodicityof the expression under the integral; and (ii) for Z ≡ H(1,2), C = C1,2. Theother recursion formula (6) may be derived for J by differentiating in thearbitrary phase shift χ, and is therefore not applicable to H(1,2).

6.5 Asymptotics

We noted that there is no Taylor expansion of H(1,2)n at x → 0. However,

we may try to find the asymptotic expansion or at least the biggest term.Because of the symmetries, it is sufficient to consider only H

(1)n and n ≥ 0.

For n = 0 we may use the Green’s function at pr → 0, i.e., take p = 0 in theHelmholtz equation and compare to the Poisson equation Green’s function

GPoisson(r) = − log r

2π+ const = − log(pr/2)

2π+ const2 ≈

i

4H

(1)0 (pr)

from where H(1)0 (x) ≈ i(2/π) log(x/2) + const . We may neglect the constant

at very small x.For n ≥ 1, write out (22) explicitly, while changing the integration vari-

able so that θ = it on the interval θ ∈ [i∞, 0] and θ = π − it on the intervalθ ∈ [π, π − i∞]:

H(1)n (x) = − i

π

∫ ∞0

e−x sinh t+nt dt+1

π

∫ π

0

eix sin θ−inθ dθ−i(−1)n

π

∫ ∞0

e−x sinh t−nt dt

The middle term is finite for x = 0 and thus does not contribute to theinfinite value. Regarding the last term we notice that for x, t > 0 we havesinh t > t, e−x sinh t−nt < e−(x+n)t and therefore∫ ∞

0

e−x sinh t−nt dt <1

x+ n<

1

n(finite)

In the first term, change the variable to u = x sinh t or t = log[(u+

√u2 + x2)/x

]:∫ ∞

0

e−x sinh t+nt dt =1

xn

∫ ∞0

e−u(u+√u2 + x2)n√

u2 + x2du

We may set x = 0 in the integral because it is finite:∫ ∞0

e−x sinh t+nt dt ≈ 1

xn

∫ ∞0

e−u(2u)n

udu = (n− 1)!

(x2

)−n16

Collecting everything, we have for the biggest term

H(1)n (x) ≈

−i(

2

π

)log

(2

x

)+ const , n = 0

−i(n− 1)!

π

(2

x

)n, n > 0

(25)

As expected, it is purely imaginary because the real part (= Jn(x)) is finite.

Asymptotics of H(1,2)n (x) at x → ∞ is obtained in the same way as we

did for Jn(x) above in equation (20). This time, we have only one sta-tionary phase point at θ0 = arccos(n/x) ≈ π/2 on contour C1 (or θ0 =2π − arccos(n/x) ≈ 3π/2 on contour C2). The value at large x is therefore

H(1,2)n (x) ≈

√2

πxexp

[±i(x− nπ

2− π

4

)], x→∞ (26)

which makes it obvious that they represent outward and inward propagatingwaves. The non-oscillating contributions from intervals θ ∈ [+i∞, 0] andθ ∈ [π, π − i∞] contribute a smaller amount since eix sin θ−inθ ≈ e−x|θ| withintegral (from both intervals) ≈ 2/(πx).

6.6 Application to some integrals

Let us show that we can “straighten” our contours:

C1 → C ′1 = {+i∞+ π/2,−i∞+ π/2}C2 → C ′2 = {−i∞+ 3π/2,+i∞+ 3π/2} or {−i∞− π/2,+i∞− π/2}

Let us consider only C1 here (C2 is done analogously, or just use symmetry(24)).

The difference in integration is C1 − C ′1 = C+1 + C−1 , over two short

contours with X → +∞:

C+1 = {iX, iX + π/2}, C−1 = {−iX + π/2,−iX + π}

The integrated function is analytic, so integration over a closed contour giveszero.

In the integral over C+1 we can change the variable to θ = χ− iX which

is then integrated in [0, π/2]. The asymptotic behaviour of the exponent ofthe function under integral in (22) on C+

1 is

ix sinχ− inχ = ix sin(θ + iX)− in(θ + iX) ≈ −x2eX−iθ

17

The integral over C+1 will vanish only if the real part of the above expression

becomes large and negative. This is equivalent to arg x − θ ∈ [−π/2, π/2],and C+

1 is covered if arg x ∈ [0, π/2].Analogously, for C−1 , the exponent of the function under integral after

change θ = χ+ iX (where θ ∈ [π/2, π]) behaves as

ix sinχ− inχ = ix sin(θ − iX)− in(θ − iX) ≈ x

2eX+iθ

The integral over C−1 will vanish if arg x+ θ ∈ [π/2, 3π/2], and C−1 is coveredagain if arg x ∈ [0, π/2].

Thus, contributions from C±1 vanish for arg x ∈ [0, π/2] and integrationover C1 may be changed to integration over C ′1. Analogously, C2 may bechanged to C ′2 if arg x ∈ [−π/2, 0]. After change χ = −iξ + π/2, we get an

alternative definition of H(1)n :

H(1)n (x) =

(−i)n+1

π

∫ +∞

−∞eix cosh ξ−nξ dξ

Using symmetry (24), we arrive at the combined formula

H(1,2)n (x) =

(∓i)n+1

π

∫ +∞

−∞e±ix cosh ξ−nξ dξ (27)

We see that these integrals actually converge in wider ranges arg x ∈ [0, π]and arg x ∈ [−π, 0], respectively, for which the original definitions (22,23) donot always converge. Using symmetry (24), or by inverting the integration(ξ → −ξ) we may show that in both formulas we may switch +nξ ↔ −nξ inthe exponent.

The above integrals may occur in a form such that u = sinh ξ. Then

cosh ξ =√u2 + 1

dξ =d sinh ξ

cosh ξ=

du√u2 + 1

e±ξ = cosh ξ ± sinh ξ =√u2 + 1± u

and integration limits stay as [−∞,+∞]:

H(1,2)n (x) =

(∓i)n+1

π

∫ +∞

−∞

(√u2 + 1 + u)ne±ix

√u2+1 du√

u2 + 1(28)

18

where (a) the exponent n may be also changed to −n and/or (b)√u2 + 1+u

may be changed to√u2 + 1− u in either formula.

These integrals may occur, in particular, in calculations of spacetime-domain propagators in relativistic quantum field theory (QFT) (Greiner andReinhardt , 2009, p. 68).

6.7 Modified Bessel functions of the second kind

When the argument is imaginary, we have the modified Bessel functionsof the second kind (Basset or Macdonald functions):

Kn(x) =

{π2in+1H

(1)n (ix) −π < arg x ≤ π

2π2(−i)n+1H

(2)n (−ix) −π

2< arg x ≤ π

(29)

From either of expressions (27) for H(1,2) we get:

Kn(x) =1

2

∫ +∞

−∞e−x cosh ξ±nξ dξ

This formula confirms the symmetry K−n(x) = Kn(x). From (28):

Kn(x) =

∫ +∞

−∞

(√u2 + 1± u)±ne−x

√u2+1 du

2√u2 + 1

where any choice of signs is valid.

6.7.1 Another integral representation

We can find another integral representation of Kn(x) by shifting the integra-tion line in ξ by an imaginary number (this is equivalent to going back tothe χ-plane and shifting the contour C ′1 again to {+i∞,−i∞}). Let us takeξ = ζ − iπ/2. Then cosh ξ = cos(iξ) = cos(iζ + π/2) = − sin(iζ) = −i sinh ζso we have

Kn(x) =1

2

∫ +∞+iπ2

−∞+iπ2

eix sinh ζ±nζ∓inπ/2 dζ =(∓i)n

2

∫ +∞+iπ2

−∞+iπ2

eix sinh ζ±nζ dζ

We want to shift the integration contour back to the real axis. Let us seewhat is going on at the ends of this integral:

2(±i)nKn(x) =

∫ +∞

−∞eix sinh ζ±nζ dζ+

∫ −X−X+iπ

2

eix sinh ζ±nζ dζ+

∫ X+iπ2

X

eix sinh ζ±nζ dζ

19

We are interested only in asymptotic behavior X → +∞. If the second andthird integral vanish, we can do the aforementioned shift. Let us check if thisis true.

In the second integral, change variable to ζ = −X + iθ:

second integral ≈ −i∫ π

2

0

e(−xeX/2)[ie−iθ]∓nX±inθ dθ

In the third integral, change variable to ζ = X + iθ:

third integral ≈ i

∫ π2

0

e(−xeX/2)[−ieiθ]±nX±inθ dθ

We are in luck again and both of these integrals vanish for x > 0 and n = 0because <{ie−iθ} > 0 and <{−ieiθ} > 0 on the interval θ ∈ [0, π/2]. So weget another integration rule

K0(x) =1

2

∫ +∞

−∞eix sinh ζ dζ (30)

which may be written as

K0(x) =

∫ +∞

0

cos(x sinh ζ) dζ

This formula is also used in QFT (Greiner and Reinhardt , 2009, p. 72), andin the last form, it is presented in Abramowitz and Stegun (1965, equation(9.6.21)). Note that unfortunately there is no “wiggle room” either in arg x(it has to be x > 0) or n (it has to be n = 0).

What happens when x > 0 but n 6= 0? Consider the second integral, inwhich for concreteness choose the lower sign:

2nd integral ≈ −i∫ π

2

0

e−(x/2)eX sin θ+nX−i[(x/2)eX cos θ+nθ] dθ

The biggest contribution will be for θ ≈ 0, so we can just write

2nd integral ≈ −ienX−i(x/2)eX∫ ∞0

e−(x/2)eXθ dθ = −ie

nX−i(x/2)eX

(x/2)eX=

2

ixe(n−1)X−i(x/2)e

X

Analogously, we find

3rd integral ≈ 2i

xe−(n+1)X+i(x/2)eX

20

and finally

2(−i)nKn(x) =

∫ +∞

−∞eix sinh ζ−nζ dζ+

2

ix

[e(n−1)X−i(x/2)e

X − e−(n+1)X+i(x/2)eX]

The second term is infinite for |n| > 1 and finite, but rapidly oscillating, forn = ±1. Thus, the shift to the real axis is only strictly valid for n = 0.

Let us consider case n = 1 more closely. The above formula gives twoequivalent presentations (taking n = ±1):

K1(x) = ± i2

∫ +∞

−∞eix sinh ζ∓ζ dζ +

e∓i(x/2)eX

x

In physical situations, we might be able to get rid of the oscillating termby shifting the contour into the complex domain just a very bit so that theexponent vanishes.

On the other hand, from recursion relation (5) (which is also valid forH(1,2)), symmetry in n and definition (29), we have K ′0(x) = −K1(x). Thus,we find

K1(x) = −1

2

d

dx

∫ +∞

−∞eix sinh ζ dζ = −1

2

∫ +∞

−∞i sinh ζ eix sinh ζ dζ

= − i4

[∫ +∞

−∞eix sinh ζ+ζ dζ −

∫ +∞

−∞eix sinh ζ−ζ dζ

]This is the average of the two versions of the above formula, with the neglectof the oscillating term.

6.7.2 Asymptotics

Asymptotics for K are obtained directly from its definition (29) and asymp-totics of H(1,2) (25,26):

Kn(x) ≈

log

(2

x

)+ const , n = 0

(n− 1)!

2

(2

x

)n, n > 0

x→ 0

and

Kn(x) ≈√

π

2xe−x, x→∞

21

References

Abramowitz, M., and I. A. Stegun (1965), Handbook of Mathematical Func-tions with Formulas, Graphs, and Mathematical Tables, Dover.

Greiner, W., and J. Reinhardt (2009), Quantum Electrodynamics, 4 ed.,Springer, doi:10.1007/978-3-540-87561-1.

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