Bessel functions

The Bessel function Jν(z) of the first kind of order ν is defined by

∞ (z/2)ν  − z2  z ν X (−1)k z 2k J (z) = F ; − = . (1) ν Γ(ν + 1) 0 1 ν + 1 4 2 Γ(ν + k + 1) k! 2 k=0 For ν ≥ 0 this is a solution of the Bessel differential equation

z2y00(z) + zy0(z) + z2 − ν2
y(z) = 0, ν ≥ 0. (2)

For ν∈ / {0, 1, 2,...} we have that J−ν(z) is a second solution of the differential equation (2) and the two solutions Jν(z) and J−ν(z) are clearly linearly independent. For ν = n ∈ {0, 1, 2,...} we have

∞ ∞ z −n X (−1)k z 2k z −n X (−1)k z 2k J (z) = = , −n 2 Γ(−n + k + 1) k! 2 2 Γ(−n + k + 1) k! 2 k=0 k=n since 1 = 0 for k = 0, 1, 2, . . . , n − 1. Γ(−n + k + 1) This implies that

∞ ∞ z −n X (−1)k z 2k z −n X (−1)n+k z 2(n+k) J (z) = = −n 2 Γ(−n + k + 1) k! 2 2 Γ(k + 1) (n + k)! 2 k=n k=0 ∞ z n X (−1)k z 2k = (−1)n = (−1)nJ (z). 2 Γ(n + k + 1) k! 2 n k=0

This implies that Jn(z) and J−n(z) are linearly dependent for n ∈ {0, 1, 2,...}. A second linearly independent solution can be found as follows. Since (−1)n = cos nπ, we see that Jν(z) cos νπ − J−ν(z) is a solution of the differential equation (2) which vanishes when ν = n ∈ {0, 1, 2,...}. Now we define

J (z) cos νπ − J (z) Y (z) := ν −ν , (3) ν sin νπ where the case that ν = n ∈ {0, 1, 2,...} should be regarded as a limit case. By l’Hopital’s rule we have   n   1 ∂Jν(z) (−1) ∂J−ν(z) Yn(z) = lim Yν(z) = − . ν→n π ∂ν ν=n π ∂ν ν=n n This implies that Y−n(z) = (−1) Yn(z) for n ∈ {0, 1, 2,...}. The function Yν(z) is called the Bessel function of the second kind of order ν. Using the definition (1) we find that

  ∞ k ∂Jν(z) z  z n X (−1) ψ(n + k + 1) z 2k = J (z) ln − , ∂ν n 2 2 (n + k)! k! 2 ν=n k=0

1 where d Γ0(z) ψ(z) = ln Γ(z) = . dz Γ(z) For ν∈ / {0, 1, 2,...} we have

∞ k ∂J−ν(z) z  z −ν X (−1) ψ(−ν + k + 1) z 2k = −J (z) ln + . ∂ν ν 2 2 Γ(−ν + k + 1) k! 2 k=0 Now we use (−1)n (−1)n lim (z + n)Γ(z) = =⇒ Γ(z) ∼ for z → −n z→−n n! (z + n) n! and (−1)n 0 Γ (z) − 2 1 ψ(z) = ∼ (z+n) n! = − for z → −n. Γ(z) (−1)n z + n (z+n) n! This implies that ψ(z) lim = (−1)n+1n! for n = 0, 1, 2,.... z→−n Γ(z) Hence ∞ X (−1)kψ(−ν + k + 1) z 2k lim ν→n Γ(−ν + k + 1) k! 2 k=0 n−1 ∞ X (n − k − 1)! z 2k X (−1)kψ(−n + k + 1) z 2k = (−1)n + k! 2 Γ(−n + k + 1) k! 2 k=0 k=n n−1 ∞ X (n − k − 1)! z 2k z 2n X (−1)kψ(k + 1) z 2k = (−1)n + (−1)n . k! 2 2 Γ(k + 1) (n + k)! 2 k=0 k=0 This implies that

  n−1 ∂J−ν(z) z  z −n X (n − k − 1)! z 2k = −J (z) ln + (−1)n ∂ν −n 2 2 k! 2 ν=n k=0 ∞ z n X (−1)kψ(k + 1) z 2k + (−1)n . 2 (n + k)! k! 2 k=0

n Finally we use the fact that J−n(z) = (−1) Jn(z) to conclude that

n−1 2 z  1 z −n X (n − k − 1)! z 2k Y (z) = J (z) ln − n π n 2 π 2 k! 2 k=0 ∞ 1 z n X (−1)k z 2k − [ψ(n + k + 1) + ψ(k + 1)] π 2 (n + k)! k! 2 k=0 for n ∈ {0, 1, 2,...}. Compare with the theory of Frobenius for linear second differential equations.

2 In the theory of second order linear differential equations of the form

y00 + p(z)y0 + q(z)y = 0,

two solutions y1 and y2 are linearly independent if and only if

y (z) y (z) 1 2 0 0 W (y1, y2)(z) := = y1(z)y (z) − y (z)y2(z) 6= 0. 0 0 2 1 y1(z) y2(z)

This determinant is called the Wronskian of the solutions y1 and y2. It can (easily) be shown that this determinant of Wronski satisfies the differential equation

W 0(z) + p(z)W (z) = 0.

This result is called Abel’s theorem or the theorem of Abel-Liouville. In the case of the Bessel differential equation we have p(z) = 1/z, which implies that 1 c W 0(z) + W (z) = 0 =⇒ W (y , y )(z) = z 1 2 z for some constant c. Now we have

Theorem 1. 2 sin νπ 2 W (J ,J )(z) = − and W (J ,Y )(z) = . (4) ν −ν πz ν ν πz

For ν ≥ 0 this implies that Jν(z) and J−ν(z) are linearly independent if ν∈ / {0, 1, 2,...} and that Jν(z) and Yν(z) are linearly independent for all ν ≥ 0.

Proof. Note that

0 0 W (Jν,Yν)(z) = Jν(z)Yν (z) − Jν(z)Yν(z) J 0 (z) cos νπ − J 0 (z) J (z) cos νπ − J (z) = J (z) ν −ν − J 0 (z) ν −ν ν sin νπ ν sin νπ J (z)J 0 (z) − J 0 (z)J (z) W (J ,J )(z) = − ν −ν ν −ν = − ν −ν . sin νπ sin νπ Now we use the definition (1) to obtain

∞ ∞ X (−1)k zν+2k X (−1)k(ν + 2k) zν+2k−1 J (z) = · =⇒ J 0 (z) = · ν Γ(ν + k + 1) k! 2ν+2k ν Γ(ν + k + 1) k! 2ν+2k k=0 k=0 and ∞ X (−1)m z−ν+2m J (z) = · −ν Γ(−ν + m + 1) m! 2−ν+2m m=0 ∞ X (−1)m(−ν + 2m) z−ν+2m−1 =⇒ J 0 (z) = · . −ν Γ(−ν + m + 1) m! 2−ν+2m m=0

3 Hence we have

 0 0  z W (Jν,J−ν)(z) = z Jν(z)J−ν(z) − Jν(z)J−ν(z) ∞ ∞ X X (−1)k+m(−ν + 2m) z2k+2m = · Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 22k+2m k=0 m=0 ∞ ∞ X X (−1)k+m(ν + 2k) z2k+2m − · Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 22k+2m k=0 m=0 ∞ ∞ X X (−1)k+m(2ν + 2k − 2m) z2k+2m = − · . Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 22k+2m k=0 m=0 This implies that 2ν 2ν 2 sin νπ lim z W (Jν,J−ν)(z) = − = − = − . z→0 Γ(ν + 1)Γ(−ν + 1) νΓ(ν)Γ(1 − ν) π

Using the definition (1) we obtain

∞ ∞ d d X (−1)k z2ν+2k X (−1)k(2ν + 2k) z2ν+2k−1 [zνJ (z)] = · = · dz ν dz Γ(ν + k + 1) k! 2ν+2k Γ(ν + k + 1) k! 2ν+2k k=0 k=0 ∞ X (−1)k z2ν+2k−1 = · = zνJ (z). Γ(ν + k) k! 2ν+2k−1 ν−1 k=0 Hence we have d [zνJ (z)] = zνJ (z) ⇐⇒ zJ 0 (z) + νJ (z) = zJ (z). (5) dz ν ν−1 ν ν ν−1 Similarly we have

∞ ∞ d d X (−1)k z2k X (−1)k 2k z2k−1 z−νJ (z) = · = · dz ν dz Γ(ν + k + 1) k! 2ν+2k Γ(ν + k + 1) k! 2ν+2k k=0 k=0 ∞ X (−1)k+1 z2k+1 = · = −z−νJ (z). Γ(ν + k + 2) k! 2ν+2k+1 ν+1 k=0 Hence we have d z−νJ (z) = −z−νJ (z) ⇐⇒ zJ 0 (z) − νJ (z) = −zJ (z). (6) dz ν ν+1 ν ν ν+1 0 Elimination of Jν(z) from (5) and (6) gives 2ν J (z) + J (z) = J (z) ν−1 ν+1 z ν and elimination of Jν(z) from (5) and (6) gives

0 Jν−1(z) − Jν+1(z) = 2Jν(z).

4 Special cases

For ν = 1/2 we have from the definition (1) by using Legendre’s duplication formula for the gamma function

r ∞ k r ∞ k r x X (−1) x2k 2x X (−1) 2 J (x) = = x2k = sin x, x > 0 1/2 2 Γ(k + 3/2) k! 2 π (2k + 1)! πx k=0 k=0 and for ν = −1/2 we have r ∞ r ∞ r 2 X (−1)k x2k 2 X (−1)k 2 J (x) = = x2k = cos x, x > 0. −1/2 x Γ(k + 1/2) k! 2 πx (2k)! πx k=0 k=0 Note that the definition (3) implies that r 2 r 2 Y (x) = −J (x) = − cos x and Y (x) = J (x) = sin x, x > 0. 1/2 −1/2 πx −1/2 1/2 πx

Integral representations

First we will prove Theorem 2. ν Z 1 1 z  izt 2 ν−1/2 Jν(z) = √ e (1 − t ) dt, Re ν > −1/2. (7) Γ(ν + 1/2) π 2 −1 Proof. We start with ∞ Z 1 X (iz)n Z 1 eizt(1 − t2)ν−1/2 dt = tn(1 − t2)ν−1/2 dt. n! −1 n=0 −1 Note that the latter integral vanishes when n is odd. For n = 2k we obtain using t2 = u Z 1 Z 1 du Z 1 t2k(1 − t2)ν−1/2 dt = 2 uk(1 − u)ν−1/2 √ = uk−1/2(1 − u)ν−1/2 du −1 0 2 u 0 Γ(k + 1/2)Γ(ν + 1/2) = B(k + 1/2, ν + 1/2) = . Γ(ν + k + 1) Now we use Legendre’s duplication formula to find that √ √ √ π Γ(2k) π Γ(2k + 1) π (2k)! Γ(k + 1/2) = · = · = · . 22k−1 Γ(k) 22k Γ(k + 1) 22k k! Hence we have ∞ Z 1 X (iz)2k Γ(k + 1/2)Γ(ν + 1/2) eizt(1 − t2)ν−1/2 dt = · (2k)! Γ(ν + k + 1) −1 k=0 ∞ √ X (−1)k z 2k = Γ(ν + 1/2) π . Γ(ν + k + 1) k! 2 k=0 This proves the theorem.

5 We also have Poisson’s integral representations Theorem 3. ν Z π 1 z  iz cos θ 2ν Jν(z) = √ e (sin θ) dθ Γ(ν + 1/2) π 2 0 1 z ν Z π = √ cos(z cos θ) (sin θ)2ν dθ, Re ν > −1/2. Γ(ν + 1/2) π 2 0 Proof. Use the substitution t = cos θ to obtain Z 1 Z 0 Z π eizt(1 − t2)ν−1/2 dt = eiz cos θ(1 − cos2 θ)ν−1/2 · (− sin θ) dθ = eiz cos θ(sin θ)2ν dθ. −1 π 0 Further we have eiz cos θ = cos(z cos θ) + i sin(z cos θ) and Z π sin(z cos θ)(sin θ)2ν dθ = 0. 0 This shows that Poisson’s integral representations follow from the integral representation (7).

Remarks: 1. The Fourier transform is defined by 1 Z ∞ F (z) := √ e−iztf(t) dt 2π −∞ with inversion formula 1 Z ∞ f(t) = √ eiztF (z) dz. 2π −∞ This implies that the Fourier transform of the function ( (1 − t2)ν−1/2, |t| ≤ 1 f(t) = 0, |t| > 1 is Γ(ν + 1/2) z −ν F (z) = √ Jν(z). 2 2 2. Instead of the substitution t = cos θ in (7) one can also use the substitution t = sin θ, which leads to slightly different forms of Poisson’s integral representations. In fact we have

ν Z π/2 1 z  iz sin θ 2ν Jν(z) = √ e (cos θ) dθ Γ(ν + 1/2) π 2 −π/2 1 z ν Z π/2 = √ cos(z sin θ) (cos θ)2ν dθ Γ(ν + 1/2) π 2 −π/2 2 z ν Z π/2 = √ cos(z sin θ) (cos θ)2ν dθ, Re ν > −1/2. Γ(ν + 1/2) π 2 0

6 Integrals of Bessel functions

The Hankel transform of a function f is defined by Z ∞ F (s) = tf(t)Jν(st) dt 0 for functions f for which the integral converges. The inversion formula is given by Z ∞ f(t) = sF (s)Jν(st) ds. 0 This pair of integrals is called a Hankel pair of order ν.

An example of such an integral is

Z ∞ µ+ν
ν  2  µ−1 −ρ2t2 Γ 2 s (µ + ν)/2 s t e Jν(st) dt = ν+1 · µ+ν · 1F1 ; − 2 , Re(µ + ν) > 0. 0 2 Γ(ν + 1) ρ ν + 1 4ρ It can be shown that the integral converges for Re(µ + ν) > 0. Now we use the definition (1) to obtain Z ∞ ∞ k ν+2k Z ∞ 2 2 X (−1) s 2 2 tµ−1e−ρ t J (st) dt = · · tµ+ν+2k−1e−ρ t dt. ν Γ(ν + k + 1) k! 2ν+2k 0 k=0 0 Using the substitution ρ2t2 = u we find that ∞ ∞ Z 2 2 Z du tµ+ν+2k−1e−ρ t dt = ρ−µ−ν−2k+1 u(µ+ν+2k−1)/2e−u √ 0 0 2ρ u 1 Z ∞ = ρ−µ−ν−2k u(µ+ν+2k−2)/2e−u du 2 0 1 µ + ν + 2k  = ρ−µ−ν−2k Γ . 2 2 Hence we have Z ∞ ∞ k µ+ν
ν+2k 2 2 1 X (−1) Γ + k s tµ−1e−ρ t J (st) dt = ρ−µ−ν 2 · ν 2 Γ(ν + k + 1) k! 2ν+2k ρ2k 0 k=0 µ+ν
ν ∞ k 2k Γ 2 s X (−1) ((µ + ν)/2)k s = µ+ν · ν · · 2k 2k 2 ρ Γ(ν + 1) 2 (ν + 1)k k! 2 ρ k=0 Γ µ+ν
sν (µ + ν)/2 s2  = 2 · · F ; − . 2ν+1Γ(ν + 1) ρµ+ν 1 1 ν + 1 4ρ2

The special case µ = ν + 2 is of special interest: in that case we have (µ + ν)/2 = ν + 1. This implies that the 1F1 reduces to a 0F0 which is an exponential function. The result is Z ∞ ν ν+1 −ρ2t2 s −s2/4ρ2 t e Jν(st) dt = 2 ν+1 · e , Re ν > −1. 0 (2ρ )

7 Hankel functions

(1) (2) The functions Hν (z) and Hν (z) are defined by (1) (2) Hν (z) := Jν(z) + iYν(z) and Hν (z) := Jν(z) − iYν(z). These functions are called Hankel functions or Bessel functions of the third kind. Note that these definitions imply that H(1)(z) + H(2)(z) H(1)(z) − H(2)(z) J (z) = ν ν and Y (z) = ν ν . ν 2 ν 2i Further we have r 2 r 2 H(1) (x) = J (x) + iY (x) = (cos x + i sin x) = eix, x > 0 −1/2 −1/2 −1/2 πx πx and r 2 r 2 H(2) (x) = J (x) − iY (x) = (cos x − i sin x) = e−ix, x > 0. −1/2 −1/2 −1/2 πx πx Similarly we have r 2 r 2 H(1) (x) = J (x) + iY (x) = (sin x − i cos x) = −i eix, x > 0 1/2 1/2 1/2 πx πx and r 2 r 2 H(2) (x) = J (x) − iY (x) = (sin x + i cos x) = i e−ix, x > 0. 1/2 1/2 1/2 πx πx

Modified Bessel functions

The modified Bessel function Iν(z) of the first kind of order ν is defined by ∞ (z/2)ν  − z2  z ν X 1 z 2k I (z) := F ; = . ν Γ(ν + 1) 0 1 ν + 1 4 2 Γ(ν + k + 1) k! 2 k=0 For ν ≥ 0 this is a solution of the modified Bessel differential equation z2y00(z) + zy0(z) − z2 + ν2
y(z) = 0, ν ≥ 0.

For ν∈ / {0, 1, 2,...} we have that I−ν(z) is a second solution of this differential equation and the two solutions Iν(z) and I−ν(z) are linearly independent. For ν = n ∈ {0, 1, 2,...} we have I−n(z) = In(z). The modified Bessel function Kν(z) of the second kind of order ν is defined by π K (z) := [I (z) − I (z)] for ν∈ / {0, 1, 2,...} ν 2 sin νπ −ν ν and Kn(z) = lim Kν(z) for n ∈ {0, 1, 2,...}. ν→n Now we have for x > 0 r 2 r 2 r π I (x) = sinh x, I (x) = cosh x and K (x) = K (x) = e−x. 1/2 πx −1/2 πx 1/2 −1/2 2x

8 A generating function

The Bessel function Jn(z) of the first kind of integer order n ∈ Z can also be defined by means of the generating function ∞ 1  X exp z t − t−1
= J (z)tn. (8) 2 n n=−∞ In fact, the series on the right-hand side is a so-called Laurent series at t = 0 for the function at the left-hand side. Using the Taylor series for the exponential function we obtain ∞ j ∞ ∞ 1  zt  z  X 1 zt X (−1)k  z k X exp z t − t−1
= exp · exp − = · = a tn. 2 2 2t j! 2 k! 2t n j=0 k=0 n=−∞ For n ∈ {0, 1, 2,...} we have ∞ ∞ X 1 z n+k (−1)k z k z n X (−1)k z 2k a = · = = J (z) n (n + k)! 2 k! 2 2 (n + k)! k! 2 n k=0 k=0 and ∞ X 1 z j (−1)j+n z n+j a = · = (−1)nJ (z) = J (z). −n j! 2 (n + j)! 2 n −n j=0 This proves (8).

If t = eiθ, then we have 1 eiθ − e−iθ t − t−1
= = i sin θ 2 2 1  =⇒ exp x t − t−1
= eix sin θ = cos(x sin θ) + i sin(x sin θ). 2 Hence we have ∞ ∞ ix sin θ X inθ X cos(x sin θ) + i sin(x sin θ) = e = Jn(x)e = Jn(x) [cos(nθ) + i sin(nθ)] . n=−∞ n=−∞ n Since J−n(x) = (−1) Jn(x) this implies that ∞ ∞ X X cos(x sin θ) = Jn(x) cos(nθ) = J0(x) + 2 J2k(x) cos(2kθ) n=−∞ k=1 and ∞ ∞ X X sin(x sin θ) = Jn(x) sin(nθ) = 2 J2k+1(x) sin(2k + 1)θ. n=−∞ k=0 For θ = π/2 this implies that ∞ ∞ X k X k cos x = J0(x) + 2 (−1) J2k(x) and sin x = 2 (−1) J2k+1(x). k=1 k=0 For θ = 0 we also have ∞ X 1 = J0(x) + 2 J2k(x). k=1

9 The generating function (8) can be used to prove that

∞ X Jn(x + y) = Jk(x)Jn−k(y). k=−∞

The proof is as follows:

∞ X 1  J (x + y)tn = exp (x + y) t − t−1
n 2 n=−∞ 1  1  = exp x t − t−1
· exp y t − t−1
2 2 ∞ ∞ ∞ ∞ ! X k X m X X n = Jk(x)t · Jm(y)t = Jk(x)Jn−k(y) t . k=−∞ m=−∞ n=−∞ k=−∞

We can also find an integral representation for the Bessel function Jn(x) of the first kind of integral order n starting from the generating function

∞ ix sin θ X inθ e = Jn(x)e . n=−∞

We use the orthogonality property of the exponential function, id est ( Z π 0, k ∈ Z, k 6= 0 eikθ dθ = −π 2π, k = 0.

Hence we have

Z π ∞ Z π ix sin θ −inθ X ikθ −inθ e · e dθ = Jk(x) e · e dθ = 2π · Jn(x). −π k=−∞ −π

This implies that Z π Z π 1 i(x sin θ−nθ) 1 Jn(x) = e dθ = cos (x sin θ − nθ) dθ. 2π −π π 0 The special case n = 0 reads

1 Z π 2 Z 1 cos(xt) J0(x) = cos (x sin θ) dθ = √ dt. 2 π 0 π 0 1 − t

10