Question 6.33: Prove that if z≠0, then e^1/2a(z-1/z) = ∑n=-∞^∞ Jn(a)z^n whe...
Prove that if z \neq 0, then
e^{1 / 2 \alpha(z-1 / z)}=\sum\limits_{n=-\infty}^{\infty} J_{n}(\alpha) z^{n}
where
J_{n}(\alpha)=\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} \cos (n \theta-\alpha \sin \theta) d \theta \quad n=0,1,2, \ldots
The point z=0 is the only finite singularity of the function e^{1 / 2 \alpha(z-1 / z)} and it follows that the function must have a Laurent series expansion of the form
e^{1 / 2 \alpha(z-1 / z)}=\sum\limits_{n=-\infty}^{\infty} J_{n}(\alpha) z^{n} (1)
which holds for |z|>0. By equation (6.7), page 174, the coefficients J_{n}(\alpha) are given by
J_{n}(\alpha)=\frac{1}{2 \pi i} \oint\limits_{C} \frac{e^{1 / 2 \alpha(z-1 / z)}}{z^{n+1}} d z (2)
where C is any simple closed curve having z=0 inside.
Let us, in particular, choose C to be a circle of radius 1 having center at the origin; that is, the equation of C is |z|=1 or z=e^{i \theta}. Then (2) becomes
\begin{aligned} J_{n}(\alpha) & =\frac{1}{2 \pi i} \int\limits_{0}^{2 \pi} \frac{e^{1 / 2 \alpha\left(e^{i \theta}-e^{-i \theta}\right)}}{e^{i(n+1) \theta}} i e^{i \theta} d \theta=\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} e^{i \alpha \sin \theta-i n \theta} d \theta \\ & =\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} \cos (\alpha \sin \theta-n \theta) d \theta+\frac{i}{2 \pi} \int\limits_{0}^{2 \pi} \sin (\alpha \sin \theta-n \theta) d \theta=\frac{1}{2 \pi} \int\limits_{0}^{2 \pi} \cos (n \theta-\alpha \sin \theta) d \theta \end{aligned}
using the fact that I=\int_{0}^{2 \pi} \sin (\alpha \sin \theta-n \theta) d \theta=0. This last result follows since, on letting \theta=2 \pi-\phi, we find
I=\int\limits_{0}^{2 \pi} \sin (-\alpha \sin \phi-2 \pi n+n \phi) d \phi=-\int\limits_{0}^{2 \pi} \sin (\alpha \sin \phi-n \phi) d \phi=-I
so that I=-I and I=0. The required result is thus established.
The function J_{n}(\alpha) is called a Bessel function of the first kind of order n.
For further discussion of Bessel functions, see Chapter 10.