Complex Function Expansion

If z is a complex variable $\displaystyle z\in \mathbb{C}$

Prove that:

$\displaystyle e^{\frac{1}{2}a(z-1/z)}=\sum _{n=-\infty }^{\infty } J_n(a)z^n$

where:

$\displaystyle J_n(a)=\frac{1} {2\pi } \int _0^{2\pi }\text {cos} ({n*$\theta $} - a *{sin}(\theta )) d\theta$

Re: Complex Function Expansion

If you asked me that question 25 years ago, it would have been a cinch. All that I recall is that the function has an infinite (in both directions i.e. -inf to inf on the sum, as YOU have. One must apply the Residue Theorem judiciously. I shall try off line ans see if I can put it together. MAYBE, I''l be back!