Power series expansion of $e^{\frac{c}{2}(z-1/z)}$

Show that

$$e^{\frac{c}{2}(z-1/z)}=\sum_{n=-\infty}^{\infty}a_nz^n$$

where

$$a_n:=\frac{1}{2\pi}\int_0^{2\pi}\cos(n\theta-c\sin(\theta))d\theta$$

• This is an incredibly inconvenient - but known and understood - reference to the $J$-Bessel function. From that, my answer here also answers this question. – davidlowryduda May 17 '14 at 5:49
• @mixedmath In the link that you sent me to, I believe you have a mistake/typo, you prove that $$e^{c/2(z-z^{-1})}=\sum J_n(x)z^n$$ which is contrary to your claim, however, thanks, I have been able to prove it – user115850 May 17 '14 at 6:39
• Since you know markup properly, can you please edit $z-1/z$ as $z-\frac{1}{z}$ or $\frac{z-1}{z}$ as appropriate? Thanks! – MattAllegro May 17 '14 at 7:04
• @MattAlegro, it can't mean anything but the former – vonbrand May 17 '14 at 10:16