<s>[INST] How to prove that $\frac{1}{2\pi}\int^{2\pi}_0 e^{\cos \theta}\,d\theta = \sum\limits^\infty_{n=0}\frac{1}{(n!2^n)^2}$ 
How to prove that $$\frac{1}{2\pi}\int^{2\pi}_0 e^{\cos \theta}\,d\theta = \sum\limits^\infty_{n=0}\frac{1}{(n!2^n)^2}.$$

It seems Laurent expansion doesn't work well here. I visited similar questions, maybe expand $1=\cos^2(\theta)+\sin^2(\theta)$ or some different arguments? Still not working to me.
 [/INST] Taylor series of $$e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$$
For $$x = \cos \theta$$
$$e^{\cos \theta} = \sum\limits_{n=0}^{\infty}\frac{\cos^n\theta}{n!}$$
Let's integrate
$$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{n=0}^{\infty} \frac{1}{2\pi n!}\int\limits_{0}^{2\pi}\cos^n\theta \ d\theta$$
But for even $n$ we have: $$\int\limits_{0}^{2\pi} \cos^n \theta = \frac{2 \pi}{2^{2k}}\frac{(2k)!}{(k!)^2}$$
and zero for odd $n$ (See here why), 
Therefore:
$$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{2 \pi}{2^{2k}}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2^{2k}}\frac{(1}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{(2^k k!)^2}$$
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