# 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.

• Did you try with Bessel function? Aug 10, 2018 at 16:37
• I would use the substitution $u=\cos\theta$ and use Integration by Parts Aug 10, 2018 at 16:38
• expand $e^{\cos\theta}$ as $\sum \cos^n\theta/n!$ and integrate termwise. Aug 10, 2018 at 16:45
• @LordSharktheUnknown Crap my brain jams recently. Thank you. Aug 10, 2018 at 16:54

## 2 Answers

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: $$\frac{1}{2\pi}\int\limits_{0}^{2\pi} e^{\cos \theta} \ d \theta = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2\pi (2k)!}\frac{2 \pi}{2^{2k}}\frac{(2k)!}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{2^{2k}}\frac{(1}{(k!)^2} = \sum\limits_{k=0}^{\infty} \frac{1}{(2^k k!)^2}$$

• Very detailed. Thank you. Aug 10, 2018 at 16:55
• @MonkeyKing welcome !! Aug 10, 2018 at 16:56

Hint: With Bessel's integrals form and also series form $$J_0(x)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}e^{-ix\sin\tau}\ d\tau=\sum_{k=0}^\infty(-1)^n\dfrac{1}{(k!)^2}\left(\dfrac{x}{2}\right)^{2k}$$ set $x=i$ and you will find the result. Also modified Bessel functions $I_0(x)$ has the same case.