$\int_0^{2\pi}e^{\cos x}\cos(\sin x)dx$

Question

$$\int_0^{2\pi}e^{\cos x}\cos(\sin x)dx$$

I tried Integration by parts but failed. Wolfram alpha gives answer in decimal points which are same as of $2\pi$. Any hints or suggestions will be helpful.

It's the real part of $\int_{0}^\pi e^{e^{ix}},dx$. Not sure if that helps... — Thomas Andrews, Jul 19 '14 at 18:36
Duplicate of integrate $\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$ and of this.The last link doesn't use complex analysis. — Git Gud, Jul 19 '14 at 18:41

score 22 · Accepted Answer · answered Jul 19 '14 at 18:35

Note that

$$e^{\cos x}\cos (\sin x) = \operatorname{Re} \left(e^{\cos x}(\cos (\sin x) + i \sin (\sin x))\right) = \operatorname{Re} e^{\cos x + i \sin x}.$$

So we can transform the integral into a standard contour integral over the unit circle writing $z = e^{ix}$, which gives us $dx = \frac{dz}{iz}$, and the integral becomes

$$\operatorname{Re} \int_{\lvert z\rvert = 1} \frac{e^z}{iz}\,dz = \operatorname{Re} \frac{2\pi i e^0}{i} = 2\pi$$

by Cauchy's integral formula.

$\int_0^{2\pi}e^{\cos x}\cos(\sin x)dx$

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