Evaluate $\int_0^{\pi} e^{a\cos(t)}\cos(a\sin t)dt$ [closed]

Solution 1:

By exploiting $\cos\theta=\frac{1}{2}\left(e^{i\theta}+e^{-i\theta}\right)$ and the Taylor series of the exponential function we have: $$\begin{eqnarray*}I(a)&=&\int_{0}^{\pi}e^{a\cos t}\cos(a\sin t)\,dt\\&=&\frac{1}{2}\int_{0}^{\pi}\exp\left(a e^{it}\right)+\exp(ae^{-it})\,dt\\&=&\frac{1}{2}\sum_{n\geq 0}\int_{0}^{\pi}\frac{a^n(e^{nit}+e^{-nit})}{n!}\,dt\end{eqnarray*} $$ but the innermost integral always vanishes, unless $n=0$. It follows that $I(a)$ is constant, and $$ I(0) = \pi $$ is trivial.

Solution 2:

The integral may be expressed at

$$\frac12 \operatorname{Re} \int_0^{2 \pi} dt \, e^{a e^{i t}} $$

The integral may be written in complex form as, letting $z=e^{i t}$:

$$-i \oint_{|z|=1} dz \frac{e^{a z}}{z} $$

which by Cauchy's integral theorem is simply $2 \pi$. Thus, the sought-after integral is $\pi$.