compute the following integral using Cauchy Integral Formula

Question

Prove that $\int_{0}^{\pi}{e^{k\cos t}\cos (k\sin t)}=\pi$. Using Cauchy Integral Formula. But I don't know how. I want to rewrite the integral as a line integral first.

Answer 1

First of all, note that one may exploit symmetry in the integral and rewrite as

$$\frac12 \int_0^{2 \pi} dt \, e^{k \cos{t}} \, \cos{(k \sin{t})}$$

Now sub $z=e^{i t}$, $dt = -i dz/z$, and note that

$$e^{k \cos{t}} \, \cos{(k \sin{t})} = \Re{\left [ e^{k z}\right]}$$

(To see this, expand $e^{k e^{i t}}$ into real and imaginary parts.) The integral is therefore equal to

$$\Re{\left [-\frac{i}{2} \oint_{|z|=1} dz\, \frac{e^{k z}}{z}\right]}$$

By Cauchy's theorem/residue theorem, the integral is equal to $i 2 \pi$ times the residue at the pole $z=0$, which is $1$. The result to be shown follows.