# Interesting Integral

• Apr 24th 2009, 04:08 PM
TheEmptySet
Interesting Integral
Show that for $a \ne 0, a \in \mathbb{R}$ that

$\int_{0}^{\pi}\cos(a\sin(\theta))e^{a\cos(\theta)} d\theta=\pi$
• Apr 24th 2009, 04:56 PM
NonCommAlg
Quote:

Originally Posted by TheEmptySet

Show that for $a \ne 0, a \in \mathbb{R}$ that $\int_{0}^{\pi}\cos(a\sin(\theta))e^{a\cos(\theta)} d\theta=\pi$

it's nice but probably not hard enough! (Wink)

Spoiler:

$e^{ae^{i \theta}}=e^{a \cos \theta}e^{ia\sin \theta}=e^{a \cos \theta}(\cos(a \sin \theta) + i \sin(a \sin \theta)).$ therefore: $e^{a \cos \theta} \cos(a \sin \theta) = \text{Re}(e^{ae^{i \theta}})=\text{Re} \left(\sum_{n=0}^{\infty}\frac{a^ne^{in \theta}}{n!} \right)=\sum_{n=0}^{\infty} \frac{a^n \cos(n \theta)}{n!}.$

hence $\int e^{a \cos \theta} \cos(a \sin \theta) \ d \theta = \theta + \sum_{n=1}^{\infty} \frac{a^n \sin(n \theta)}{n \cdot n!} +C.$ your definite integral is just a simple result of what we have now!
• Apr 24th 2009, 05:22 PM
TheEmptySet
Wow that is a very nice solution. (Clapping)Better than this, but here is another way....

Consider the integral where $\gamma$ is the unit circle

$\int_\gamma \frac{e^{az}}{z}dz$ in the complex plane. Then by cauchy's integral formula or the residue theorem.

$\int_\gamma \frac{e^{az}}{z}dz=2\pi i$

Parameterize the curve with $z=e^{i\theta}$ for $-\pi \le \theta \le \pi$ and then $dz=ie^{i\theta}d\theta=izd\theta$=

$\int_{-\pi}^{\pi} \frac{e^{ae^{i\theta}}}{z}izd\theta=i\int_{-\pi}^{\pi}e^{a\cos(\theta) +ia\sin(\theta)}d\theta=i\int_{-\pi}^{\pi}e^{a\cos(\theta)}[\cos(a\sin(\theta)+i\sin(a\sin(\theta))]d\theta=$

$i\int_{-\pi}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)}d\the ta-\int_{-\pi}^{\pi}\sin(a\sin(\theta))e^{a\cos(\theta)}d\th eta$

So we know that this equals $2\pi i$ setting immaginary parts equal we get

$2\pi =\int_{-\pi}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)}d\the ta$

Since this is an even function we get (Finally)

$\pi =\int_{0}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)} d\theta$