# Evaluating $\int_{-\pi}^{\pi} \cos(e^{it})dt$

We have to calculate:

$$\displaystyle \int_{-\pi}^{\pi} \cos(e^{it})dt.$$

Is there something more promising one could try instead of a subsitution $u=e^{it}$?

-
You don't happen to recognize a contour integral around a circle? –  Guess who it is. Jul 23 '12 at 19:04
If you want brute force instead, you could use the Taylor series expansion of $\cos$. –  copper.hat Jul 23 '12 at 19:08
Oh, I see, so we have a path $\gamma:[-\pi,\pi]\rightarrow\mathbb{C}$, $\gamma(t)=e^{it}$ and we can rewrite it as: $\frac{1}{i}\int_{\gamma}\frac{cos(z)}{z}dz$ and use Cauchy's integral formula. –  Chris Jul 23 '12 at 19:10

Actually the Taylor series approach is fairly straightforward too...

$\int_{-\pi}^{\pi} \cos(e^{it}) \,dt = \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} (-1)^n \frac{e^{2int}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \int_{-\pi}^{\pi} \frac{e^{2int}}{(2n)!} \, dt = 2 \pi$.

The exchange of summation and integration is justified by uniform convergence on the unit circle. All terms except the constant disappear.

-
Shouldn't there be alternating signs? Not that it really matters, because the Taylor series essentially gives you the Fourier series, and you are still just looking for $2\pi$ times the degree $0$ coefficients. –  Thomas Andrews Jul 23 '12 at 20:45
@ThomasAndrews: Thanks, just fixed the omission... –  copper.hat Jul 23 '12 at 20:57