How to Apply Cauchy Integral Theorem

Question

I have a question about complex analysis.

Let C be the unit circle z = e^jθ described from θ = -π to θ = π where α is a real constant. I need to prove this equation:

I thought that somehow I should rearrange the integrant to apply cauchy integral theorem. Thus I need singularity offspring of it.

Then I made variable displacement like this:

Now, I have the singularity but I'am stuck. How to go on with that?

Answer 1

We have

\begin{align} \int_0^\pi e^{a\cos \theta} \cos(a\sin \theta)\, d\theta &= \frac{1}{2}\int_{-\pi}^{\pi} e^{a\cos \theta} \cos(a\sin \theta)\, d\theta \\ &= \frac{1}{2}\operatorname{Re} \int_{-\pi}^{\pi} e^{ae^{i\theta}}\, d\theta\\ &= \frac{1}{2}\operatorname{Re} \oint_C e^{az} \frac{dz}{iz} \tag{*}\\ \end{align}

Since $e^{az}/i$ is analytic inside and on $C$, and $0$ lies inside $C$, Cauchy's integral formula gives

$$\oint_C e^{az} \frac{dz}{iz} = 2\pi i\cdot \frac{e^{az}}{i}\bigg|_{z = 0} = 2\pi.$$

Hence, by $(*)$,

$$\int_0^\pi e^{a\cos \theta}\cos(a\sin \theta)\, d\theta = \pi.$$