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I have a question about complex analysis.

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

enter image description here

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:

enter image description here

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

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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 $(*)$,

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

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  • $\begingroup$ Thanks for the elegant solution! I am just wondering if my approach is going anywhere, you think? $\endgroup$ – mehmet Mar 7 '15 at 22:18
  • $\begingroup$ Hi @mehmet, your approach may work, but more work needs to be shown in order for me to get a clear idea where you're heading. Was your approach an attempt to solve this using real variables? $\endgroup$ – kobe Mar 7 '15 at 22:35
  • $\begingroup$ Actually yes, the integration becomes real integration I suppose. Is it a right path for this problem? Or I should just focus on complex integration? I can not estimate the consequences of it thinking as real integration. $\endgroup$ – mehmet Mar 8 '15 at 6:37
  • $\begingroup$ Since the question was to prove the equation using complex analysis, why not follow my method? The way you have written the integral (after substitution), it becomes less obvious how to express it in terms of a contour integral over $C$, which I believe was part of the point of the exercise. $\endgroup$ – kobe Mar 8 '15 at 6:44
  • $\begingroup$ Yeah I think you are right. It would be difficult to adapt it contour integral so my thought would result wrongly. I am following your solutio. I have gained an elegant view and I appreciate you. (: $\endgroup$ – mehmet Mar 8 '15 at 7:19

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