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QUESTION


Let $C$ denote the unit circle $z=e^{i\theta}$\ \ $(-\pi&lt;
\theta\le\pi)$. Show that for any real constant $a$
$$\int_C{{e^{az}\over z}dz}=2\pi i$$ and then by writing the
integral in terms of $\theta$  derive the formula
$$\int_0^\pi{e^{a\cos \theta} \cos(a\sin\theta)}d\theta=\pi.$$



ANSWER

 In (*), $f(z)=e^{az}$, $b=0$. Thus $\int_C{e^az\over z}=2\pi
ie^0=2\pi i.$ As $C$ is the unit circle that can be parameterized
as $\{e^{i\theta}|-\pi&lt;\theta\le \pi\}$ We put
$z=e^{i\theta}=\cos\theta+i\sin \theta$ and get $\int_{-\pi}^\pi
e^{a(\cos\theta+i\sin \theta)}d\theta=2\pi.$ Hence
$\int_{-\pi}^{\pi}e^{a\cos\theta}e^{ia\sin\theta}d\theta=2\pi.$
Thus $$\int_{-\pi}^{\pi}e^{a\cos\theta}\cos(a\sin\theta)d\theta +
i\int_{-\pi}^{\pi}e^{a\cos\theta}\sin(a\sin\theta)d\theta =2\pi$$
Now equate real parts and use the fact that the first integrand is
even to get the solution.




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