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I'm trying to compute the following integral using complex analysis tools:

$$ \int_0^{2\pi} e^{e^{i\theta}} \, d\theta. $$

It was suggested to me to use the substitution $z = e^{i\theta}$, but that results in $dz = ie^{i\theta}$ and I don't see how everything can be substituted in to convert to $z$.

Anyone able to help?

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2 Answers 2

If you substitute $z = e^{i\theta}$, then you get $dz = ie^{i\theta}\,d\theta = iz\,d\theta$, and so $d\theta = \dfrac{dz}{iz}$.

Then, we get $\displaystyle\int_{0}^{2\pi}e^{e^{i\theta}}\,d\theta = \oint_{|z| = 1}\dfrac{e^z}{iz}\,dz$, which is easy if you know Cauchy's integral formula.

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The value of the integral is just $2\pi$. This happens because: $$e^{x} = \sum_{n=0}^{+\infty}\frac{x^n}{n!}$$ and $$\int_{0}^{2\pi}e^{mi\theta}d\theta$$ equals zero for any integer $m$, except $m=0$.

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