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I came across the following integral which seems pretty interesting when I evaluated it with Woflram Alpha:

$$ I = \int_0^{2 \pi} \frac{a e^{i\theta} + b}{c e^{i\theta} + d} e^{i n \theta} d\theta $$

I found the following:

\begin{equation} I = \begin{cases} \displaystyle \frac{2\pi b}{d} &\quad n = 0, \\ 0 & \quad n \neq 0. \end{cases} \end{equation}

Does anyone know of a method to arrive at this result?

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    What happens when $d=0$? – 2017-02-27
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    @UmbertoP. When $d = 0$ the result is $$\frac{2\pi a}{c}$$ – 2017-02-27
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    $n \in\ \color{#f00}{\large ?}$. – 2017-02-27

2 Answers 2

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It is enough to exploit the orthogonality relation $$ \int_{0}^{2\pi}e^{-mi\theta}e^{ni\theta}\,d\theta = 2\pi\,\delta(m,n) $$ and expand $\frac{ae^{i\theta}+b}{c e^{i\theta}+d}$ as a geometric series in $e^{i\theta}$ or $e^{-i\theta}$, according to $|c|>|d|$ or $|d|>|c|$.
If $|c|=|d|$ we have a singular integral since $e^{i\pi}+1=0$.

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By some algebraic manuplation you can turn it into a contour integral in complex,I guess plane and use cauchy integral formula.