Complex Fourier series

Question

I need to find the complex Fourier series of this function, and I'm having problems calculating these integers:

$$|a|<1$$

$$x\in [-\pi,\pi]$$

$$f(x)=\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}$$

$$a_0=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}dx$$

$$b_n=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}\sin(nx)dx$$

Answer 1

Note: The following answers the question before it was edited and changed.

Let $z = e^{ix}$ (so $|z|=1$) then

$$ f(x) = \frac{1-a^2}{(1-a z)(1-a z^{-1})} = \frac{a}{z-a} + \frac{1}{1- a z}. $$

As a function of $z$ this has the following Laurent expansion on the unit circle:

$$ \frac{a}{z}\sum_{k=0}^{\infty} \frac{a^k}{z^k} + \sum_{k=0}^{\infty} a^k z^k = 1 + \sum_{k=1}^{\infty}a^k(z^k + z^{-k}) = 1 + 2 \sum_{k=1}^{\infty} a^k \cos(k x) $$

This gives the Fouries series of $f$ without calculating any integral.

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The same can be done for the latest form of $f$.

$$ f(x) = \frac{2-a (z + z^{-1})}{2(1-a z)(1 - a z^{-1})} = \frac{a}{2 (z-a)}+\frac{1}{2 (1 - a z)}+\frac{1}{2} $$

The Fourier series is now apparent given the previous result:

$$ f(x) = 1 + \sum_{k=1}^{\infty} a^k \cos(k x) $$