I need help finding the Fourier coefficients of:
$f(x) =\begin{cases} \sum_{n=0}^\infty{\frac{e^{inx}}{1+n^2}} & \text{if } x\neq 2k\pi \\0& \text{if } x= 2k\pi \end{cases}$
And my main problem is that I know how to find the coefficients for each case separately, but how do I reach a final answer for the whole function?
On
According to Mathematica $$\sum_{n=0}^\infty{1\over 1+n^2}={1+\pi\,{\rm coth}(\pi)\over2}\doteq 2.07667\ne0\ .$$ Therefore your function $f$ differs from the function $$g(x):=\sum_{n=0}^\infty {e^{inx}\over 1+n^2}$$ only at the points $x=2k\pi$, $\ k\in{\mathbb Z}$. It follows that $f$ has the same Fourier coefficients as $g$, namely $$c_n=\cases{{1\over 1+n^2}\quad&$(n\geq0)$ \cr 0&$(n<0)$\cr}\quad .$$
The fourier coefficients of a function depend only on its class in $L^2([0,2\pi])$ so you can think of $f$ to be the series $\sum_{n\geq 0} \dfrac{e^{inx}}{1+n^2}$ since they differ only on the set (of null Lebesgue measure) $2\pi\mathbb Z$.
If you don't really know much about Lebesgue theory, you want to compute the fourier coefficients in terms of the integrals : $c_n(f)=\frac{1}{2\pi}\int_{0}^{2\pi} f(t)e^{-inx}dx$. However you should know that the integral of a function doen't change if you modify its values in a finite number of points.
So the two integrals, with $f$ or with the series, are exactly the same : $c_n(f)=c_n(x\mapsto \sum_{n\geq 0} \frac{e^{inx}}{1+n^2})$.