Let $f(z)=\frac{\cos(\pi z)}{z^2 \sin(\pi z)}$. I want to compute the following integral $$\int_{\gamma_n}f(z)\;dz$$, where $\gamma_n$ is the circle $|z|=n+\frac{1}{2}$.
My work so far: first tougth is to use the residue theorem, I know that the poles of the function are exactly the integers, and i know how to calculate the residues for $\{\pm 1,\cdot...\pm n\}$. But im having trouble to find the residue in $z=0$ since it's not a simple pole. Any help would be appreciated
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Since $0$ is a zero of order $3$ of $z^2\sin(\pi z)$, since $0$ is not a zero of $\cos(\pi z)$, and since $f$ is an odd function, then the Laurent series of $f$ near $0$ is of the form$$\frac{a_0}{z^3}+\frac{a_1}z+a_2z+a_3z^3+\cdots$$and $\operatorname{res}_{z=0}f(z)=a_1$. Since whe have$$\frac{a_0}{z^3}+\frac{a_1}z+a_2z+a_3z^3+\cdots=f(z)=\frac{1-\frac{\pi^2}2z^2+\frac{\pi^4}{24}z^4-\cdots}{\pi^3z^3-\frac{\pi^3}6z^5+\frac{\pi^5}{120}z^7-\cdots},$$you have$$a_0+a_1z^2+a_3z^4+a_3z^6+\cdots=\frac{1-\frac{\pi^2}2z^2+\frac{\pi^4}{24}z^4-\cdots}{\pi^3-\frac{\pi^3}6z^2+\frac{\pi^5}{120}z^4-\cdots}.$$So,\begin{align*}1-\frac{\pi^2}2z^2+\frac{\pi^4}{24}z^4-\cdots&=\left(a_0+a_1z^2+a_2z^4+\cdots\right)\left(\pi^3-\frac{\pi^3}6z^2+\frac{\pi^5}{120}z^4-\cdots\right)\\&=a_0\pi^3+\left(a_1\pi^3-a_0\frac{\pi^3}6\right)z^2+\cdots\end{align*}and therefore $a_0=\frac1{\pi^3}$. So, since $a_1\pi^3-a_0\frac{\pi^3}6=-\frac{\pi^2}2$, it is easy to compute $a_1\left(=-\frac\pi3\right)$.
Since $f(z)$ is holomorphic in $\mathbb{D}(0,n+1/2)/\{0,\pm 1,\cdots,\pm n\}$, you can apply the Cauchy's residue theorem, as you rightly said.
The poles are $\{0,\pm 1,\cdots,\pm n\}$ (note that not all the integers as you said, be careful). All of them except $0$ are simple poles so their residues are
$$Res(f,k)=\lim_{z\to k}(z-k)\frac{\cos(\pi z)}{z^2\sin(\pi z)}=_{L'Hopital}\lim_{z\to k}\frac{\cos(\pi z)-(z-k)sin(\pi z)}{2z\sin(\pi z)+z^2\pi\cos(\pi z)}=\frac{\cos(\pi k)}{k^2\pi\cos(\pi k)}=\frac{1}{k^2\pi}$$ The case for $z=-k$ is just the same.
Now lets approach your problem with the pole $z=0$. Note that $g(z)=z^2\sin(\pi z)$ has a zero in $z=0$. That zero has order three because the numerator of $f$ is not zero at $z=0$ and $g(0)=g'(0)=g''(0)=0$ but $g'''(0)\not=0$. The easiest way I know to calculate it is to find the term $a_{-1}$ of the Laurent's series of $f(z)$ around $0$.
Note that $$\sin(\pi z)=\pi z-\frac{\pi^3 z^3}{3!}+\frac{\pi^5 z^5}{5!}+O(z^5)\\ \cos(\pi z)=1-\frac{\pi^2 z^2}{2!}+\frac{\pi^4 z^4}{4!}+O(z^6)$$ And because the zero at $z=0$ has order $3$ the Laurent series is like this $$f(z)=a_{-3}\frac{1}{z^3}+a_{-2}\frac{1}{z^2}+a_{-1}\frac{1}{z^1}+a_{0}\frac{1}{z}+a_{1}z+O(z^2)$$ So we have the following: $$z^2(\pi z-\frac{\pi^3 z^3}{3!}+\frac{\pi^5 z^5}{5!}+O(z^5))(a_{-3}\frac{1}{z^3}+a_{-2}\frac{1}{z^2}+a_{-1}\frac{1}{z^1}+a_{0}\frac{1}{z}+a_{1}z+O(z^2))=1-\frac{\pi^2 z^2}{2!}+\frac{\pi^4 z^4}{4!}+O(z^6)$$ Since we are only interested in $a_{-1}$, we multiplie and equalize the first terms, and we get that: $$\pi a_{-3}=1\implies a_{-3}=\frac{1}{\pi}\\ a_{-2}\pi=0\implies a_{-2}=0\\ -a_{-3}\frac{\pi^3}{3!}+a_{-1}\pi=-\frac{\pi^2}{2}\implies a_{-1}=-\frac{\pi}{3}$$
Now we are ready to apply the residues theorem: $$\int_{\gamma_n}f(z)\;dz=2\pi i(Res(f,0)+\sum_{k=1}^{n}Res(f,k)+\sum_{k=1}^{n}Res(f,-k))=-\frac{2\pi^2 i}{3}+\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k^2}$$