I have a question regarding a integral I just can't find the right solution to. It's this beauty:
$$\int_\gamma \frac{\cos(\pi z)}{z^2+1} dz$$ where $\gamma = 3 \cdot e^{it}$.
I used the Cauchy Integral Formula (see here) with $n=0$, $a=1$ and $f(z) = \frac{\cos(\pi z)}{z+1}$ and it got me a result of $-\pi i$. However, this integral should be zero according to my professor. Any help is appreciated!
On
You don't need any advanced Complex Analysis theorems to compute that integral. Let $f(z)=\frac{\cos(\pi z)}{z^2+1}$. In order to compute $\oint_\gamma f(z)\,\mathrm dz$, all that I will need is that $f$ is integrable and even. In fact, $\oint_\gamma f(z)\,\mathrm dz$ is equal to $\int_0^{2\pi}f(3e^{it})3ie^{it}\,\mathrm dt$, which, in turn, is equal to $3i$ times$$\int_0^\pi f(3e^{it})e^{it}\,\mathrm dt+\int_\pi^{2\pi}f(3e^{it})e^{it}\,\mathrm dt.\tag1$$But\begin{align}\int_\pi^{2\pi}f(3e^{it})e^{it}\,\mathrm dt&=\int_0^\pi f\left(3i^{i(t+\pi)}\right)e^{i(t+\pi)}\,\mathrm dt\\&=\int_0^\pi f\left(-3e^{it}\right)\left(-e^{it}\right)\,\mathrm dt\\&=-\int_0^\pi f\left(3e^{it}\right)e^{it}\,\mathrm dt,\end{align}since $f$ is an even function. So, $(1)$ is equal to $0$, and therefore $\oint_\gamma f(z)\,\mathrm dz=0$.
It is not clear to me as to how you applied CIF for you function $f$. To apply CIF you need an analytic function.
Poles are at $\pm i$ and the residues are $\cos (\pi i)=\frac {e^{-1}+e^{1}}{2 i}$ and $\cos (-\pi i)=\frac {e^{-1}+e^{1}}{ -2i }$. So the integral is $0$ by Residue Theorem.