Evaluating the integral $\int_{0}^\infty \frac{z\sin (z)}{z^2+a^2}$ for $a>0$, we can rewrite:
$$\int_{0}^\infty \frac{z\sin (z)}{z^2+a^2}=\frac{1}{2}\Im\int_{-\infty}^\infty\frac{ze^{iz}}{z^2+a^2}dz=2\pi i *\text{Res}\,_{ia}\frac{ze^{iz}}{z^2+a^2}$$
I know that you can create a counter integration $\gamma_C:=\gamma_M+\gamma_{HF}$ where $\gamma_M$ goes from $-M$ to $M$ and $\gamma_{HF}$ goes from M along a half circle back to $-M$. So you get $$\lim_{M \to \infty}\left|\int_{\gamma_{HF}}\frac{ze^{iz}}{z^2+a^2}dz\right|=0$$ and therefore $$\int_{\gamma_{C}}\frac{ze^{iz}}{z^2+a^2}dz=\int_{\gamma_{M}}\frac{ze^{iz}}{z^2+a^2}dz$$ Why can you apply the residue formula $$\int_{-\infty}^\infty \frac{p(x)}{q(x)}=2\pi i \sum_{\mathbb{a\in H}}Res(\frac{p}{q};a)$$ The grade of $p(z)$ should be at least bigger than the grade of $q(z)$ plus $2$. But here it isn't the case.
No, it is not the case here. But when we are dealing with integration of function of the type $\frac{p(x)e^{ix}}{q(x)}$, then having $\deg q(x)>\deg p(x)$ is enough. And that's the case here. This can be proved using, not a path which consists of a line segment plus a half-circle, but a rectangular path instead.