Real integral using a contour integral I am going to calculate $\int_{-\infty}^{\infty}\dfrac{x \sin \pi x}{x^2+2x+5}dx$
So I have to compute the following limit
$\lim_{R \to \infty}\int_{C_1}\dfrac{z \sin \pi z}{z^2+2z+5}dz$ where $C_1$ is the semicircle with centre 0 and radius R and $C_1$ goes from $-R$ to $R$.
How to show this limit is $0$. Should this be considered as 
$\lim_{R \to \infty}\int_{0}^{\pi}\dfrac{Re^{i\theta} \sin(\pi Re^{i\theta})}{Re^{2i\theta}+2Re^{i\theta}+5}Re^{i\theta}i \ d\theta$.
But I am heading nowhere. Please help. I have exam today!!
 A: By the ML-lemma
$$\left|\int_{C_1}\frac{ze^{i\pi z}}{z^2+2z+5}dz\right|\le \pi R\frac{Re^{-\pi\,\text{Im}\,z}}{R^2-2R-5}\xrightarrow[R\to\infty]{}0$$
since $\;\text{Im}\,z\to\infty\;$ when $\;R\to\infty\;$
Please observe that
$$\left|e^{i\pi z}\right|=\left|e^{i\pi\left(\text{Re}\,z+i\,\text{Im}\,z\right)}\right|=e^{-\pi\,\text{Im}\,z}\cdot\left|e^{\pi i\,\text{Re}\,z}\right|=e^{-\pi\,\text{Im}\,z}$$
As commented, when dealing with $\;\sin ax,\,\cos ax\;$ in real integrals, you better take the exponential $\;e^{iaz}\;$ .
A: You cannot prove that your limit is zero simply because it isn't. The sine function grows really fast in absolute value along the imaginary axis (as a $\sinh$, that is no wonder). You may circumvent this issue by writing $\sin(z)$ as $\text{Im}\left(e^{iz}\right)$ and replacing the integrand function with a more well-behaving one. Anyway, your integral equals:
$$ \int_{-\infty}^{+\infty}\frac{z \sin(\pi z)}{(z+1)^2+2^2}\,dz = \int_{-\infty}^{+\infty}\frac{(1-z)\sin(\pi z)}{z^2+2^2}\,dx=-\int_{-\infty}^{+\infty}\frac{z\sin(\pi z)}{z^2+2^2}\,dz $$
or:
$$ -\frac{1}{2}\int_{-\infty}^{+\infty}\sin(\pi z)\left(\frac{1}{z-2i}+\frac{1}{z+2i}\right)\,dz=\color{red}{-\frac{\pi}{e^{2\pi}}}.$$
