Evaluate $\int_{-\infty}^{\infty} \frac{x \sin x}{x^2+4} \, \mathrm{d}x$ using contour integration 
Problem.src) Evaluate $\displaystyle \int_{-\infty}^{\infty} \frac{x \sin x}{x^2+4} \, \mathrm{d}x $

I know I am supposed to split it up like this

and $\Gamma(R)$ tends to zero and the other tends to my integral as $R$ tends to infinity?
I compute the residue at $2i$ which I think is $\frac{\sin(2i)}{2}$ ?
But I am a little stuck as to what to do now, I have never seen an example of this type of integral involving $\sin(x)$.
Usually we use the ML lemma for these types of problems, but because of the $x$ on top do I need to use Jordan's lemma?
Do I have to use $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ ?
I am quite confused, any help would be appreciated.
Thanks.
 A: HINT:
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+4}dx=\text{Im}\left(\int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2+4}dx\right)$$
Then, evaluate
$$\text{Im}\left(\oint_C\frac{ze^{iz}}{z^2+4} dz\right)=2\pi i\,\text{Im}\left(\text{Res}\left(\frac{ze^{iz}}{z^2+4}\right),z=i2\right)$$

By request, we show that $\int_{C_R}\frac{ze^{iz}}{z^2+4} dz \to 0$ as $R\to \infty$.  On $C_R$, we have $z=Re^{it}$ so that $dz=iRe^{it}dt$, with $0\le t\le \pi$.   Then, for $R>2$ we have
$$\begin{align}
\left| \int_{C_R}\frac{ze^{iz}}{z^2+4} dz\right|&=\left| \int_0^{\pi}\frac{Re^{it}e^{iRe^{it}}}{R^2e^{i2t}+4} iRe^{it}dt\right|\\\\
&=\left| \int_0^{\pi}\frac{R^2e^{iR\cos t}e^{-R\sin t}}{R^2e^{i2t}+4} dt\right|\\\\
&\le\int_0^{\pi}\left|\frac{R^2e^{-R\sin t}}{R^2e^{i2t}+4} \right|dt\\\\
&=\int_0^{\pi}\frac{R^2e^{-R\sin t}}{|R^2e^{i2t}+4|} dt\\\\
&\le \frac{2R^2}{R^2-4}\int_0^{\pi/2}e^{-R\sin t} dt\\\\
&\le\frac{2R^2}{R^2-4}\int_0^{\pi/2}e^{-R(2t/\pi)} dt\\\\
&=\frac{2R^2}{R^2-4}\frac{\pi}{2}\frac{1-e^{-R}}{R}\\\\
&\to 0\,\,\text{as} R\to \infty
\end{align}$$
A: Just to be a little bit more general, let's look at the integral 
$$
J[a,b]=\int_{-\infty}^{\infty}\frac{e^{i a x}}{x^2+b^2}dx
$$
where $a>0$ and $\Im[{b}]>0$.
it's clear that we can write our orignal integral as 
$$
I=-\Re[\partial_a J[a,b]]_{a=1,b=2}
$$
Consider now the complex function 
$$
f(z)=\frac{e^{i a z}}{z^2+b^2}
$$
Let's apply Cauchy's theorem to a large semicircle in the upper half plane as depicted in your question (This allowed, because the integral vanishs as $1/z$ for $z\rightarrow \infty$). We obtain
$$
J[a,b]=\int_{SC} dz f(z)=\int_{-\infty}^{\infty}f(x)dx=-2\pi i \times \text{res}[z=i b]=2\pi i \times\frac{e^{-a b}}{2 i b}=\pi \frac{e^{-a b}}{b}
$$
Here $SC$ stands for semicircle, 
therefore 
$$
I=\pi \Re[e^{-ab}]_{a=1,b=2}=\frac{\pi}{e^2}
$$
The advantage of this (maybe a little bit over-the-top looking) method is that you can now calclulate arbritary integrals of the form $\int_{-\infty}^{\infty}\frac{ x^n e^{i a x}}{(x^2+b^2)^m}$  (with $m>n$ for convergence reasons)by just taking appropriate derivatives w.r.t to  $a$ and $b$ and then seperate into real and imaginary parts if you are interested in the trigonometric cases!
