Showing $\int_0^\infty \frac{x\sin(2x)}{x^2+3}dx=\frac{\pi}{2}e^{-2\sqrt3}$ Things are about to get real, prepare your mind!

$$f(z)=\frac{ze^{2iz}}{z^2+3}$$
Now this has two singular points, $\pm\sqrt3i$
$$f(z)=\frac{ze^{2iz}}{(z-(-\sqrt3i))(z-\sqrt3i)}$$ and hence both poles are of order $1$.
We can find $\operatorname{Res}_{z=\sqrt3i}f(z)=\frac{(\sqrt3i)e^{2i(\sqrt3i)}}{(\sqrt3i)+\sqrt3i}=e^{-2\sqrt3}/2$
Now I want to show that:
$$\int_0^\infty \frac{x\sin(2x)}{x^2+3}dx=\frac{\pi}{2}e^{-2\sqrt3}$$
Apparently this is pretty much immediate from my work above it, how?
 A: We first note that the integrand $\frac{x\sin 2x}{x^2+3}$ of the integral of interest $I$ is an even function around $x=0$.  Therefore we can write
$$I=\frac12 \int_{-\infty}^{\infty}\frac{x\sin 2x}{x^2+3}$$
Next, let $C_R$ be a closed contour that is comprised of (i) that part of the real axis $|x|\le R$ and (ii) a semi-circle in the upper-half plane such that $|z|=R$ and $\text{Im}(z)>0$.  We then note that 
$$\lim_{R\to \infty}\text{Im}\left(\oint_{C_R} \frac{z\,e^{ i2z}}{z^2+3}dz\right)=\int_{-\infty}^{\infty}\frac{x\sin 2x}{x^2+3}dx+\lim_{R\to \infty}\int_{|z|=R,\text{Im}(z)>0}\frac{z\,e^{ i2z}}{z^2+3}dz \tag 1$$
We can easily see that the limit of last integral on the right-hand side of $(1)$  vanishes by using Jordan's Lemma.
Therefore, we have that 

$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{I=\frac12\text{Im}\left(2\pi i\text{Res}\left(\frac{z\,e^{ i2z}}{z^2+3},z=i\sqrt{3}\right)\right)=\frac{\pi}{2}e^{-2\sqrt{3}}}
\end{align}$$

A: You are close to the answer.  Remember from the Residue Theorem that
$$\text{Res}_{z=z_0}f(z)=\frac{1}{2\pi i}\int_Cf(z)\,dz.$$
Also you can use the fact that $e^{2iz}=\cos(2z)+i\sin(2z).$  Make this substitution into your $f(z)$ function and integrate.  Since your residue is real valued, the imaginary part of the integral is zero.
