Definite integral calculation with poles at $0$ and $\pm i\sqrt{3}$ $$\int_0^\infty \frac{\sin(2\pi x)}{x(x^2+3)} \, dx$$
I looked at $\frac{e^{2\pi i z}}{z^{3}+3z}$, also calculated the residues, but they don't get me the right answer. I used that $\int_{-\infty}^\infty f(z)\,dz = 2\pi i (\sum \operatorname{Res} z_r) + \pi i \operatorname{Res}_0$, but my answer turns out wrong when I check with wolframalpha.
Residue for $0$ is $1$, for $z=\sqrt{3}i$ it's $-\frac{e^{-2\pi}}{2}$ . . .
In a worse attempt I forgot $2\pi$ and used $z$ only (i.e. $\frac{e^{iz}}{z^{3}+3z}$) and the result was a little closer, but missing a factor of 2 and and $i$.
 Can anyone see the right way? Please do tell. 

Resolving $\frac{1}{x(x^2+1)}$  into partial fractions yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin (2 \pi x)}{x\left(x^{2}+3\right)} d x &=\frac{1}{3}\left( \int_{0}^{\infty} \frac{\sin (2 \pi x)}{x} dx-\int_{0}^{\infty} \frac{x^{2} \sin (2 \pi x)}{x^{2}+3} dx \right)\\
&=\frac{1}{6}\left(\pi-\int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x\right)
\end{aligned}
$$
By Jordan’s Lemma,
$$
\begin{aligned}
& \int_{-\infty}^{\infty} \frac{x \cos (2 \pi x)}{x^{2}+3} d x+i \int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x \\
=& 2 \pi i \operatorname{Res}\left(\frac{z e^{2 \pi zi}}{z^{2}+3}, z=\sqrt 3 i\right)\\
=& 2 \pi i \frac{\sqrt{3} i e^{-2 \sqrt{3} \pi}}{2 \sqrt{3} i} \\
=& \pi ie^{-2 \sqrt{3} \pi},
\end{aligned}
$$
using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where }  \gamma_{1}(t)=t+i 0(-R \leq t \leq R)  \textrm{ and } \gamma_{2}(t)=R e^{i t}  (0<t<\pi) $$
Comparing the imaginary parts of both sides gives
$$\int_{-\infty}^{\infty} \frac{x \sin (2 \pi x)}{x^{2}+3} d x = \pi e^{-2 \sqrt{3} \pi} $$
Hence $$\int_{0}^{\infty} \frac{\sin (2 \pi x)}{x\left(x^{2}+3\right)} d x = \frac{\pi}{6}\left(1-e^{-2 \sqrt{3} \pi}\right) $$
