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. 
 A: $$
\begin{align}
\int_0^\infty\frac{\sin(2\pi x)}{x(x^2+3)}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\sin(2\pi x)}{x(x^2+3)}\,\mathrm{d}x\tag{1}\\
&=\frac12\int_{i-\infty}^{i+\infty}\frac{\sin(2\pi z)}{z(z^2+3)}\,\mathrm{d}z\tag{2}\\
&=\frac12\int_{i-\infty}^{i+\infty}\frac{e^{i2\pi z}-e^{-i2\pi z}}{2iz(z^2+3)}\,\mathrm{d}z\tag{3}\\
&=\frac1{4i}\int_{\gamma_+}\frac{e^{i2\pi z}}{z(z^2+3)}\mathrm{d}z
-\frac1{4i}\int_{\gamma_-}\frac{e^{-i2\pi z}}{z(z^2+3)}\mathrm{d}z\tag{4}\\
&=\frac{2\pi i}{4i}\frac{e^{-2\pi\sqrt3}}{i\sqrt3(i\sqrt3+i\sqrt3)}
+\frac{2\pi i}{4i}\left(\frac13+\frac{e^{-2\pi\sqrt3}}{-i\sqrt3(-i\sqrt3-i\sqrt3)}\right)\tag{5}\\
&=\frac\pi6\left(1-e^{-2\pi\sqrt3}\right)\tag{6}
\end{align}
$$
where $\gamma_+$ passes from $i-R$ to $i+R$ then circles back counterclockwise around the upper half-plane, and where $\gamma_-$ passes from $i-R$ to $i+R$ then circles back clockwise around the lower half-plane.
$\gamma_+$ contains the singularity at $i\sqrt3$.
$\gamma_-$ contains the singularities at $0$ and  $-i\sqrt3$.
Explanation of steps
$(1)$ integrand is an even function.
$(2)$ there are no singularities for the integrand in the rectangle with corners $i-R,i+R,R,-R$ and the integral over the ends of the rectangle vanishes as $R\to\infty$.
$(3)$ write $\sin(2\pi z)=\dfrac{e^{i2\pi z}-e^{-i2\pi z}}{2i}$
$(4)$ use contour $\gamma_+$ for $e^{i2\pi z}$ and $\gamma_-$ for $e^{-i2\pi z}$ so that the integrand vanishes over the large circles in the upper and lower half-planes.
$(5)$ use residues to compute the integrals over $\gamma_+$ and $\gamma_-$.
$(6)$ simplification.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large%
\int_{0}^{\infty}{\sin\pars{2\pi x} \over x\pars{x^{2} + 3}}\,\dd x}=
\half\int_{-\infty}^{\infty}{\sin\pars{2\pi x} \over x\pars{x^{2} + 3}}\,\dd x
=\pi\int_{-\infty}^{\infty}
{1 \over x^{2} + 3}\,\pars{\half\int_{-1}^{1}\expo{2\pi\ic k x}\,\dd k}\,\dd x
\\[3mm]&={\pi \over 2}\int_{-1}^{1}\dd k
\int_{-\infty}^{\infty}{\expo{2\pi\ic kx} \over x^{2} + 3}\,\dd x
\\[3mm]&={\pi \over 2}\int_{-1}^{1}\bracks{%
\Theta\pars{-k}\pars{-2\pi\ic}\,
{\expo{2\pi\ic k\pars{-\ic\root{3}}} \over -2\ic\root{3}}
+\Theta\pars{k}\pars{2\pi\ic}\,
{\expo{2\pi\ic k\pars{\ic\root{3}}} \over 2\ic\root{3}}}\,\dd k
\\[3mm]&={\pi \over 2}\int_{-1}^{1}{\pi \over \root{3}}\,
\expo{-2\pi\root{3}\verts{k}}\,\dd k
={\root{3} \over 3}\,\pi^{2}\int_{0}^{1}\expo{-2\pi\root{3}k}\,\dd k
\\[3mm]&={\root{3} \over 3}\,\pi^{2}\,{\expo{-2\pi\root{3}} - 1 \over -2\pi\root{3}}
=\color{#00f}{\large{1 \over 6}\,\pi\pars{1 - \expo{-2\pi\root{3}}}}
\end{align}
A: We make the following lemma:

Lemma. Suppose $f(z)$ is holomorphic near $z = z_0$. Fix $\theta_0 \in (0, 2\pi)$. If $\gamma_\varepsilon$ denotes a counter-clockwise oriented arc of angle $\theta_0$ on the circle of radius $\varepsilon$ centered at $z_0$, then
$$ \lim_{\varepsilon\to0} \int_{\gamma_\varepsilon} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta=i\theta_0 f(z_0).$$


Proof. By the substitution $\zeta = z_0 + \varepsilon e^{i\theta}$, we have
$$\begin{align*}
\left| \int_{\gamma_\varepsilon} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta - i\theta_0 f(z_0)\right|
&= \left| i \int_{\theta_1}^{\theta_1+\theta_0} f(z_0 + \varepsilon e^{i\theta})\;d\theta - i\theta_0 f(z_0)\right| \\
& \leq \int_{\theta_1}^{\theta_1+\theta_0} \left| f(z_0 + \varepsilon e^{i\theta}) - f(z_0) \right| \;d\theta,
\end{align*}$$
which clearly goes to zero when $\varepsilon \to 0$.

As a corollary, if $f(z)$ has a simple pole at $z = z_0$ then with the same notation as in the Lemma, we have
$$ \lim_{\varepsilon\to0} \int_{\gamma_\varepsilon} f(\zeta) \;d\zeta=i\theta_0 \operatorname{Res} \{ f(z), z_0 \}.$$
Now let $C$ be the upper-semicircular contour of radius $R \gg 1$ with a small semicircular indent of radius $\varepsilon \ll 1$ at the origin. Let us write $C$ as
$$ C = \Gamma_{R} + L_{\varepsilon,R} - \gamma_\varepsilon,$$
where $\Gamma_R$ and $\gamma_\varepsilon$ denote counter-clockwise oriented arcs corresponding to the outer circle and the inner circle of $C$, respectively, and $L_{\varepsilon,R}$ denote the remaining union of two lines on $C \cap \Bbb{R}$. Now for
$$ f(z) = \frac{e^{2\pi i z}}{z(z^2 + 3)}, $$
the integral in question, which we denote as $I$, is equal to
$$I = \frac{1}{2i} \lim_{{\varepsilon \to 0 \atop R \to \infty}} \int_{L_{\varepsilon, R}} f(z) \; dz.$$
Now, by Cauchy integration formula,
$$ \oint_{C} f(z) \; dz = 2\pi i \operatorname{Res} \{ f(z), i\sqrt{3} \}. $$
This means that
$$ \int_{L_{\varepsilon, R}} f(z) \; dz = 2\pi i \operatorname{Res} \{ f(z), i\sqrt{3} \} + \int_{\gamma_\varepsilon} f(z) \; dz - \int_{\Gamma_R} f(z) \; dz. $$
Taking limit $\varepsilon \to 0$ and $R \to \infty$, the integral $\int_{\Gamma_R} f(z) \; dz$ vanishes by Jordan's lemma. Thus by our lemma,
$$ \lim_{{var\epsilon \to 0 \atop R \to \infty}} \int_{L_{\varepsilon, R}} f(z) \; dz = 2\pi i \operatorname{Res} \{ f(z), i\sqrt{3} \} + \pi i \operatorname{Res} \{ f(z), 0 \}.$$
But since
$$ \operatorname{Res} \{ f(z), i\sqrt{3} \} = \left. (z-i\sqrt{3})f(z) \right|_{z=i\sqrt{3}} = -\frac{1}{6}e^{-2\pi \sqrt{3}} $$
and
$$ \operatorname{Res} \{ f(z), 0 \} = \left. z f(z) \right|_{z=0} = \frac{1}{3}, $$
we have
$$ I = \frac{1}{2i} \left[ 2\pi i \left(-\frac{1}{6}e^{-2\pi \sqrt{3}}\right) + \pi i \left(\frac{1}{3} \right) \right] = \frac{\pi}{6}\left(1 - e^{-2\pi \sqrt{3}} \right). $$
A: Given that to solve  Integral
$$\int_{0}^{\infty}\frac{\sin(2\pi x) }{x(x^2+3)}dx$$
Assume that $$\mathcal{I}(a)=\int_{0}^{\infty}\frac{\sin(ax)}{x(x²+1)}dx \to (1)$$ here $a$ is positive real
Differentiating $(1) $ we get
$$\mathcal{I}'=\int_{0}^{\infty}\frac{\cos(ax) }{x^2+3}dx$$
Now it is easy to use complex analysis by taking function $$f(z) =\frac{e^{\iota az}}{z^2+3}$$
and using Jordan inequality we will get
$$\mathcal{I}'(a) =\frac{\pi}{2\sqrt{3}}e^{-\sqrt{3}a}$$
Now integrate the expression
$$\mathcal{I}(a) =-\frac{\pi}{6}e^{-a\sqrt{3}}+C$$
Where $C $ is constant of integration
Since $\mathcal{I}(0) =0$ then
$C=\frac{\pi}{6}$
Combine every thing put the value of a and we are done
A: 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 $$\int_{0}^{\infty} \frac{\sin (2 \pi x)}{x\left(x^{2}+3\right)} d x = \frac{\pi}{6}\left(1-e^{-2 \sqrt{3} \pi}\right) $$
