Evaluation of integral $\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$ I'm trying to evaluate the following integral:
$$\mathcal{J}=\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$
Well there are $3$ poles , one lying on the real line the other on the upper half plane and the other on the lower half plane. The residue at $z=i$ is $\displaystyle -\frac{3}{4e}$ .
I'm integrating on a contour that looks like a semicircle on the upper half plane and has a branch about the origin .
Well I considered the function $\displaystyle f(z)=\frac{e^{iz}+1}{z\left ( z+1 \right )^2}$ which is clearly analytic expect for the poles. Hence if $\gamma$ denotes the contour , then: 
$$\oint_{\gamma}f(z)\,{\rm d}z=\oint_{\gamma}\frac{e^{iz}+1}{z\left ( z^2+1 \right )^2}\,{\rm d}z =\oint_{\gamma}\frac{e^{iz}}{z\left ( z^2+1 \right )^2}\,{\rm d}z+\oint_{\gamma}\frac{{\rm d}z}{z\left ( z^2+1 \right )^2}=-2\pi i \frac{3}{4e}+2\pi i = 2\pi i \left ( 1-\frac{3}{4e} \right )=i \left (2\pi - \frac{3\pi}{2e} \right )$$
Hmm... applying the classical method I get that:
$$\int_{-\infty}^{\infty}f(x)\,{\rm d}x =2\pi - \frac{3\pi}{2e}$$ 
which is almost correct except for that $2$ in front of $\pi$. Where I have gone wrong?
P.S: I used the very obvious that the integrand is even.
 A: Depict carefully the path of integration: it is a semicircle in the upper half plane with a bulge at $z=0$ and a keyhole around $z=i$. This gives that you have to compute the residues of $f(z)=\frac{e^{iz}}{z(z^2+1)^2}$ at $z=0$ and $z=i$, but to consider only half the residue at $z=0$:
$$\mathcal{J}=\frac{1}{2}\text{Im}\int_{-\infty}^{+\infty}\frac{e^{iz}}{z(z^2+1)^2}\,dz = \frac{1}{2}\text{Im}\left(2\pi i\operatorname{Res}(f(z),z=i)+\pi i\operatorname{Res}(f(z),z=0)\right)$$
so:
$$\mathcal{J} = \frac{1}{2}\text{Im}\left(2\pi i\cdot \frac{-3}{4e}+\pi i\right)=\frac{1}{2}\left(\pi-\frac{3\pi}{2e}\right)=\color{red}{\frac{\pi}{2}\left(1-\frac{3}{2e}\right)}.$$
A: Another way to evaluate this integral is to use Parseval's theorem, which states that for functions $f$ and $g$ with respective Fourier transforms $F$ and $G$, we have
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$
Here $f(x) = \sin{x}/x$ and $g(x) = (1+x^2)^{-2}$.  Thus, $F(k) = \pi$ when $k \in [-1.1]$ and $0$ otherwise and $G(k) = (\pi/2) (|k|+1) e^{-|k|}$.  The integral is then 
$$\frac{\pi}{8} \int_{-1}^1 dk \, (|k|+1) e^{-|k|} = \frac{\pi}{4} \int_0^1 dk (k+1) e^{-k} $$
$$\int_0^1 dk \, e^{-k} = 1-e^{-1}$$
$$\int_0^1 dk \,k \,  e^{-k} = -e^{-1}+ \int_0^1 dk \, e^{-k} = 1-2 e^{-1}$$
Therefore, the integral is equal to
$$\frac{\pi}{4} \left (2-\frac{3}{e} \right ) = \frac{\pi}{2} \left (1-\frac{3}{2 e} \right )$$
A: Another approach: Parameterize the integral as
$$I(a)=\int_{0}^{\infty}\frac{\sin ax}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$
Take the Laplace transform, find your partial fractions, and take the inverse transform:
$$\begin{align*}\mathcal{L}_s\{I(a)\}&=\int_0^\infty\int_{0}^{\infty}\frac{\sin ax}{x\left ( 1+x^2 \right )^2}e^{-as}\,{\rm d}a\,{\rm d}x\\
&=\int_0^\infty\frac{\mathcal{L}_s\{\sin ax\}}{x\left ( 1+x^2 \right )^2}\,{\rm d}x\\
&=\int_0^\infty\frac{x}{x\left ( 1+x^2 \right )^2\left(s^2+x^2\right)}\,{\rm d}x\\
&=\int_0^\infty\frac{{\rm d}x}{\left ( 1+x^2 \right )^2\left(s^2+x^2\right)}\\
&=-\frac{1}{(s^2-1)^2}\int_0^\infty\left(\frac{1}{1+x^2}-\frac{s^2-1}{(1+x^2)^2}-\frac{1}{s^2+x^2}\right)\,{\rm d}x\\
&=-\frac{1}{(s^2-1)^2}\left(\frac{\pi}{2}-\frac{\pi(s^2-1)}{4}-\frac{\pi}{2s}\right)\\
&=\frac{\pi}{4}\left(\frac{2}{s}-\frac{2}{s+1}-\frac{1}{(s+1)^2}\right)\\
I(a)&=\frac{\pi}{4}{\mathcal{L}^{-1}}_a\left\{\frac{2}{s}-\frac{2}{s+1}-\frac{1}{(s+1)^2}\right\}\\
&=\frac{\pi}{4}(2+-2e^{-a}-ae^{-a})\end{align*}$$
Finally, since $\mathcal{J}=I(1)$, you have
$$\mathcal{J}=\frac{\pi}{2}-\frac{3\pi}{4e}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin\pars{x} \over
x\pars{1 + x^{2}}^{2}}\,\dd x} =
{1 \over 2}\,\Im\int_{-\infty}^{\infty}{\expo{\ic x} - 1 \over
x\pars{x - \ic}^{2}\pars{x + \ic}^{2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Im\braces{2\pi\ic\,\lim_{x\ \to\ \ic}\,\,\,
\totald{}{x}\bracks{\expo{\ic x} - 1 \over
x\pars{x + \ic}^{2}}}
\\[5mm] = &\
\bbx{{\pi \over 2} - {3\pi \over 4\expo{}}} \approx 0.7040 \\ &
\end{align}
