Value of the integral $\int_{\mathbb{R}} \frac{x\sin {(\pi x)}}{(1+x^2)^2}$ How do we evaluate the integral $$I=\displaystyle\int_{\mathbb{R}} \dfrac{x\sin {(\pi x)}}{(1+x^2)^2}$$
I have wasted so much time on this integral, tried many substitutions $(x^2=t, \ \pi x^2=t)$.
Wolfram alpha says $I=\dfrac{e^{-\pi} \pi^2}{2}$, but I don't see how.
How do I calculate it using any of the methods taught in real analysis, and not complex analytical methods?
 A: Consider
$$
\mathcal{I}(y,t)=\int_{-\infty}^{\infty}\frac{\cos xt}{x^2+y^2}\ dx=\frac{\pi e^{-yt}}{y}\quad;\quad\text{for}\ t>0.\tag1
$$
Differentiating $(1)$ with respect $t$ and $y$ yields
\begin{align}
\frac{\partial^2\mathcal{I}}{\partial y\partial t}=\int_{-\infty}^{\infty}\frac{2xy\sin xt}{(x^2+y^2)^2}\ dx&=\pi te^{-yt}\\
\int_{-\infty}^{\infty}\frac{x\sin xt}{(x^2+y^2)^2}\ dx&=\frac{\pi te^{-yt}}{2y}.\tag2
\end{align}
Putting $y=1$ and $t=\pi$ to $(2)$ yields
$$
\large\color{blue}{\int_{-\infty}^{\infty}\frac{x\sin\pi x}{(x^2+1)^2}\ dx=\frac{\pi^2 e^{-\pi}}{2}}.
$$
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
Use a semicircle in the upper complex plane since the function
$\expo{\ic\pi x}$ will assure the integral convergence ( the contribution of the upper arc will vanishes out when its radius goes to $\ds{\infty}$ ):  

In the case described above, the integral has a $\ul{double}$ pole at $\ds{x = \ic}$:
  \begin{align}
I&\equiv\color{#66f}{\large%
\int_{\mathbb R}{x\sin\pars{\pi x} \over \pars{1 + x^{2}}^{2}}\,\dd x}
=\Im\int_{-\infty}^{\infty}{x\expo{\ic\pi x} \over \pars{1 + x^{2}}^{2}}\,\dd x
\\[3mm]&=\Im\braces{2\pi\ic\lim_{x\ \to\ \ic}\totald{}{x}
\bracks{\pars{x - \ic}^{2}{x\expo{\ic\pi x} \over \pars{1 + x^{2}}^{2}}}}
\\[3mm]&=2\pi\,\Re\braces{\lim_{x\ \to\ \ic}\totald{}{x}
\bracks{{x\expo{\ic\pi x} \over \pars{x + \ic}^{2}}}}
=\color{#66f}{\large\half\,\expo{-\pi}\pi^{2}}
\end{align}

A: Note 
$$ \int_0^\infty e^{-xt}\sin tdt=\frac{1}{1+x^2} $$
and hence
$$ \frac{d}{dx}\int_0^\infty e^{-xt}\sin tdt=-\frac{2x}{(1+x^2)^2}. $$
Also
$$ \int_0^\infty\frac{\cos(\pi x)}{1+x^2}dx=\frac{1}{2}\pi e^{-\pi}. $$
So 
\begin{eqnarray}
I&=&2\int_0^\infty \frac{x\sin(\pi x)}{(1+x^2)^2}dx=-\int_0^\infty \sin(\pi x) \left(\frac{d}{dx}\int_0^\infty e^{-xt}\sin tdt\right)dx\\
&=&-\sin(\pi x)\int_0^\infty e^{-xt}\sin tdt\bigg|_{x=0}^{x=\infty}+\pi\int_0^\infty\cos(\pi x)\left(\int_0^\infty e^{-xt}\sin tdt\right)dx\\
&=&\pi\int_0^\infty\frac{\cos(\pi x)}{1+x^2}dx=\frac{1}{2}\pi^2e^{-\pi}
\end{eqnarray}
