Value of the integral $\int_{\mathbb{R}} \frac{x\sin {(\pi x)}}{(1+x^2)^2}$

Question

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?

Answer 1

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 $$ \large\color{blue}{\int_{-\infty}^{\infty}\frac{x\sin\pi x}{(x^2+1)^2}\ dx=\frac{\pi^2 e^{-\pi}}{2}}. $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}

Answer 3

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}