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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?

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  • $\begingroup$ Do you know residue theorem? In any event, you can integral by part and bring the integral to the form which has been used as an example in above wiki page. $\endgroup$ – achille hui Jul 27 '14 at 17:46
  • $\begingroup$ @achillehui Sorry..but i do not know residue theorem, or for that matter any of the complex analytical methods to solve integrals. $\endgroup$ – pkwssis Jul 27 '14 at 17:49
  • $\begingroup$ Reduce to an ODE via the introduction of a suitable parameter $\endgroup$ – Julien Godawatta Jul 27 '14 at 17:51
  • $\begingroup$ @Lac You mean differentiation under the integral sign? $\endgroup$ – pkwssis Jul 27 '14 at 17:52
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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 $$ \large\color{blue}{\int_{-\infty}^{\infty}\frac{x\sin\pi x}{(x^2+1)^2}\ dx=\frac{\pi^2 e^{-\pi}}{2}}. $$

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

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  • $\begingroup$ OP wrote: "How do i calculate it using any of the methods taught in real analysis, and not complex analytical methods?" $\endgroup$ – Start wearing purple Jul 27 '14 at 18:10
  • $\begingroup$ @O.L. I guess there is not any possibility to escape from complex methods. Indeed, I didn't read completely the whole question. Sorry. Thanks. $\endgroup$ – Felix Marin Jul 27 '14 at 18:13
  • $\begingroup$ @Felix Don't worry, the requirement for real methods was edited in after this answer was posted. (+1) $\endgroup$ – Brad Jul 27 '14 at 18:35
  • $\begingroup$ @Brad It's a good new to know that. Thanks. $\endgroup$ – Felix Marin Jul 27 '14 at 18:55
  • $\begingroup$ I'm learning contour integration. I just want to know, can we also use Cauchy integral $$ \oint_C \frac{f(z)}{(z-z_o)^{n+1}}\ dz=\frac{2\pi i}{n!}f^{(n)}(z_o), $$ where $f(z)=\Im\left[\dfrac{ze^{i\pi z}}{(z+i)^2}\right], z_o=i,$ and $n=1$, right? $\endgroup$ – Tunk-Fey Jul 28 '14 at 11:46
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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}

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