# 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?

• 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. – achille hui Jul 27 '14 at 17:46
• @achillehui Sorry..but i do not know residue theorem, or for that matter any of the complex analytical methods to solve integrals. – pkwssis Jul 27 '14 at 17:49
• Reduce to an ODE via the introduction of a suitable parameter – Julien Godawatta Jul 27 '14 at 17:51
• @Lac You mean differentiation under the integral sign? – pkwssis Jul 27 '14 at 17:52

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


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}

• OP wrote: "How do i calculate it using any of the methods taught in real analysis, and not complex analytical methods?" – Start wearing purple Jul 27 '14 at 18:10
• @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. – Felix Marin Jul 27 '14 at 18:13
• @Felix Don't worry, the requirement for real methods was edited in after this answer was posted. (+1) – Brad Jul 27 '14 at 18:35
• @Brad It's a good new to know that. Thanks. – Felix Marin Jul 27 '14 at 18:55
• 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? – Tunk-Fey Jul 28 '14 at 11:46

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}