# Calculating the integral $\int_{-\infty}^{\infty} x\sin(x)/(x^2+1)^2 dx$

I need help calculation the integral $$\int_{-\infty}^{\infty} x\sin(x)/(x^2+1)^2 dx$$

I am getting confused on how to integrate around $$\pm i$$. I think I'll have to use $$\gamma = Re^{i\theta}$$ where $$|R|\leq n$$ for $$n\geq1$$

• Yes, half a circle is a good idea. Prove that when $R\to\infty$ the integral goes to zero. You don't need anything around $-i$, that will only be more work.
– Mark
Commented Jan 4, 2019 at 16:22
• You could try solving $\int_{-\infty}^{\infty} \frac{xsin(\alpha{x})}{(x^{2}+1)^{2}}dx=-\frac{d}{d\alpha}\int_{-\infty}^{\infty} \frac{cos(\alpha{x})}{(x^{2}+1)^{2}}dx$ and then let $\alpha\rightarrow1$. It might be easier. Commented Jan 4, 2019 at 16:50
• The result should be $$\frac{\pi}{2e}$$ Commented Jan 4, 2019 at 17:12

Let us define $$g(a) = \int_{0}^{+\infty}\frac{x\sin(ax)}{(x^2+1)^2}\,dx$$ for $$a\in\mathbb{R}^+$$. Our integral is clearly $$2\,g(1)$$ by parity. We have $$\mathcal L g(s) = \int_{0}^{+\infty}\frac{x^2}{(s^2+x^2)(1+x^2)^2}\,dx = \frac{\pi}{4(1+s)^2}$$ by partial fraction decomposition, and $$g(a) = \frac{\pi a}{4}e^{-a}$$ by the inverse Laplace transform. It follows that the original integral equals $$\color{blue}{\frac{\pi}{2e}}$$.

$$\int_{-\infty}^\infty\frac{x\sin x}{(x^2+1)^2}\mathrm dx=\mathfrak I\left\{\int_{-\infty}^\infty\frac{xe^{ix}}{(x^2+1)^2}\mathrm dx\right\}$$

Let us compute this integral. It is given by a contour in the complex plane that runs along the real axis. Since the exponential on top decays as $$|x|\to\infty$$ in the upper half plane, this is where we choose to close the contour. Therefore we have a semicircular contour in the upper half plane, of the form $$Re^{i\theta}$$ for $$R>1$$ and $$\theta\in[0,\pi]$$. Let us denote this contour by $$\gamma_R$$.

$$\lim_{R\to\infty}\int_{\gamma_R} \frac{xe^{ix}}{(x^2+1)^2}\mathrm dx=\int_{-\infty}^{\infty}\frac{xe^{ix}}{(x^2+1)^2}\mathrm dx+\lim_{R\to\infty}\int_0^\pi\frac{Re^{i\theta}e^{iRe^{i\theta}}}{(R^2e^{2i\theta}+1)^2}iRe^{i\theta}\mathrm d\theta\tag1$$

Now we want to show that this second integral is zero in the limit as $$R\to\infty$$.

\begin{align}\left|\int_0^\pi \frac{Re^{i\theta}e^{iRe^{i\theta}}}{(R^2e^{2i\theta}+1)^2}iRe^{i\theta}\mathrm d\theta\right|&\le\int_0^\pi\left|\frac{Re^{i\theta}e^{iRe^{i\theta}}}{(R^2e^{2i\theta}+1)^2}iRe^{i\theta}\right|\mathrm d\theta\\&\le\int_0^\pi\left|\frac{R^2e^{-R\sin \theta}}{(R^2-1)^2}\right|\mathrm d\theta\end{align}

$$\int_0^\pi e^{-R\sin\theta}\mathrm d\theta$$ is finite, and in the limit as $$R\to\infty$$, the prefactor tends to zero, so the second integral in $$(1)$$ is zero.

So we have $$\int_{-\infty}^{\infty}\frac{xe^{ix}}{(x^2+1)^2}\mathrm dx=\lim_{R\to\infty}\int_{\gamma_R} \frac{xe^{ix}}{(x^2+1)^2}\mathrm dx\tag2$$

We evaluate the RHS of $$(2)$$ by residue calculus. The contour encloses a pole at $$x=i$$, of order $$2$$. We evaluate the residue at the pole.

$$\text{Res}\left(\frac{xe^{ix}}{(x^2+1)^2},i\right)=\lim_{x\to i}\frac{d}{dx}\left(\frac{xe^{ix}(x-i)^2}{(x^2+1)^2}\right)=\lim_{x\to i}\frac{d}{dx}\left(\frac{xe^{ix}}{(x+i)^2}\right)=\frac1{4e}$$after some calculations.

Therefore the RHS of $$(2)$$ is $$2\pi i\cdot \text{Res}\left(\frac{xe^{ix}}{(x^2+1)^2},i\right)=\frac{\pi i }{2e}$$

Since the integral we wish to calculate is the imaginary part of the LHS of $$(2)$$, the final answer is $$\frac{\pi}{2e}.$$