# Compute $\int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$

Compute $$\int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$

Of course we have $$\int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx$$

I think that firstly I should compute $\int \left( \frac{x}{1+x^2} \right) ^2 dx$ but I don't have idea.

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As the denominator contains $x^2+1$ we put $x=\tan y\implies dx=\sec^2ydy$

For $x^2-1$ with $x\ge1$ we need to put $x=\sec y$

and for $1-x^2$ with $x\le1$ we need to put $x=\sin y$

$$\implies \int\frac{x^2dx}{(x^2+1)^2}=\int\frac{\tan^2y\sec^2ydy}{\sec^4y}=\int\sin^2ydy=\frac12\int(1-\cos2y)dy=\frac{y}2-\frac{\sin2y}4+C$$

If it were indefinite integral, we had to replace back $y$ with $x$

$y=\arctan x$ and $\sin2y=\frac{2\tan y}{1+\tan^2y}=\frac{2x}{1+x^2}$

For definite integral, $x=0,y=\arctan 0=0$ and $x=\infty,y=\arctan \infty=\frac\pi2$

So, the required definite integral will be $$2\left[\frac{y}2-\frac{\sin2y}4+C\right]_0^{\frac\pi2}=2\cdot\frac\pi4=\frac\pi2$$

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This integral may be evaluated using the residue theorem. The integrand has poles at $x=\pm i$; if we close with a contour in the upper half plane, then we need only worry about the pole at $x=i$. The integral is therefore

$$i 2 \pi \left [\frac{d}{dx} \frac{x^2}{(x+i)^2} \right ]_{x=i} = i 2 \pi\left [\frac{i 2 x}{(x+i)^3} \right ]_{x=i} = i 2 \pi \frac{-i}{4} = \frac{\pi}{2}$$

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$i$ is missing in the numerator. Should read $i 2 \pi\left [\frac{2 i x}{(x+i)^3} \right ]_{x=i}$. Editor doesn't let changing less than 6 chars. –  karakfa Sep 3 '13 at 18:43
@karakfa: thanks for pointing out the typo. –  Ron Gordon Sep 3 '13 at 18:45
This takes a little more to justify it. You need to describe the path of integration: use the semicircle of radius $R$ together with the $x$-axis from $-R$ to $R$. Then let $R\to\infty$, and show that the integral on the semicircle goes to $0$. –  Eric Jablow Sep 3 '13 at 21:32
@EricJablow: sure of course. One must also demonstrate the convergence of improper integrals at each endpoint as a limit of $\int_{-R}^R \cdot$ as $R \to \infty$. But you knew that already and are making the same remarks to all the other answerers, right? But seriously, my purpose here is to demonstrate the beauty and simplicity of residues - it is hard to do that when one feels the need to justify convergence each and every single time one wishes to use residues. At some point, the training wheels come off. Yes, the integral there vanishes as $\pi/R$. Justified. –  Ron Gordon Sep 3 '13 at 21:35
I tend to think of the posters here as students, and it seemed as though the OP was unfamiliar with complex analysis. –  Eric Jablow Sep 4 '13 at 0:13

$$\frac{\pi}{t}=\int_{-\infty}^{\infty} \frac{dx}{t^2x^2+1}\Rightarrow\frac{\pi}{t^2}=\int_{-\infty}^{\infty}\frac{2tx^2}{(t^2x^2+1)^2}\,dx\Rightarrow \frac{\pi}{2}=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2}\,dx$$

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$$F(a)=\int-\frac{1}{1+ax^2}dx=-\frac{\arctan(\sqrt{a}x)}{\sqrt a}+c$$

$$\int_{-\infty}^{+\infty}-\frac{1}{1+ax^2}dx=-\frac{\pi}{\sqrt a}$$

so we take the derivaitve of $-\frac{\pi}{\sqrt a}$ with respect to $"a"$ then put a=1 so

$$\int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{\pi}{2}$$

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HINT:

$$I=\int\frac{x^2dx}{(x^2+1)^2}=\int x\cdot \frac x{(x^2+1)^2}dx$$

Integrating by parts, $$I=x\int \frac x{(x^2+1)^2}dx-\int\left(\frac{d(x)}{dx}\cdot \frac x{(x^2+1)^2}dx\right)dx$$

$$\text{As }\int \frac x{(x^2+1)^2}dx=\frac12\int\frac{d(x^2+1)}{(x^2+1)^2}=-\frac1{2(x^2+1)}$$

$$I=-x\cdot\frac1{2(x^2+1)}+\int \frac1{2(x^2+1)}dx=-\frac x{2(x^2+1)}+\frac{\arctan x}2+C$$

Can you take it from here?

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I know I'm a bit late for the party, but here is another solution: \begin{align} 2\int_{0}^\infty \frac{x^2}{(1+x^2)^2}\mathrm dx &= \frac22\int_{-\infty}^\infty \frac{y^{\frac{2-1}{2}}}{(1+y)^2}\mathrm dx \\ &= B\bigl(\tfrac12 + 1, 2-\tfrac12 -1\bigr) = B\bigl(\tfrac32,\tfrac12\bigr) \\ &= \frac{\Gamma\bigl(\tfrac32\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma(2)} \\ &= \frac{\sqrt\pi \cdot \frac12 \sqrt\pi}{1} = \frac\pi 2 \end{align} The first step is the substitution $x^2 \mapsto y$, the second and third step are the use of well-known (and very easy to proof) representations of the Beta function.

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We can use Parseval's theorem! If $F$ is the Fourier transform of $f$, and $G$ is the Fourier transform of $g$, then $$\int_{-\infty}^\infty\overline{f(t)}g(t)\,dt= \frac{1}{2\pi}\int_{-\infty}^\infty\overline{F(x)}G(x)\,dx.$$

We have the Fourier transform pairs

1. $\displaystyle e^{-|t|}\longmapsto\frac{2}{1+\omega^2}$
2. $\displaystyle \frac{d}{dt}f(t) \longmapsto i\omega F(\omega)$

It is clear from those that $$f(t)=\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\longmapsto \frac{ix}{1+x^2}=F(x),$$ and conveniently $\overline{F(x)}F(x)=\frac{x^2}{(1+x^2)^2}$ is the integrand that we want.

Using this and Parseval's theorem, we see that

$$\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx= 2\pi\int_{-\infty}^\infty\left(\frac{1}{2}\frac{d}{dt}\left(e^{-|t|}\right)\right)^2dt,$$ and since the integrand is even, one can write $$\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}dx= \pi\int_0^\infty\left(\frac{d}{dt}\left(e^{-t}\right)\right)^2dt= \pi\int_0^\infty e^{-2t}dt=\frac{\pi}{2}.$$

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