Calculating the integral $\int_{-\infty}^{\infty} \frac{x^2}{(1 + x^2)^3}dx$

Question

The above integral appeared while trying to calculate the Fisher-information of a Cauchy-distributed sample in my statistics homework. I plugged it into Wolfram Alpha, which gives the answer $\frac \pi 8$, and it's also able to find the indefinite integral, but I'm unable to infer what method it used.

So far, I have tried substituting a couple things, such as $(1 + x^2)$, $x^2$, or $(1 + x^2)^3$, but nothing seems to have worked. I also tried integrating by parts, integrating the $x^2$ on the top and differentiating the denominator, but this also doesn't work out. I also tried splitting the integral by writing the $x^2$ in the nominator as $(1 + x^2) - 1$, but this isn't fruitful either.

I am of course aware that I'm supposed to get an $\arctan$ somewhere in there, but I have no idea how to sneak it into the expression.

I'm sure someone among the userbase here has faced a similar integral before, can I maybe get a hint on how to proceed? What is the trick here?

Answer 1

You can compute it using residues. The function$$\begin{array}{rccc}f\colon&\Bbb C\setminus\{\pm i\}&\longrightarrow&\Bbb C\\&z&\mapsto&\dfrac{z^2}{(z^2+1)^3}\end{array}$$is a rational function whose numerator has degree $2$ and whose denominator has degree $6$; in particular, $\lim_{z\to\infty}zf(z)=0$. Furthermore, it has one and only one singularity in the upper half-plane, located at $i$. So\begin{align}\int_{-\infty}^\infty f(x)\,\mathrm dx&=2\pi i\operatorname{res}_{z=i}f(z)\\&=2\pi i\times\left(\frac{-i}{16}\right)\\&=\frac\pi8.\end{align}

Answer 2

Let $I_n=\int\frac{dx}{(x^2+1)^n}.$

$$I_n=\frac x{(x^2+1)^n}+2n\int\frac{x^2dx}{(x^2+1)^{n+1}}=\frac x{(x^2+1)^n}+2n(I_n-I_{n+1}).$$

Answer 3

My answer using elementary techniques:

$$I=\int\frac{x^2}{(1+x^2)^3}\mathrm dx$$ Substitute $t=\arctan(x)$ hence $x = \tan(t)$ and $dt = \frac{dx}{1+x^2} $, giving $$I=\int\frac{\tan^2(t)}{(1+\tan^2(t))^2}\mathrm dt$$ Since $\sec^2(t)=1+\tan^2(t)$: $$I=\int\frac{\tan^2(t)}{\sec^4(t)}\mathrm dt$$ $$I=\int\frac{\sin^2(t)}{\cos^2(t)}{\cos^4(t)}\mathrm dt$$ $$I=\int\sin^2(t)\cos^2(t)\mathrm dt$$ $$I=\int\sin^2(t)(1-\sin^2(t)\mathrm dt$$ $$I=\int\sin^2(t)dt-\int\sin^4(t)\mathrm dt$$

The rest of it is standard

Answer 4

Because the integrand is even, your integral can be written as $\int_{-\infty}^\infty (...)=2\int_0^\infty (...)$. With this in mind consider the general integral $$F(p,q)=2\int_0^\infty \frac{x^{p}}{(1+x^2)^q}\mathrm dx$$ Substitute $t=x^2$, giving $$F(p,q)=\int_0^\infty \frac{t^{\frac{p-1}{2}}}{(1+t)^q}\mathrm dt$$ Recall the following integral representation of the Beta function: $$\mathrm B(u,v)=\int_0^\infty \frac{t^{u-1}}{(1+t)^{u+v}}\mathrm dt$$

From which we conclude $$F(p,q)=\mathrm B\left(\frac{p+1}{2}~,~q-\frac{p+1}{2}\right)$$

Hence, your integral is $$\int_{-\infty}^\infty \frac{x^2}{(1+x^2)^3}\mathrm dx=2\int_{0}^\infty \frac{x^2}{(1+x^2)^3}\mathrm dx \\ =F(2,3) \\ =\mathrm B\left(\frac{3}{2},\frac{3}{2}\right) \\ =\frac{\Gamma(3/2)\Gamma(3/2)}{\Gamma(3/2+3/2)} \\ =\frac{\left(\frac{1}{2}\Gamma(1/2)\right)^2}{\Gamma(3)} \\ =\boxed{\frac{\pi}{8}}$$

Answer 5

$$\begin{align*} &\int_{-\infty}^\infty \frac{x^2}{(1+x^2)^3} \, dx \\ &= 2 \int_0^\infty \frac{x^2}{(1+x^2)^3} \, dx \tag1 \\ &= 4 \int_{-1}^1 \frac{\frac{(1-y)^2}{(1+y)^2}}{\left(1+\frac{(1-y)^2}{(1+y)^2}\right)^3(1+y)^2} \, dy \tag2\\ &= \frac12 \int_{-1}^1 \frac{(1-y^2)^2}{(1+y^2)^3} \, dy \\ &= \int_0^1 \frac{(1-y^2)^2}{(1+y^2)^3} \, dy \tag1 \\ &= \frac12 \sum_{n=0}^\infty (-1)^n (n+2) (n+1) \int_0^1 (1-y^2)^2 y^{2n} \, dy \tag3 \\ &= \frac12 \sum_{n=0}^\infty (-1)^n (n+2)(n+1) \left(\frac1{2n+1} - \frac2{2n+3} + \frac1{2n+5}\right) \\ &= \frac38 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} + \frac14 \sum_{n=0}^\infty \frac{(-1)^n}{2n+3} + \frac38 \sum_{n=0}^\infty \frac{(-1)^n}{2n+5} \tag4 \\ &= \left\{\frac38 \sum_{n=0}^\infty - \frac14 \sum_{n=1}^\infty + \frac38 \sum_{n=2}^\infty\right\} \frac{(-1)^n}{2n+1} \tag5 \\ &= \frac38 \arctan(1) - \frac14 \left(\arctan(1) - 1\right) + \frac38 \left(\arctan(1) - 1 + \frac13\right) = \boxed{\frac\pi8} \tag6 \end{align*}$$

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• $(1)$ : symmetry
• $(2)$ : substitute $x=\dfrac{1-y}{1+y}$
• $(3)$ : exploit the Maclaurin series of $\dfrac1{1-y}$
• $(4)$ : partial fractions
• $(5)$ : shift indices on the latter sums
• $(6)$ : recall the Maclaurin series for $\arctan$