# Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$.

I have managed to solve it in one way, but I became very interested in this failed attempt.

$$\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x - \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x$$

We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $$-\frac{\pi}{4}$$ but I don't know how to break these down.

I am currently really interested in proving this $$\int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4}$$

Anyway, here's my trivial solution using $$u = \frac1x$$:

\begin{align} \int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\ & = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\ & = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\ & = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\ & = 0 \end{align}

I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $$\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$$. Please share them if you do come up with any.

For the first part of the question we can substitute $$x=\frac{1}{t}$$ in order to get: $$I=\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x=\int^0_\infty \frac{\ln\left(\frac{1}{t}\right)}{t^2\left(1+\frac{1}{t^2}\right)^3}\frac{-dt}{t^2}=\int_0^\infty \frac{t^2 \ln\left(\frac{1}{t}\right)}{(1+t^2)^3}dt=-I$$ So we just saw that $$I=-I\Rightarrow I=0$$

I am currently really interested in proving this $$\int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4}$$

Well now that we showed that both integrals are equal it's enough to compute only one of it. $$\Omega=\int_0^\infty \frac{\ln x}{(1+x^2)^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2} \ln(\tan t)\cos^2 t\,\mathrm dt =\frac12\int_0^\frac{\pi}{2} \ln(\tan t) (1+\cos(2t))dt$$ Now we will split into two integrals and show that the first one vanishes using the following property of the definite integrals: $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$ $$J=\int_0^\frac{\pi}{2}\ln(\tan t)dt=\int_0^\frac{\pi}{2} \ln(\cot t)dt=-\int_0^\frac{\pi}{2} \ln (\tan t)dt=-J\Rightarrow J=0$$ $$\Rightarrow \Omega=\frac12 \int_0^\frac{\pi}{2}\ln(\tan t)\cos (2t)\mathrm dt=\frac12 \int_0^\frac{\pi}{2} \ln(\tan t)\left(\frac12 \sin(2t)\right)' \mathrm dt=$$ $$=\frac14 \underbrace{\ln(\tan t)\sin(2t)\bigg|_0^\frac{\pi}{2}}_{=0}-\frac14\int_0^\frac{\pi}{2} \frac{\sec^2 t}{\tan t}\sin(2t)\mathrm dt=-\frac12 \int_0^\frac{\pi}{2} dt=-\frac{\pi}{4}$$

• Nice, much quicker. I should've just done that from the start. Commented Dec 29, 2018 at 18:39

$$\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx \stackrel{(*)}{=}\frac{\pi(1-\alpha^2)}{16\cos\frac{\pi \alpha}{2}}$$ $$(*)$$: we use the substitution $$\frac{1}{1+x^2}=u$$, the Beta function and the reflection formula for the $$\Gamma$$ function. This holds for any $$\alpha$$ such that $$-3<\text{Re}(\alpha)<3$$, and since the RHS is an even function, the origin is a stationary point, i.e. $$\color{red}{0}=\frac{d}{d\alpha}\left.\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx\right|_{\alpha=0}\stackrel{\text{DCT}}{=}\int_{0}^{+\infty}\frac{x^{2}\log x}{(1+x^2)^3}\,dx.$$

I found a way to directly evaluate both of those integrals!

We first use the well known result from $$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$$ Evaluate Integral $$\int_0^\infty \frac{\ln(x)}{x^2 +\alpha^2} {\rm d}x = \frac{\pi}{2\alpha} \ln(\alpha)$$ and use differentiation under the integral sign.

$$\int_0^\infty \frac{-2\alpha\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x = \frac{\pi}{2} \frac{1-\ln(\alpha)}{\alpha^2} \\ \implies \int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x = \frac{\pi}{4\alpha^3} (\ln(\alpha)-1)$$

And again, to get $$\int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^3} {\rm d}x = \frac{\pi}{16\alpha^5}(3\ln(\alpha) - 4)$$

And so setting $$\alpha = 1$$ gives us the immediate result $$\int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4}$$

Another method to evaluate this is to use the integral from the 2015 MIT Integration Bee, and is also how I first came across this integral.

From this result (solved by using the substitution $$u=\frac1x$$) $$\int_0^\infty \frac{1}{(1+x^2)(1+x^\alpha)} {\rm d}x = \frac\pi4$$

we will get, by differentiating with respect to $$\alpha$$,

$$\int_0^\infty \frac{x^\alpha \ln(x)}{(1+x^2)(1+x^\alpha)^2} {\rm d}x = 0$$

And finally setting $$\alpha = 2$$.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$

With $$\ds{\Re\pars{\mu} > - 1}$$ and $$\ds{\Re\pars{\nu} > 0}$$:

\begin{align} I_{\mu\nu} & \equiv \bbox[10px,#ffd]{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{1 + x^{2}}^{\nu}}\,\dd x} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 4}\int_{0}^{\infty} {x^{\mu/2 - 1/2}\ln\pars{x} \over \pars{1 + x}^{\nu}}\,\dd x \\[5mm] & = \left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{\infty} {x^{\alpha + \mu/2 - 1/2} \over \pars{1 + x}^{\nu}}\,\dd x \,\right\vert_{\ \alpha\ =\ 0} \\[5mm] & \stackrel{x + 1\ \mapsto\ x}{=}\,\,\, \left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{\infty} {\pars{x - 1}^{\alpha + \mu/2 - 1/2} \over x^{\nu}}\,\dd x \,\right\vert_{\ \alpha\ =\ 0} \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{0} {\pars{1/x - 1}^{\alpha + \mu/2 - 1/2} \over \pars{1/x}^{\nu}}\, \pars{-\,{\dd x \over x^{2}}} \right\vert_{\ \alpha\ =\ 0} \\[5mm] & = \left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{1} x^{\nu - \alpha - \mu/2 - 3/2}\pars{1 - x}^{\alpha + \mu/2 - 1/2}\, \dd x\,\right\vert_{\ \alpha\ =\ 0} \\[5mm] & = {1 \over 4}\,\partiald{}{\alpha}\bracks{% \Gamma\pars{\nu - \alpha - \mu/2 - 1/2}\Gamma\pars{\alpha + \mu/2 + 1/2} \over \Gamma\pars{\nu}}_{\ \alpha\ =\ 0} \\[5mm] & = \bbx{{\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{\nu - \mu/2 - 1/2} \over 4\Gamma\pars{\nu}} \bracks{H_{\mu/2 - 1/2} - H_{\nu - \mu/2 - 3/2}}} \end{align}

$$\ds{H_{z}}$$ is a Harmonic Number.

$$\begin{array}{|c|c|}\hline \ds{\mu \setminus \nu} & \ds{I_{\mu\nu}} \\ \hline \ds{0 \setminus 2} & \ds{-\,{\pi \over 4}} \\ \hline \ds{0 \setminus 3} & \ds{-\,{\pi \over 4}} \\ \hline \end{array}$$

Note that $$\ds{\bbx{I_{\mu,\mu + 1} = 0}}$$ because, in such a case, $$\ds{H_{\mu/2 - 1/2} = H_{\nu - \mu/2 - 3/2}}$$.

$$\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x= \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x$$ Then change in first integral $$x=\frac{1}{t}$$.