1

While evaluating some integrals I came across this one that got me stuck: $$I=\int_0^\infty \frac{\ln^2(x)}{1+x^2}dx=\frac{\pi^3}{8}$$ The result is from wolframalpha since I wasn't able to compute it.

Firstly I thought of series expansion of $\frac{1}{1+x^2}$, but it's wrong since we are not integrating in $(0,1)$ but in $(0,+\infty)$ so the series doesn't converge.

Then I tried with the substitution $x=\tan(t)$, and this way the integral reduces to

$$I=\int_0^\frac{\pi}{2} \ln^2(\tan(t)) dt$$

but got stuck again. Can anybody provide a solution?

Zima
  • 4,125

3 Answers3

6

Hint: To exploit the series expansion, make a substitution over part of the integration interval.

$$\int_1^\infty \frac{\ln^2(x)}{1+x^2}\,dx \stackrel{x\mapsto\tfrac1x}= \int_0^1 \frac{\ln^2\left(\frac1x\right)}{1+\frac1{x^2}}\,\frac{dx}{x^2} = \int_0^1 \frac{\ln^2(x)}{1+x^2} \, dx\\ \implies I = 2\int_0^1 \frac{\ln^2(x)}{1+x^2}\,dx$$

user170231
  • 25,320
3

You can use Feynman’s trick to evaluate. Let \begin{eqnarray} I(s)&=&\int_0^\infty \frac{x^{s-1}}{1+x^2}dx. \end{eqnarray} By Mellin's transform table from here, you have $$ I(s)=\frac{\pi}{2\sin(\frac12\pi s)}. $$ It is easy to get $$ I''(1)=\frac{\pi^3}{8}. $$

xpaul
  • 47,821
2

$$ I=\frac{\partial^2}{\partial a^2}\left.I(a)\right|_{a=0} $$ where $I( a)=\int_0^{\infty} \frac{x^a}{1+x^2} d x$ is transformed to a beta function via $x=\tan \theta$. $$ I(a)=\int_0^{\infty} \tan ^a \theta d \theta= \int_0^{\infty} \sin ^a\theta \cos ^{-a} \theta d \theta= \frac{1}{2} B\left(\frac{1-a}{2}, \frac{1-a}{2}\right)=\frac{\pi}{2} \sec \left(\frac{\pi a}{2}\right) $$ Then $$ I=\frac{\partial^2}{\partial a^2}\left[ \frac{\pi}{2} \sec \left(\frac{\pi a}{2}\right)\right] _{a=0}=\frac{\pi^3}{8} $$

Lai
  • 31,615