osula9a

2022-01-02

Calculate:
${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{2}\left(x\right)}{{\left(1-{x}^{2}\right)}^{2}}dx$

Stella Calderon

You're definetly on the right track with that substitution of $x=\frac{1}{t}$
Basically we have:
$I={\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{2}\left(x\right)}{{\left(1-{x}^{2}\right)}^{2}}dx={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}{\mathrm{ln}}^{2}x}{{\left(1-{x}^{2}\right)}^{2}}dx$
Now what if we add them up?
$2I={\int }_{0}^{\mathrm{\infty }}{\mathrm{ln}}^{2}x\frac{1+{x}^{2}}{{\left(1-{x}^{2}\right)}^{2}}dx$
If you don't know how to deal easily with the integral
$\int \frac{1+{x}^{2}}{{\left(1-{x}^{2}\right)}^{2}}dx=\frac{x}{1-{x}^{2}}+C$
Anyway we have, integrating by parts:
$2I=\underset{=0}{\underset{⏟}{\frac{x}{1-{x}^{2}}{\mathrm{ln}}^{2}x{\mid }_{0}^{\mathrm{\infty }}}}+2\underset{=\frac{{\pi }^{2}}{4}}{\underset{⏟}{{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}-1}dx}}$
$⇒2I=2\cdot \frac{{\pi }^{2}}{4}⇒I=\frac{{\pi }^{2}}{4}$

Fasaniu

This is a variant of TheSimpliFires

Vasquez

Here is yet another slight variation on a theme.
Let
$I={\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{2}x}{\left(1-{x}^{2}{\right)}^{2}}dx$
then
$I={\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}x}{\left(1-{x}^{2}{\right)}^{2}}dx+{\int }_{1}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{2}x}{\left(1-{x}^{2}{\right)}^{2}}dx$
$={\int }_{0}^{1}\frac{\left(1+{x}^{2}\right){\mathrm{ln}}^{2}x}{\left(1-{x}^{2}{\right)}^{2}}dx$ (1)
after a substitution of $↦1/xZ$ has been enforced in the second of the integrals.
As

differentiating with respect to x gives
$\frac{1}{\left(1-{x}^{2}{\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}n{x}^{2n-2}$
On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have
$I=\sum _{n=1}^{\mathrm{\infty }}n{\int }_{0}^{1}\left({x}^{2n-2}+{x}^{2n}\right){\mathrm{ln}}^{2}xdx$
Integrating be parts twice, we are left with
$I=\sum _{n=1}^{\mathrm{\infty }}\left(\frac{2n}{\left(2n-1{\right)}^{3}}+\frac{2n}{\left(2n+1{\right)}^{3}}\right)$
As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have
$\begin{array}{}I=\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{\left(2n-1{\right)}^{2}}+\frac{1}{\left(2n-1{\right)}^{3}}+\frac{1}{\left(2n+1{\right)}^{2}}-\frac{1}{\left(2n+1{\right)}^{3}}\right]\\ =\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{\left(2n-1{\right)}^{2}}+\frac{1}{\left(2n-1{\right)}^{3}}\right]+\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{\left(2n+1{\right)}^{2}}-\frac{1}{\left(2n+1{\right)}^{3}}\right]\\ =\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{\left(2n+1{\right)}^{2}}+\frac{1}{\left(2n+1{\right)}^{3}}\right]+\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{\left(2n+1{\right)}^{2}}-\frac{1}{\left(2n+1{\right)}^{3}}\right]\\ =2+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n+1{\right)}^{2}}\\ =2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\left(2n+1{\right)}^{2}}\\ =2\left[\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n{\right)}^{2}}\right]\\ =2\left(1-\frac{1}{4}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}\\ =\frac{3}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}\\ =\frac{3}{2}\cdot \frac{{\pi }^{2}}{6}\\ =\frac{{\pi }^{2}}{4}\end{array}$
as expected.

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