Lennie Davis

2021-12-30

How to evaluate:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{ln}\left({x}^{2}+1\right)}{{x}^{2}+1}dx$

maul124uk

Let
$I\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(a{x}^{2}+1\right)}{{x}^{2}+1}dx$
Then
${I}^{\prime }\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{x}^{2}}{\left(a{x}^{2}+1\right)\left({x}^{2}+1\right)}dx$
$=\frac{1}{\alpha -1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left[\frac{1}{{x}^{2}+1}-\frac{1}{\alpha {x}^{2}+1}\right]dx$
$=\frac{\pi }{\sqrt{\alpha }+\alpha }$
and hence
$I\left(\alpha \right)=\int \frac{\pi }{\sqrt{\alpha }+\alpha }d\alpha$
$=2\pi \mathrm{ln}\left(1+\sqrt{\alpha }\right)+C$
Clearly $I\left(0\right)=0$ implies $C=0$. Thus
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\mathrm{ln}\left({x}^{2}+1\right)}{{x}^{2}+1}dx=I\left(1\right)=2\pi \mathrm{ln}2$

encolatgehu

Letting $x=\mathrm{tan}t$ leads to
$-4{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(\mathrm{cos}t\right)dt=2\pi \mathrm{log}2$
we could consider:
$I\left(s\right)={\int }_{\mathbb{R}}{\left(1+{x}^{2}\right)}^{s}dx$, and then evaluate ${I}^{\prime }\left(-1\right)$
$I\left(s\right)={\int }_{\mathbb{R}}{\left(1+{x}^{2}\right)}^{s}dx=2{\int }_{0}^{\mathrm{\infty }}{\left(1+{x}^{2}\right)}^{s}dx={\int }_{0}^{1}{\left(1-u\right)}^{-\frac{3}{2}-s}\frac{du}{\sqrt{u}}$
$=B\left(\frac{1}{2},-\frac{1}{2}-s\right)$
Thus we established $I\left(s\right)=\sqrt{\pi }\frac{\mathrm{\Gamma }\left(-\frac{1}{2}-s\right)}{\mathrm{\Gamma }\left(-s\right)}$. We are now ready to compute the derivative:
${I}^{\prime }\left(s\right)=I\left(s\right)\left(\psi \left(-s\right)-\psi \left(-s-\frac{1}{2}\right)\right)$
and
${I}^{\prime }\left(-1\right)=I\left(-1\right)\left(\psi \left(1\right)-\psi \left(\frac{1}{2}\right)\right)$
$=\sqrt{\pi }\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(1\right)}\left(\psi \left(1\right)-\psi \left(\frac{1}{2}\right)\right)=2\pi \mathrm{log}\left(2\right)$
where $\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }$ was used, as well as a polygamma duplication identity:
$\psi \left(2s\right)=\mathrm{log}\left(2\right)+\frac{1}{2}\left(\psi \left(s\right)+\psi \left(s+\frac{1}{2}\right)\right)$
That evaluated at $s=\frac{1}{2}$ gives $\psi \left(1\right)-\psi \left(\frac{1}{2}\right)=2\mathrm{log}\left(2\right)$.

Vasquez