Wanda Kane

2022-01-06

Evaluate integral:
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{1+{x}^{2}}dx$

In general
$I\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}+{\alpha }^{2}}dx$
can be evaluated by using substitution $u=\frac{{\alpha }^{2}}{x}⇒x=\frac{{\alpha }^{2}}{u}⇒dx=-\frac{{\alpha }^{2}}{{u}^{2}}du$, then
$I\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(\frac{{\alpha }^{2}}{u}\right)}{{\left(\frac{{\alpha }^{2}}{u}\right)}^{2}+{\alpha }^{2}}\cdot \frac{{\alpha }^{2}}{{u}^{2}}du$
$={\int }_{0}^{\mathrm{\infty }}\frac{2\mathrm{ln}\alpha -\mathrm{ln}u}{{\alpha }^{2}+{u}^{2}}du$
$=2\mathrm{ln}\alpha {\int }_{0}^{\mathrm{\infty }}\frac{1}{{\alpha }^{2}+{u}^{2}}du-{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}u}{{u}^{2}+{\alpha }^{2}}du$
$=2\mathrm{ln}\alpha {\int }_{0}^{\mathrm{\infty }}\frac{1}{{\alpha }^{2}+{u}^{2}}du-I\left(\alpha \right)$
$I\left(\alpha \right)=\mathrm{ln}\alpha {\int }_{0}^{\mathrm{\infty }}\frac{1}{{\alpha }^{2}+{u}^{2}}du$
The last integral can easily be evaluated since it is a common integral. Using substitution $u=\mathrm{tan}\theta$, the integral turns out to be
$I\left(\alpha \right)=\frac{\mathrm{ln}\alpha }{\alpha }{\int }_{0}^{\frac{\pi }{2}}d\theta$
$=\frac{\pi \mathrm{ln}\alpha }{2\alpha }$
Thus
$I\left(1\right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}+1}dx=0$

Ethan Sanders

Here's another general result, consider for $|a|\le 1$:
$I\left(a,b\right)={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{a}}{{b}^{2}+{x}^{2}}dx$
$=\frac{{b}^{a-1}}{2}{\int }_{0}^{\mathrm{\infty }}\frac{{t}^{\frac{a-1}{2}}}{1+t}dt$ (1)
$=\frac{{b}^{a-1}}{2}B\left(\frac{1+a}{2},\frac{1-a}{2}\right)$ (2)
$=\frac{{b}^{a-1}}{2}\frac{\pi }{\mathrm{cos}\left(\frac{\pi }{2}a\right)}$ (3)
(1) is by subst. $x=b\sqrt{t}$
(2) is by the definition of Beta function: $B\left(a,b\right)={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{a-1}}{{\left(1+x\right)}^{a+b}}dx$
(3) is by using $B\left(a,b\right)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a+b\right)}$ and Euler's reflection formula
$\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(1-a\right)=\frac{\pi }{\mathrm{sin}\left(\pi a\right)}$
Hence,
${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{a}{\left(\mathrm{log}x\right)}^{n}}{{b}^{2}+{x}^{2}}dx=\frac{{d}^{n}}{d{a}^{n}}\left(\frac{{b}^{a-1}}{2}\frac{\pi }{\mathrm{cos}\left(\frac{\pi }{2}a\right)}\right)$
and when $b=1,n=1,a=0$, the result is 0

karton

Another general approach :
Consider
${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{b-1}}{{a}^{2}+{x}^{2}}dx$ (1)
Rewrite (1) as
${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{b-1}}{{a}^{2}+{x}^{2}}dx=\frac{1}{{a}^{2}}{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{b-1}}{1+\left(\frac{x}{a}{\right)}^{2}}dx$ (2)
Putting $x=ay⇒dx=ady$ yields
${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{b-1}}{{a}^{2}+{x}^{2}}dx={a}^{b-2}{\int }_{0}^{\mathrm{\infty }}\frac{{y}^{b-1}}{1+{y}^{2}}dy$
where
${\int }_{0}^{\mathrm{\infty }}\frac{{y}^{b-1}}{1+{y}^{2}}dy=\frac{\pi }{2\mathrm{sin}\left(\frac{b\pi }{2}\right)}$
Hence
${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{b-1}}{{a}^{2}+{x}^{2}}dx=\frac{\pi }{2}\cdot \frac{{a}^{b-2}}{\mathrm{sin}\left(\frac{b\pi }{2}\right)}$ (3)
Differentiating (3) with respect to b and setting b=1 yields
${\int }_{0}^{\mathrm{\infty }}\frac{d}{db}\left[\frac{{x}^{b-1}}{{a}^{2}+{x}^{2}}{\right]}_{b=1}dx=\frac{\pi }{2}\frac{d}{db}\left[\frac{{a}^{b-2}}{\mathrm{sin}\left(\frac{b\pi }{2}\right)}{\right]}_{b=1}$
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}+{a}^{2}}dx=\frac{\pi \mathrm{ln}a}{2\alpha }$
Thus
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}+1}dx=0$

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