Evaluate integral: \int_0^\infty \frac{\ln x}{1+x^2}dx

Wanda Kane 2022-01-06 Answered
Evaluate integral:
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\ln{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Pademagk71
Answered 2022-01-07 Author has 5116 answers

In general
\(\displaystyle{I}{\left(\alpha\right)}={\int_{{0}}^{{\infty}}}{\frac{{{\ln{{x}}}}}{{{x}^{{2}}+\alpha^{{2}}}}}{\left.{d}{x}\right.}\)
can be evaluated by using substitution \(\displaystyle{u}={\frac{{\alpha^{{2}}}}{{{x}}}}\Rightarrow{x}={\frac{{\alpha^{{2}}}}{{{u}}}}\Rightarrow{\left.{d}{x}\right.}=-{\frac{{\alpha^{{2}}}}{{{u}^{{2}}}}}{d}{u}\), then
\(\displaystyle{I}{\left(\alpha\right)}={\int_{{0}}^{\infty}}{\frac{{{\ln{{\left({\frac{{\alpha^{{2}}}}{{{u}}}}\right)}}}}}{{{\left({\frac{{\alpha^{{2}}}}{{{u}}}}\right)}^{{2}}+\alpha^{{2}}}}}\cdot{\frac{{\alpha^{{2}}}}{{{u}^{{2}}}}}{d}{u}\)
\(\displaystyle={\int_{{0}}^{{\infty}}}{\frac{{{2}{\ln{\alpha}}-{\ln{{u}}}}}{{\alpha^{{2}}+{u}^{{2}}}}}{d}{u}\)
\(\displaystyle={2}{\ln{\alpha}}{\int_{{0}}^{{\infty}}}{\frac{{{1}}}{{\alpha^{{2}}+{u}^{{2}}}}}{d}{u}-{\int_{{0}}^{\infty}}{\frac{{{\ln{{u}}}}}{{{u}^{{2}}+\alpha^{{2}}}}}{d}{u}\)
\(=2\ln\alpha\int_0^{\infty}\frac{1}{\alpha^2+u^2}du-I(\alpha)\)
\(\displaystyle{I}{\left(\alpha\right)}={\ln{\alpha}}{\int_{{0}}^{\infty}}{\frac{{{1}}}{{\alpha^{{2}}+{u}^{{2}}}}}{d}{u}\)
The last integral can easily be evaluated since it is a common integral. Using substitution \(\displaystyle{u}={\tan{\theta}}\), the integral turns out to be
\(\displaystyle{I}{\left(\alpha\right)}={\frac{{{\ln{\alpha}}}}{{\alpha}}}{\int_{{0}}^{{{\frac{{\pi}}{{{2}}}}}}}{d}\theta\)
\(\displaystyle={\frac{{\pi{\ln{\alpha}}}}{{{2}\alpha}}}\)
Thus
\(\displaystyle{I}{\left({1}\right)}={\int_{{0}}^{\infty}}{\frac{{{\ln{{x}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}={0}\)

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Ethan Sanders
Answered 2022-01-08 Author has 918 answers

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

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karton
Answered 2022-01-11 Author has 8659 answers

Another general approach :
Consider
\(\int_0^\infty\frac{x^{b-1}}{a^2+x^2}dx\) (1)
Rewrite (1) as
\(\int_0^\infty\frac{x^{b-1}}{a^2+x^2}dx=\frac{1}{a^2}\int_0^{\infty}\frac{x^{b-1}}{1+(\frac{x}{a})^2}dx\) (2)
Putting \(x=ay\Rightarrow dx=ady\) yields
\(\int_0^\infty\frac{x^{b-1}}{a^2+x^2}dx=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}dy\)
where
\(\int_0^\infty\frac{y^{b-1}}{1+y^2}dy=\frac{\pi}{2\sin(\frac{b\pi}{2})}\)
Hence
\(\int_0^\infty\frac{x^{b-1}}{a^2+x^2}dx=\frac{\pi}{2}\cdot\frac{a^{b-2}}{\sin(\frac{b\pi}{2})}\) (3)
Differentiating (3) with respect to b and setting b=1 yields
\(\int_0^\infty \frac{d}{db}[\frac{x^{b-1}}{a^2+x^2}]_{b=1}dx=\frac{\pi}{2}\frac{d}{db}[\frac{a^{b-2}}{\sin(\frac{b\pi}{2})}]_{b=1}\)
\(\int_0^\infty\frac{\ln x}{x^2+a^2}dx=\frac{\pi\ln a}{2\alpha}\)
Thus
\(\int_0^\infty\frac{\ln x}{x^2+1}dx=0\)

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