Roger Smith

2022-01-05

Can this integral be solved with contour integral or by some application of residue theorem?

eninsala06

${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx={\int }_{0}^{1}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx+{\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx$
$={\int }_{0}^{1}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx+{\int }_{0}^{1}\frac{\mathrm{log}\left({x}^{-1}+1\right)}{{x}^{2}+1}dx$
$=2{\int }_{0}^{1}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx-{\int }_{0}^{1}\frac{\mathrm{log}x}{{x}^{2}+1}dx$
For the first integral, we plug

Then it is easy to find that
${\int }_{0}^{1}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx={\int }_{0}^{1}\frac{\mathrm{log}2-\mathrm{log}\left(u+1\right)}{{u}^{2}+1}du=\frac{\pi }{4}\mathrm{log}2-{\int }_{0}^{1}\frac{\mathrm{log}\left(u+1\right)}{{u}^{2}+1}du$
and hence
${\int }_{0}^{1}\frac{\mathrm{log}\left(x+1\right)}{{x}^{2}+1}dx=\frac{\pi }{8}\mathrm{log}2$
For the second integral, we plug $x={e}^{-t}$ and we have

Elaine Verrett

If youre

karton

In this answer, the substitution $x=\frac{1-y}{1+y}$ is used to get
${\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx=\frac{\pi }{8}\mathrm{log}\left(2\right)$
We can use the substitution $x↦1/x$ to get
${\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx={\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)-\mathrm{log}\left(x\right)}{1+{x}^{2}}dx$
which implies
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx=2{\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx-{\int }_{0}^{1}\frac{\mathrm{log}\left(x\right)}{1+{x}^{2}}dx$
Therefore, we can use
${\int }_{0}^{1}{x}^{k}\mathrm{log}\left(x\right)dx=\frac{1}{k+1}{\int }_{0}^{1}\mathrm{log}\left(x\right)d{x}^{k+1}\phantom{\rule{0ex}{0ex}}=-\frac{1}{k+1}{\int }_{0}^{1}{x}^{k+1}d\mathrm{log}\left(x\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{k+1}{\int }_{0}^{1}{x}^{k}dx\phantom{\rule{0ex}{0ex}}=-\frac{1}{\left(k+1{\right)}^{2}}$
to get
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}=2{\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)}{1+{x}^{2}}dx-{\int }_{0}^{1}\frac{\mathrm{log}\left(x\right)}{1+{x}^{2}}dx$
$=\frac{\pi }{4}\mathrm{log}\left(2\right)-{\int }_{0}^{1}\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}{x}^{2k}\mathrm{log}\left(x\right)dx$
$=\frac{\pi }{4}\mathrm{log}\left(2\right)+\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{\left(2k+1{\right)}^{2}}$
$=\frac{\pi }{4}\mathrm{log}\left(2\right)+G$

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