How to solve \int^\infty_0\frac{\log(x+\frac{1}{x})}{1+x^2}dx ?

Zerrilloh6 2022-01-03 Answered
How to solve \(\displaystyle\int^{\infty}_{0}{\frac{{{\log{{\left({x}+{\frac{{{1}}}{{{x}}}}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\) ?

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Expert Answer

rodclassique4r
Answered 2022-01-04 Author has 2580 answers

You may write
\(\displaystyle\int^{\infty}_{0}{\frac{{{\log{{\left({x}+{\frac{{{1}}}{{{x}}}}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}=\int^{\infty}_{0}{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}-{\log{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\int^{\infty}_{0}{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}-\int^{\infty}_{0}{\frac{{{\log{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)
Clearly, by the change of variable \(\\ \rightarrow\frac{1}{x}\), we get
\(\displaystyle=\int^{\infty}_{0}{\frac{{{\log{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}=-\int^{\infty}_{0}{\frac{{{\log{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)
On the other hand, by the change of variable
\(\displaystyle{x}={\tan{\theta}},{\left.{d}{x}\right.}={\left({1}-{{\tan}^{{2}}\theta}\right)}{d}\theta,{1}+{{\tan}^{{2}}\theta}={\frac{{{1}}}{{{{\cos}^{{2}}\theta}}}}\)
we obtain the classic evaluation:
\(\displaystyle=\int^{\infty}_{0}{\frac{{{\log{{\left({1}+{x}^{{2}}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}=-{2}\int^{{{\frac{{\pi}}{{{2}}}}}}_{0}{\log{{\left({\cos{\theta}}\right)}}}{d}\theta=\pi{\log{{2}}}\)

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Hector Roberts
Answered 2022-01-05 Author has 3403 answers
Instead of using the trigonometric subsitution you may use the identity
\(\displaystyle{\log{{\left({1}+{x}^{{2}}\right)}}}={\int_{{0}}^{{1}}}{d}{a}{\frac{{{x}^{{2}}}}{{{1}+{a}{x}^{{2}}}}}\)
Calling your integral of interest \(\displaystyle{I}\) It follows that
\(\displaystyle{I}={\int_{{0}}^{{1}}}{d}{a}\underbrace{{{\int_{{0}}^{\infty}}{\left.{d}{x}\right.}{\frac{{{x}^{{2}}}}{{{\left({x}^{{2}}+{1}\right)}{\left({1}+{a}{x}^{{2}}\right)}}}}}}_{{\text{use Residue Theorem to calculate this integral}}}\)
\(\displaystyle=\underbrace{{{\int_{{0}}^{{1}}}{d}{a}{\frac{{\pi}}{{{2}{\left({a}+\sqrt{{{a}}}\right)}}}}}}_{{\text{use }\ {a}={b}^{{2}}\ \text{ to turn this into a standard integra}}}=\pi{\log{{\left({2}\right)}}}\)
In agreement with Olivier’s answer
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Vasquez
Answered 2022-01-11 Author has 8850 answers

\(x\rightarrow\tan x \\I=-\int^{\frac{\pi}{2}}_0\log(\sin x\cos x)dx \\=-\int^{\frac{pi}{2}}_0\log(\sin x)dx+\int^{\frac{\pi}{2}}_0\log(\cos x)dx \\=-2\int^{\frac{\pi}{2}}_0\log(\cos x)dx \\=\pi\log2\)

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