Irvin Dukes

2022-01-02

How would I go about evaluating this integral?
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}\left({x}^{2}+1\right)}{{x}^{2}+1}dx$

Anzante2m

Use $x\to \mathrm{tan}\theta$ and $dx={\mathrm{sec}}^{2}\theta d\theta$. The integral becomes
${\int }_{0}^{\frac{\pi }{2}}2\mathrm{ln}\left(\mathrm{sec}\theta \right)d\theta$
Which is
${\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{cos}\theta \right)d\theta$
And can be solved
Let
$I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{sin}\theta \right)\cdot d\theta$
$I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{cos}\theta \right)\cdot d\theta$
$2I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{sin}\theta ×\mathrm{cos}\theta \right)\cdot d\theta$
$2I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(2\mathrm{sin}\theta ×\mathrm{cos}\theta \right)-\mathrm{ln}2\cdot d\theta$
$2I={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(\mathrm{sin}2\theta \right)\cdot d\theta =I$
So,
$I=-{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}2\cdot d\theta$
$I=-\frac{\pi \mathrm{ln}2}{2}$
And your integral comes out to be
$\pi \mathrm{ln}2$

Lakisha Archer

One way to solve this problem is to use parametric integrals. Let
$I\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(\alpha {x}^{2}+1\right)}{{x}^{2}+1}dx$
Then
${I}^{\prime }\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{\left(\alpha {x}^{2}+1\right)\left({x}^{2}+1\right)}dx$
$={\int }_{0}^{\mathrm{\infty }}\left(-\frac{1}{\alpha -1}\frac{1}{\alpha {x}^{2}+1}+\frac{1}{\alpha -1}\frac{1}{{x}^{2}+1}\right)dx$
$=-\frac{1}{\alpha -1}\frac{1}{\sqrt{\alpha }}\frac{\pi }{2}+\frac{1}{\alpha -1}\frac{\pi }{2}$
$=\frac{\pi }{2}\frac{\sqrt{\alpha }-1}{\sqrt{\alpha }\left(\alpha -1\right)}$
$=\frac{\pi }{2}\left(\frac{1}{\alpha }-\frac{1}{\alpha +1}\right)$
Thus,
$I\left(\alpha \right)=\pi \mathrm{ln}\left(\sqrt{\alpha }+1\right)+C$
But $I\left(0\right)=0$ implies $C=0$. So $I\left(1\right)=\pi \mathrm{ln}2$

Vasquez

Hints As RandomVariable suggested, use $\mathrm{log}\left({x}^{2}+1\right)=\mathrm{log}\left(x+i\right)+\mathrm{log}\left(x-i\right)$, choosing the branches of the logarithm carefully. It's generally best to isolate unpleasant singular things.
Then write
${\int }_{0}^{\mathrm{\infty }}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}$
and make use of the above splitting to integrate the two parts on different contours, each time avoiding enclosing the singularity of the logarithm. The part on the semicircle vanishes.
Answer The UHP pole gives $\mathrm{log}\left(i+i\right)/2i×2\pi i$. The LHP pole gives $\mathrm{log}\left(-i-i\right)/\left(-2i\right)×-2\pi i$. Summing and halving gives the answer
$\frac{\pi }{2}\left(\mathrm{log}\left(2i\right)+\mathrm{log}\left(-2i\right)\right)$
so all that remains is choosing the right logarithm. This is easy enough, actually, and the answer is what you expect:
$\pi \mathrm{ln}2$

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