Zerrilloh6

2022-01-03

How to solve ${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{log}\left(x+\frac{1}{x}\right)}{1+{x}^{2}}dx$ ?

rodclassique4r

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

Hector Roberts

Instead of using the trigonometric subsitution you may use the identity
$\mathrm{log}\left(1+{x}^{2}\right)={\int }_{0}^{1}da\frac{{x}^{2}}{1+a{x}^{2}}$
Calling your integral of interest $I$ It follows that
$I={\int }_{0}^{1}da\underset{\text{use Residue Theorem to calculate this integral}}{\underset{⏟}{{\int }_{0}^{\mathrm{\infty }}dx\frac{{x}^{2}}{\left({x}^{2}+1\right)\left(1+a{x}^{2}\right)}}}$

Vasquez

$x\to \mathrm{tan}x\phantom{\rule{0ex}{0ex}}I=-{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(\mathrm{sin}x\mathrm{cos}x\right)dx\phantom{\rule{0ex}{0ex}}=-{\int }_{0}^{\frac{pi}{2}}\mathrm{log}\left(\mathrm{sin}x\right)dx+{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(\mathrm{cos}x\right)dx\phantom{\rule{0ex}{0ex}}=-2{\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(\mathrm{cos}x\right)dx\phantom{\rule{0ex}{0ex}}=\pi \mathrm{log}2$