find $I=\int_0^{\infty} \log{(x+1/x)}\,\frac{dx}{1+x^2} $

Question

Using $$\int_0^{\pi/2} \log\sin x\,\mathrm dx= -\dfrac{\pi}{2} \log 2$$ how to find

$$I=\int_0^{\infty} \log{(x+1/x)}\,\frac{dx}{1+x^2}. $$

Putting $x=\tan z$,

$I=\int_0^{\pi/2} (\log 2-\log(\sin(2z)))dz=\frac{\pi}{2}\log 2-1/2\int_0^{\pi} \log(\sin(u))du$ for $2z=u$

what to do next?

Answer 1

Split the second integral:

$$\int_0^{\pi}\log(\sin(u)) du = \int_0^{\pi/2}\log(\sin(u)) du + \int_{\pi/2}^{\pi}\log(\sin(u)) du$$

And use a change of variable $u=\pi-x$ for the second part.

Answer 2

You are given that ; $\int_0^\frac\pi2 \log(\sin(x))\,dx = -\frac\pi2\log 2$

$ I=\int_0^{\pi/2} (\log 2-\log(\sin(2z)))dz$

$I = \int_0^\frac\pi2\log 2\,dz-\int_0^\frac\pi2\log(\sin(2z))\,dz$

Consider $J = \int_0^\frac\pi2\log(\sin(2z))\,dz $

let $2z =w\implies dz=\frac{dw}2$

$J= \frac12\int_0^\pi\log(\sin(w))\,dw$

$J=\frac12\cdot2\int_0^\frac\pi2\log(\sin(w))\,dw$

$J = \int_0^\frac\pi2\log(\sin(w))\,dw$

$J = -\frac\pi2\log(2) $

Therefore ,$I = \log2\cdot z\big|_0^\frac\pi2+\frac\pi2\log(2)$

$I =2\cdot\frac\pi2\log(2) $

$I =\pi\log(2)$