Prove $\int_0^{\infty}\frac{\ln(1+x^2)}{x^2(1+x^2)}dx=\pi\ln\big(\frac 2 e\big)$

Question

I was having trouble with the question:

Prove that $$I:=\int_0^{\infty}\frac{\ln(1+x^2)}{x^2(1+x^2)}dx=\pi\ln\big(\frac e 2\big)$$

My Attempt

Perform partial fractions $$I=\int_0^{\infty}\frac{\ln(1+x^2)}{x^2(1+x^2)}dx=\int_0^{\infty}\frac{\ln(1+x^2)}{x^2}dx-\int_0^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx=$$ First integral $$\int_0^{\infty}\frac{\ln(1+x^2)}{x^2}dx=-\Bigg[\frac{\ln(x^2+1)}{x}\Bigg]_0^{\infty}+\int_0^{\infty}\frac{2}{x^2+1}=\pi$$ Second integral $$\int_0^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx=$$ How do you solve this integral? Thank you for your time

Answer 1

Define

$$J[a] = \int_0^\infty \frac{\ln(a^2+x^2)}{1+x^2}\:dx \implies J'[a] = \frac{2a}{a^2-1}\int_0^\infty \frac{1}{x^2+1}-\frac{1}{x^2+a^2}\:dx$$

$$J'[a] = \frac{\pi}{1+a} \implies J[a] = \pi\ln(1+a) + C$$

We can see that

$$J[0] = \int_0^\infty\frac{2\ln x}{1+x^2}\:dx = 0 = C$$

thus

$$J[1] = \int_0^\infty \frac{\ln(1+x^2)}{1+x^2}dx = \pi\ln 2$$

and we have

$$\pi - \pi \ln 2 = \pi \ln\left(\frac{e}{2}\right)$$

Answer 2

Let $$\int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx= J$$.

Substituting $x=tan(\theta)$,

$$J=-2\int_0^{\frac{π}{2}} \log{\cos{\theta}}d{\theta}$$.

Hence, $$J=π\log 2$$.