Fast calculation for $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$

Question

I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?

Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it...

Answer 1

If you use the substitution $x=1/t$, you can check directly that $$ \int_0^1\frac{\log x}{1+x^2}\,dx=-\int_1^\infty\frac{\log t}{1+t^2}\,dt. $$

Answer 2

Use the substitution $x=\tan\theta$.

$$I=\int_0^{\pi/2} \ln(\tan\theta)\,d\theta=-\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0$$

Answer 3

Here is a general approach. Consider the integral

$$ F(s) = \int_{0}^{\infty}\frac{x^{s-1}}{1+x^2}=\frac{\pi}{2\sin(s\pi/2)},\quad 0<Re(s)<2. $$

which is the Mellin transform of the function $\frac{1}{1+x^2}$. Now, our integral can be evaluated as

$$ I = \lim_{s\to 1} F'(s) = 0.$$

Note:

1. To evaluate the above integral, you can use the technique.

$$ F(s)= \int_{0}^{\infty} x^{s-1}f(x)dx \implies F'(s)= \int_{0}^{\infty} x^{s-1}\ln(x)f(x)dx .$$