Show that $\int_0^{\infty}\dfrac{\ln x}{1+x^2}\hspace{1mm}dx=0$
The problem I am facing is with the hint.
The hint says, use the substitution $u = 1/x$, it makes no sense to me.
Why would we use the substitution u = 1/x, its makes the integral strange.
Can anyone explain
On
Using the hint setting $x=\dfrac1y,dx=-\dfrac{dy}{y^2}$
$$I=\int_0^\infty\frac{\ln x}{1+x^2}dx=\int_\infty^0\frac{\ln(1/y)}{1+(1/y)^2}\left(-\frac{dy}{y^2}\right)=-\int_\infty^0\frac{-\ln y}{1+y^2}dy$$
$$=\int_\infty^0\frac{\ln y}{1+y^2}dy$$
As $\displaystyle\int_a^bf(x)dx=-\int_b^af(x)dx,$
$$I=-\int_0^\infty\frac{\ln y}{1+y^2}dy=-I$$
Hint
Change the variable $u=1/x$ for the integral
$$\int_0^1\frac{\ln x}{1+x^2}dx$$