Show $\int^{\infty}_0\frac{ln(x)}{(1+x^2)}dx=0$ using substitution $x=e^u$

Question

I have a task to show that $$\int^{\infty}_0\frac{\ln(x)}{(1+x^2)}dx=0$$

With substitution $x=e^u$

I couldn't really get to anything:

$\int^{\infty}_0\frac{\ln(x)}{(1+x^2)}dx=$

Making substitution: $x=e^u \Rightarrow dx=e^udu \Rightarrow x=0 \Leftrightarrow u=-\infty \Rightarrow x=\infty \Leftrightarrow u=\infty$

$$= \int^{\infty}_{-\infty}\frac{ue^u}{1+e^{2u}}du$$

I don't know what i'm supposed to do with this to be honest.

Any help on how to continue? When I try to evaluate the integral without the boundaries, in the integral calculator , it shows me some weird expression with some terms "$\operatorname{Li}$" that I don't think I should be getting.

Answer 1

Hint: $$ \frac{ue^u}{1+e^{2u}}$$ is an odd function.

Answer 2

$$I-\int_{0}^{\infty} \frac{\ln x}{1+x^2} dx$$ Let $x=1/t \implies dx=-dt/t^2$, we get $$I=-\int_{0}^{\infty} \frac{\ln t}{1+t^2} dt=-I \implies I=0$$

Or after OP's last step

$$J=\int_{-\infty}^{\infty} \frac{u e^u}{1+e^{2u}} du$$ Take $u=-v \implies du=-dv$ Then $$J=-\int_{\infty}^{-\infty} \frac{-v e^{-v}}{1+e^{-2v}} dv=\int_{-\infty}^{\infty} \frac{-ve^v}{e^{2v}+1}dv=-J \implies J=0.$$

Also The integral $J$ in above can be rewritten as $$J=\int_{-\infty}^{\infty} \frac{u}{e^{-u}+e^{u}} du$$ The the integral of odd integrand between symmetric limits vanishes.