Evaluate $\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt$

Question

How can we compute $$I=\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt$$ Mathematica gives $I=0$.

Answer 1

Hint

Write

$$\int_0^\infty \frac{\log(t)}{1+t^2} \, dt=\int_0^1 \frac{\log(t)}{1+t^2} \, dt+\int_1^\infty \frac{\log(t)}{1+t^2} \, dt$$ and then change the variable $t=\frac1x$ for one of the two last integrals.