Computing integral using complex analysis methods I'm trying to compute the integral
$$
\int_0^{\infty} \frac{\ln(x)}{x^2 + 1} \, dx
$$
using complex analysis methods. We haven't learned residue calculus yet though, only contour integrals up through the Cauchy integral formula.
I'm trying to make use of a half circle centered at the origin of radius $R$ and then let $R$ tend to infinity, but there is a definite singularity for the $\ln(x)$ function. Does anybody have a suggestion?

It is not necessary to use complex analysis methods to solve this problem. You only need $x\to\frac{1}{x}$. In fact, letting $u=\frac{1}{x}$ gives
$$ \int_1^\infty\frac{\ln x}{1+x^2}dx=\int_1^0\frac{\ln \frac{1}{u}}{1+\frac{1}{u^2}}(-\frac{du}{u^2})=-\int_0^1\frac{\ln u}{1+u^2}du. $$
So
$$ \int_0^\infty\frac{\ln x}{1+x^2}dx=\int_0^1\frac{\ln x}{1+x^2}dx+\int_1^\infty\frac{\ln x}{1+x^2}dx=\int_0^1\frac{\ln x}{1+x^2}dx-\int_0^1\frac{\ln u}{1+u^2}du=0. $$
