Calcualte $\displaystyle \int_0^\infty \frac{\sqrt{x} \ln(x)}{1+x^2} \, dx.$
My solution:
We take the branch cut $[-\pi/2,3\pi/2)$ and the contour of a half disk with a smaller half disk removed at the origin. I have already show that the integrals over the half disks go to zero. Thus we have two lines left $z_1(t)=t, t\in[-R,-\varepsilon]$ and $z_1(t)=t,t \in [\varepsilon,R]$. Now let us calcualate the integral over $z_1$ we just get the integral we are trying to calcualte when we let $\varepsilon \to 0$ and $R \to 0$. We can move the limit inside the integral due to DCT(I think). Now over $z_2$ we get $\int_{-R}^{-\varepsilon}\frac{\sqrt{t} \ln(t)}{1+t^2}$ we now use the $u=-t$ substitution (Is it valid here?) to get $\int_{\varepsilon}^{R}\frac{\sqrt{-u} \ln(-u)}{1+u^2}=\int_{\varepsilon}^{R}\frac{\sqrt{u}i (\ln(|u|)+i\pi)}{1+u^2}$. Then we can see that the imaginary part of the residue will be equal to our desire integral. Thus we are done as the the residue is $\pi^2/2i(1+i)/\sqrt{2}$ so the result is $\pi^2/(2\sqrt{2})$