Contour int.

1. Aug 8, 2006

dextercioby

Is it just me, or i can't get the residues right...?

$$\int_{0}^{\infty} \frac{\ln^{2}z}{1+z^{2}} \ dz$$

I get $\frac{\pi^{3}}{2}$ , but the result is $$\frac{\pi^{3}}{8} [/itex]. I make a substitution $z=e^{t}$. And then convert to a contour integral closing it in the upper half-plane, where i have the poles [tex] i\frac{\pi}{2} + k\pi$$ k in N.

Daniel.

2. Aug 8, 2006

shmoe

I don't see that the contour over the upper semi-circle will be going to zero. The sum of the residues in the upper half plane is divergent no?

You can take a different approach, instead consider

$$\int_{-\infty}^{\infty}\frac{\log^{2}{z}}{1+z^2}dz$$

and look at a contour from -R to R with a small semicircle of radius r at the origin in the upper half plane and a large one of radius R in the upper half plane as well. As r->0 and R->infinity, these will contribute 0. From (-infinity,0) replace log(z) with log|z|+Pi*i and expand the square. One of the integrals is the one you are after, one can be evaluated as an arctan, one will essentially be the integral of log(z)/(1+z^2) from 0 to infinity (this last integral is zero, use the same contour to prove this). Finally you enclose just one simple pole at z=i.