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[tex] \int_{0}^{\infty} \frac{\ln^{2}z}{1+z^{2}} \ dz [/tex]

I get [itex]\frac{\pi^{3}}{2} [/itex] , but the result is [tex]\frac{\pi^{3}}{8} [/itex].

I make a substitution [itex] z=e^{t} [/itex]. 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 [/tex] k in N.

So please help me.

Daniel.