Saturday, 25 February 2017

integration - Complex integral with Residues Theorem

I've been going crazy with this complex integral I have to estimate with the Residues Theorem. I'm obviously missing a sign or something else, but I fear I may be commiting a conceptual mistake.



0dxlogxx21



I choose to solve this integral by computing the following complex integral:



γlog2zz21dz=lim
+\int_{2\pi}^{0}d\theta i \epsilon e^{i\theta} \frac{\log^2 (\epsilon e^{i\theta})}{(\epsilon e^{i\theta})^2-1} +\int_{2\pi}^{0}d\theta i \epsilon e^{i\theta} \frac{\log^2 (1+\epsilon e^{i\theta})}{(1+\epsilon e^{i\theta})^2-1}] + \int_{\tilde \gamma}dz \frac{\log^2 z}{z^2-1}




Where \gamma is the "keyhole" path, and \tilde\gamma is the circle centered in z=0 with R radius. If R\rightarrow\infty, then:



=\int_{0}^{\infty} dz \frac{\log^2 (x)}{x^2-1}-\int_{0}^{\infty} dx \frac{(\log (x)+2\pi i)^2}{x^2-1}
=-4\pi i \int_{0}^{\infty}dx\frac{\log (x)}{x^2-1}+ 4 \pi^2 \int_{0}^{\infty} dx \frac{1}{x^2-1}=-4\pi i \int_{0}^{\infty}dx\frac{\log (x)}{x^2-1}



since \int_{0}^{\infty} dx \frac{1}{x^2-1}=0.



I can estimate the complex integral with the Residues theorem:



\int_{\gamma} \frac{\log^2 z}{z^2-1}dz=2\pi i Res[f(z), -1]=i \pi^3




and this means that:



\int_{0}^{\infty}dx\frac{\log (x)}{x^2-1}=-\frac{\pi^2}{4}



which is wrong, since the correct result should be \frac{\pi^2}{4}. Is there something wrong with my procedure?

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