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.
$\int_{0}^{\infty}dx \frac{\log x}{x^2-1}$
I choose to solve this integral by computing the following complex integral:
$\int_{\gamma} \frac{\log^2 z}{z^2-1}dz=\lim_{\epsilon\rightarrow 0}[ \int_{0}^{R} dx \frac{\log^2 (x+i\epsilon)}{(x+i\epsilon)^2-1}+\int_{R}^{0} dx \frac{\log^2 (x-i\epsilon)}{(x-i\epsilon)^2-1}+ $
$+\int_{2\pi}^{0}d\theta i \epsilon e^{i\theta} \frac{\log^2 (\epsilon e^{i\theta})}{(\epsilon e^{i\theta})^2-1} +\int_{2\pi}^{0}d\theta i \epsilon e^{i\theta} \frac{\log^2 (1+\epsilon e^{i\theta})}{(1+\epsilon e^{i\theta})^2-1}] + \int_{\tilde \gamma}dz \frac{\log^2 z}{z^2-1}$
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?
No comments:
Post a Comment