# How to find integral of $\int_0^\infty \frac{\ln ^2z} {1+z^2}\mathrm{d}z$?

How do I find the value of $$\int_{0}^{\infty} \frac{(\ln z)^2}{1+z^2}\mathrm{d}z$$ without using contour integration, - using the usual special functions, e.g., zeta/gamma/beta/etc.

Thank you,

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$\ln (z^2)$ or $(\ln z)^2$? –  Yoni Rozenshein Mar 18 '13 at 10:31
@YoniRozenshein edited –  catenspiel Mar 18 '13 at 10:47
@catenspiel $\ln(z)^2$ doesn't help much, write $\ln^2 z$ or $(\ln z)^2$ –  Cortizol Mar 18 '13 at 10:55

Here's another way to go: $$\begin{eqnarray*} \int_0^\infty dz\, \dfrac{\ln ^2z} {1+z^2} &=& \frac{d^2}{ds^2} \left. \int_0^\infty dz\, \dfrac{z^s} {1+z^2} \right|_{s=0} \\ &=& \frac{d^2}{ds^2} \left. \frac{\pi}{2} \sec\frac{\pi s}{2} \right|_{s=0} \\ &=& \frac{\pi^3}{8}. \end{eqnarray*}$$ The integral $\int_0^\infty dz\, z^s/(1+z^2)$ can be handled with the beta function. See some of the answers here, for example.

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You can integrate by substitution with $z = e^u$. This yields $$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \int_{-\infty}^{+\infty}\frac{u^2e^u}{1+e^{2u}}du = 2\int_0^\infty \frac{u^2e^{-u}}{1+e^{-2u}}du$$

Now, expand the series: $$\frac{u^2e^{-u}}{1+e^{-2u}} = \sum_{n=0}^\infty (-1)^n u^2e^{-u(2n+1)}$$

Interverting the $\int$ and $\sum$ (use Fubini's theorem), we have $$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = 2\Gamma(3)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3} = 4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}$$

The last sum can be computed using Fourier series, yielding $$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \frac{\pi^3}{8}$$

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I didn't see that - my apologies (+1) –  Ron Gordon Mar 18 '13 at 12:35