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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

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

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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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} &\color{#f00}{\int_{0}^{\infty}{\ln^{2}\pars{z} \over 1 + z^{2}}\,\dd z} = \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}{z^{\mu} \over 1 + z^{2}}\,\dd z \end{align}

With $\ds{z \equiv \pars{{1 \over t} - 1}^{1/2}}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\ln^{2}\pars{z} \over 1 + z^{2}}\,\dd z} = \half\,\lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{0}^{1}t^{-\mu/2 - 1/2}\pars{1 - t}^{\mu/2 - 1/2}\,\dd t \\[3mm] = &\ \half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{% \Gamma\pars{-\mu/2 + 1/2}\Gamma\pars{\mu/2 + 1/2} \over \Gamma\pars{1}} = \half\,\ \overbrace{% \lim_{\mu \to 0}\partiald[2]{\bracks{\pi\csc\pars{\pi\bracks{\mu + 1}/2}}}{\mu}} ^{\ds{=\ {\pi^{3} \over 4}}} = \color{#f00}{{\pi^{3} \over 8}} \end{align}

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