Find $\int_0^\infty \frac{\ln ^2z} {1+z^2}{d}z$

Question

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

Thank you.

Answer 1

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 $$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \frac{\pi^3}{8}$$

Answer 2

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.

Answer 3

With $ \int_0^\infty \frac{2y\ln y }{(y^2+z^2)(y^2+1)}{dy}= \frac{\ln^2 z}{z^2-1}$ \begin{align} &\int_0^\infty\frac{\ln^2 z}{1+z^2}dz\\ =&\ \int_0^\infty \frac1{1+z^2} \left(\int_0^\infty\frac{2(z^2-1)y\ln y}{(y^2+z^2)(y^2+1)}\ dy \right) dz\\ =& \int_0^\infty\int_0^\infty \frac{2y\ln y}{(y^2+z^2)(y^2-1)}-\frac{4y\ln y}{(1+z^2)(y^4-1)}\ dz \ dy\\ =& \ \pi\int_0^\infty \frac{\ln y}{y^2-1} {dy}- 2\pi\int_0^\infty \frac{y\ln y}{y^4-1}\overset{y^2\to y} {dy}\\ =&\ \frac\pi2 \int_0^\infty \frac{\ln y}{y^2-1}\ dy = \frac\pi4 \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy\\ = &\ \frac{\pi^2}4\int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^3}{8} \end{align}

Answer 4

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\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}

Answer 5

\begin{align}J&=\int_0^\infty \frac{\ln^2 x}{1+x^2}dx\\ K&=\int_0^\infty\int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}dxdy\\ &\overset{u(x)=xy}=\int_0^\infty\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}dudy\\ &=\frac{1}{2}\int_0^\infty \ln^2 u\left[\frac{\ln\left(\frac{1+y^2}{u^2+y^2}\right)}{u^2-1}\right]_{y=0}^{y=\infty}du\\ &=\int_0^\infty \frac{\ln^3 u}{u^2-1}du=\int_0^1 \frac{\ln^3 u}{u^2-1}du+\underbrace{\int_1^\infty \frac{\ln^3 u}{u^2-1}du}_{z=\frac{1}{u}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}du\\ &=2\int_0^1 \frac{\ln^3 u}{u-1}du-2\underbrace{\int_0^1 \frac{u\ln^3 u}{u^2-1}du}_{z=u^2}=2\int_0^1 \frac{\ln^3 u}{u-1}du-\frac{1}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz=\frac{15}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz\\ &=\frac{15}{8}\times 6\zeta(4)=\boxed{\frac{\pi^4}{8}} \end{align} On the other hand, \begin{align}K&=\int_0^\infty\int_0^\infty\frac{\ln^2 x}{(1+x^2)(1+y^2)}dxdy+\int_0^\infty\int_0^\infty\frac{\ln^2 y}{(1+x^2)(1+y^2)}dxdy+\\& 2\underbrace{\int_0^\infty\int_0^\infty\frac{\ln x\ln y}{(1+x^2)(1+y^2)}dxdy}_{=0}\\ &=2J\underbrace{\int_0^\infty \frac{1}{1+x^2}dx}_{=\frac{\pi}{2}}=\boxed{\pi J} \end{align} Therefore,

\begin{align}\boxed{J=\frac{\pi^3}{8}}\end{align}

NB: I assume, \begin{align}\int_0^1 \frac{\log^3 x}{x-1}dx=6\zeta(4)=\frac{\pi^4}{15}\end{align}