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I am trying to evaluate this integral. $$ I=\int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2 $$ Note $$ \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}, \ |x| < 1. $$ I was trying to do use this series expansion but wasn't sure how to go about it because of the square of the logarithm. And also it seems than we then will have $I\propto \int_0^\infty x^{n-3/2}dx$ which will diverge. Thanks

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    $\begingroup$ I'd guess that the best way to proceed is to find an appropriate contour in the complex plane and integrate using residues. $\endgroup$ – user61527 Mar 29 '14 at 2:19
  • $\begingroup$ @T.Bongers I am looking for a solution, thanks though. $\endgroup$ – Jeff Faraci Mar 29 '14 at 2:25
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Integrating by parts,

$$ \begin{align} \int_{0}^{\infty} \frac{\ln^{2}(1+x)}{x^{3/2}} \ dx &= - \frac{2 \ln^{2}(1+x)}{\sqrt{x}} \Bigg|^{\infty}_{0} + 4 \int_{0}^{\infty} \frac{\ln (1+x)}{(1+x) \sqrt{x}} \ dx \\ &=4 \int_{0}^{\infty} \frac{\ln (1+x)}{(1+x) \sqrt{x}} \ dx . \end{align} $$

Now let $x = u^{2}$.

Then

$$\begin{align} \int_{0}^{\infty} \frac{\ln^{2}(1+x)}{x^{3/2}} \ dx &= 8 \int_{0}^{\infty} \frac{\ln (1+u^{2})}{1+u^{2}} \ du \\ &= 8 (\pi \ln 2) \tag{1} \\ &= 8 \pi \ln 2. \end{align}$$

$(1)$ Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$

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  • $\begingroup$ Thanks, very clear. No questions. +10 $\endgroup$ – Jeff Faraci Mar 29 '14 at 3:04
  • $\begingroup$ Your editing makes this problem active and also makes me reminisce that this is the first problem I answered on Math SE. Thanks. $\ddot\smile$ $\endgroup$ – Tunk-Fey Aug 8 '14 at 16:07
  • $\begingroup$ @Tunk-Fey Since I just linked to this question, I thought I'd improve the formatting of my answer. Sometimes I wish you could make minor edits to an answer without bumping the question. $\endgroup$ – Random Variable Aug 8 '14 at 16:33
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{3/2}}\,\dd x = 8\pi\ln\pars{2}: \ {\large ?}}$

With $\ds{t \equiv {1 \over x + 1}\quad\iff\quad x = {1 \over t} - 1}$: \begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{3/2}}\,\dd x}= -2\int_{x = 0}^{x \to \infty}\ln^{2}\pars{1 + x}\,\dd\pars{x^{-1/2}} \\[3mm]&=2\int_{0}^{\infty}x^{-1/2}\bracks{2\ln\pars{1 + x}\,{1 \over 1 + x}}\,\dd x =4\int_{1}^{0}\pars{1 - t \over t}^{-1/2}\ln\pars{1 \over t}t \pars{-\,{\dd t \over t^{2}}} \\[3mm]&=-4\int_{0}^{1}\ln\pars{t}t^{-1/2}\pars{1 - t}^{-1/2}\,\dd t =-4\lim_{\mu \to -1/2}\totald{}{\mu}\int_{0}^{1}t^{\mu}\pars{1 - t}^{-1/2}\,\dd t \\[3mm]&=-4\lim_{\mu \to -1/2}\totald{{\rm B}\pars{\mu + 1,1/2}}{\mu} -4\lim_{\mu \to -1/2}\totald{}{\mu} \bracks{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}} \\[3mm]&=-4\Gamma\pars{\half}\lim_{\mu \to -1/2}\braces{% {\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + 3/2}}\,\bracks{\Psi\pars{\mu + 1} - \Psi\pars{\mu + {3 \over 2}}}} \\[3mm]&=-4\Gamma^{2}\pars{\half}\, {\Psi\pars{1/2} - \Psi\pars{1} \over \Gamma\pars{1}} =-4\pars{\root{\pi}}^{2}\,{\bracks{-2\ln\pars{2} - \gamma} -\pars{-\gamma} \over 1} \\[3mm]&=\color{#00f}{\large 8\pi\ln\pars{2}} \end{align}

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  • $\begingroup$ You always have nice solutions with $\partial_\mu$ or $\partial_\nu$ with these special functions. Very helpful to me....Thanks a lot!!!! $\endgroup$ – Jeff Faraci Mar 29 '14 at 5:13
  • $\begingroup$ @Jeff They are very useful. Thanks. $\endgroup$ – Felix Marin Mar 29 '14 at 7:51
  • $\begingroup$ You are an expert on special functions! $\endgroup$ – xpaul Aug 8 '14 at 16:11
  • $\begingroup$ @xpaul Thanks. They are quite useful. For that reasons people call them 'specials'. $\endgroup$ – Felix Marin Aug 8 '14 at 17:58
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Using Richard Feynman's favorite method, the method of differentiation under the integral sign. $$ \begin{align} I(\alpha)&=\int_0^\infty\frac{\ln^2(1+\alpha x)}{x^{\frac{3}{2}}}dx\\ \frac{dI(\alpha)}{d\alpha}&=\int_0^\infty\frac{2x\ln(1+\alpha x)}{x^{\frac{3}{2}}(1+\alpha x)}dx\\ I'(\alpha)&=2\int_0^\infty\frac{\ln(1+\alpha x)}{\sqrt{x}(1+\alpha x)}dx. \end{align} $$ Let $\,x=t^2\;\Rightarrow\;dx=2t\,dt$, then $$ \begin{align} I'(\alpha)&=2\int_0^\infty\frac{\ln(1+\alpha t^2)}{t(1+\alpha t^2)}\cdot2t\,dt\\ &=4\int_0^\infty\frac{\ln(1+\alpha t^2)}{(1+\alpha t^2)}dt. \end{align} $$ To solve the integral part, again we use the method of differentiation under the integral sign. $$ \begin{align} I(\beta)&=\int_0^\infty\frac{\ln(1+\alpha\beta t^2)}{(1+\alpha t^2)}dt\\ \frac{dI(\beta)}{d\beta}&=\int_0^\infty\frac{\alpha t^2}{(1+\alpha\beta t^2)(1+\alpha t^2)}dt\\ I'(\beta)&=\int_0^\infty\left[\frac{1}{(\beta-1)(1+\alpha t^2)}-\frac{1}{(\beta-1)(1+\alpha\beta t^2)}\right]dt\\ &=\frac{1}{\beta-1}\int_0^\infty\left[\frac{1}{1+\alpha t^2}-\frac{1}{1+\alpha\beta t^2}\right]dt. \end{align} $$ Note that $$ \int_0^\infty\frac{1}{1+k y^2}dy=\frac{\pi}{2\sqrt{k}}. $$ Therefore $$ \begin{align} I'(\beta)&=\frac{1}{\beta-1}\left(\frac{\pi}{2\sqrt{\alpha}}-\frac{\pi}{2\sqrt{\alpha\beta}}\right)\\ &=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{\sqrt{\beta}-1}{\sqrt{\beta}(\beta-1)}\right)\\ &=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{\left(\sqrt{\beta}-1\right)}{\sqrt{\beta}(\beta-1)}\cdot\frac{\left(\sqrt{\beta}+1\right)}{\left(\sqrt{\beta}+1\right)}\right)\\ &=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{1}{\sqrt{\beta}\left(\sqrt{\beta}+1\right)}\right)\\ \frac{dI(\beta)}{d\beta}&=\frac{\pi}{2\sqrt{\alpha}}\left(\frac{1}{\sqrt{\beta}}-\frac{1}{\sqrt{\beta}+1}\right)\\ I(\beta)&=\frac{\pi}{2\sqrt{\alpha}}\int\left(\frac{1}{\sqrt{\beta}}-\frac{1}{\sqrt{\beta}+1}\right)d\beta\\ &=\frac{\pi}{2\sqrt{\alpha}}\left(2\sqrt{\beta}-2\sqrt{\beta}+2\ln\left(\sqrt{\beta}+1\right)+\text{C}_1\right)\\ &=\frac{\pi}{2\sqrt{\alpha}}\left(2\ln\left(\sqrt{\beta}+1\right)+\text{C}_1\right).\\ \end{align} $$ For $\beta=0$ implying $I_\beta(0)=0$, then $\text{C}_1=0$ and $$ I_\beta(1)=\int_0^\infty\frac{\ln(1+\alpha t^2)}{(1+\alpha t^2)}dt=\frac{\pi\ln 2}{\sqrt{\alpha}}. $$ Now, plug in $I_\beta(1)$ to $I'(\alpha)$. $$ \begin{align} I'(\alpha)&=4\cdot\frac{\pi\ln 2}{\sqrt{\alpha}}\\ \frac{dI(\alpha)}{d\alpha}&=4\pi\ln 2\cdot\alpha^{-\frac{1}{2}}\\ I(\alpha)&=4\pi\ln 2\int\alpha^{-\frac{1}{2}}\,d\alpha\\ &=(4\pi\ln 2)\left(2\alpha^{\frac{1}{2}}+\text{C}_2\right) \end{align} $$ For $\alpha=0$ implying $I_\alpha(0)=0$, then $\text{C}_2=0$. Thus $$ I_\alpha(1)=\int_0^\infty\frac{\ln^2(1+x)}{x^{\frac{3}{2}}}dx=\boxed{\color{blue}{\large8\pi\ln 2}} $$

$$$$


$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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  • $\begingroup$ clever solution. Thanks for the new approach. $\endgroup$ – Jeff Faraci Apr 4 '14 at 0:53
  • $\begingroup$ @Jeff You're welcome. Unfortunately, I failed to earn my first bounty. $\endgroup$ – Tunk-Fey Apr 4 '14 at 7:46
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There is another way to solve. Let $$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x)}{x^{\frac{3}{2}}}dx. $$ Note $I=I(1,1)$. It is easy to see that $$ \frac{\partial^2I(\alpha,\beta)}{\partial\alpha\partial\beta}=\int_0^\infty\frac{\sqrt{x}}{(1+\alpha x)(1+\beta x)}dx=\frac{\pi}{\alpha\sqrt{\beta}+\sqrt{\alpha}\beta} $$ and hence $$ \frac{\partial I(\alpha,1)}{\partial\alpha}=\int_0^1 \frac{\pi}{\alpha\sqrt{\beta}+\sqrt{\alpha}\beta}d\beta=2\pi\frac{\log(1+\frac{1}{\sqrt{\alpha}})}{\sqrt{\alpha}} $$ and $$ I(1,1)=\frac{\partial I(\alpha,1)}{\partial\alpha}=2\pi\int_0^1\frac{\log(1+\frac{1}{\sqrt{\alpha}})}{\sqrt{\alpha}}d\alpha=8\pi\ln 2. $$

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