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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
    Commented Mar 29, 2014 at 2:19
  • $\begingroup$ @T.Bongers I am looking for a solution, thanks though. $\endgroup$ Commented Mar 29, 2014 at 2:25

4 Answers 4

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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 $\endgroup$ Commented Mar 29, 2014 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
    Commented Aug 8, 2014 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. $\endgroup$ Commented Aug 8, 2014 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!!!! $\endgroup$ Commented Mar 29, 2014 at 5:13
  • $\begingroup$ @Jeff They are very useful. Thanks. $\endgroup$ Commented Mar 29, 2014 at 7:51
  • $\begingroup$ You are an expert on special functions! $\endgroup$
    – xpaul
    Commented Aug 8, 2014 at 16:11
  • $\begingroup$ @xpaul Thanks. They are quite useful. For that reasons people call them 'specials'. $\endgroup$ Commented Aug 8, 2014 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. $\endgroup$ Commented Apr 4, 2014 at 0:53
  • $\begingroup$ @Jeff You're welcome. Unfortunately, I failed to earn my first bounty. $\endgroup$
    – Tunk-Fey
    Commented Apr 4, 2014 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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