Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$

Solution 1:

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

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{3/2}}\,\dd x = 8\pi\ln\pars{2}: \ {\large ?}}$

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}

Solution 3:

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 $$ I_\alpha(1)=\int_0^\infty\frac{\ln^2(1+x)}{x^{\frac{3}{2}}}dx=\boxed{\color{blue}{\large8\pi\ln 2}} $$

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