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

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

Solutions Collecting From Web of "Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$"

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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$\ds{\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{3/2}}\,\dd x = 8\pi\ln\pars{2}:
\ {\large ?}}$

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}

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

$$$$


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

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
$$ I(1,1)=\frac{\partial I(\alpha,1)}{\partial\alpha}=2\pi\int_0^1\frac{\log(1+\frac{1}{\sqrt{\alpha}})}{\sqrt{\alpha}}d\alpha=8\pi\ln 2. $$