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$$I=\int_{0}^{\infty}{\operatorname{arccot}^2(\sqrt{1+x})\over 1+x}\mathrm dx$$

$$J=4\int_{0}^{\infty}{\operatorname{arccot}^2[(1+x)^2 ]\over 1+x}\mathrm dx$$

Q1: Show that $I=J\color{red}?$

Q2: How can we prove that $$I=\pi C-{7\over 4}\zeta(3)\color{red}?$$

Where $C$ is Catalan's constant


$u=\sqrt{1+x}\implies 2udu=dx$

$$I=2\int_{1}^{\infty}{\operatorname{arccot}^2(u)\over u}\mathrm du$$

$v=(1+x)^2\implies dv=2(1+x)dx$

$$J=2\int_{1}^{\infty}{\operatorname{arccot}^2(v)\over v}\mathrm dv$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty}{\mrm{arccot}^{2}\pars{\root{1 + x}} \over 1 + x}\,\dd x \,\,\,\stackrel{2\,\mrm{arccot}\pars{\!\!\root{1 + x}\!\!}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x \\[5mm] = &\ \left.{1 \over 2}\,\Re\int_{x = 0}^{x = \pi/2} {-\ln^{2}\pars{z} \over \pars{1 - z^{2}}\ic/\pars{2z}}\,{\dd z \over \ic z} \right\vert_{\ z\ =\ \exp\pars{\ic x}} = \left.\Re\int_{x = 0}^{x = \pi/2} {\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z\, \right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ -\,\Re\int_{1}^{0} {\bracks{\ln\pars{y} + \pi\ic/2}^{\,2} \over 1 + y^{2}}\,\ic\,\dd y - \Re\int_{0}^{1} {\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x \\[5mm] = &\ -\pi\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x - \int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x \\[5mm] = &\ -\pi\ \underbrace{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x} _{\ds{\substack{\ds{=\ {\large 0}}}\ \mbox{because}\\ \mbox{it}\ changes\ its\ sign\\ \mbox{under}\ x\ \to\ 1/x}}\ +\ \pi\ \underbrace{\int_{1}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x} _{\ds{\substack{{\large =\ G}\\[1mm]\color{#f00}{\Large\S}:\ Catalan\ Constant}}}\ -\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x} _{\ds{{7 \over 4}\,\zeta\pars{3}}}\ \\[5mm] = &\ \bbx{\pi\,G - {7 \over 4}\,\zeta\pars{3}} \end{align}

$\ds{\color{#f00}{\large\S}}$: See Catalan Constant Page.

The last integral is evaluated as follows: \begin{align} \int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x & = {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x + {1 \over 2}\int_{-1}^{0}{\ln^{2}\pars{-x} \over 1 - x}\,\dd x \\[5mm] & \stackrel{\mbox{IBP}}{=}\,\,\, -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x -\int_{-1}^{0}\mrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x \\[5mm] & \stackrel{\mbox{IBP}}{=}\,\,\, \int_{0}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x + \int_{-1}^{0}\mrm{Li}_{3}'\pars{x}\,\dd x = \mrm{Li}_{3}\pars{1} - \mrm{Li}_{3}\pars{-1} \\[5mm] & = \zeta\pars{3} - \bracks{-\,{3 \over 4}\,\zeta\pars{3}} = {7 \over 4}\,\zeta\pars{3} \end{align}

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You have already showed that $I=J$: they are associated with the same integral, i.e.

$$ I=2\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du = 4 \int_{0}^{\pi/4}\frac{t^2}{\sin(2t)}\,dt=\frac{1}{2}\int_{0}^{\pi/2}\frac{z^2}{\sin z}\,dz\tag{1}$$ The claim now follows from the Eisenstein series for $\frac{1}{\sin z}$:

$$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right) \tag{2}$$ by multiplying both sides by $z^2$, performing integration over $(0,\pi/2)$ and rearranging.
An efficient alternative is to use a step of integration by parts to get $I=\int_{0}^{\pi/2}z\log\tan\frac{z}{2}\,dz$, then exploit the Fourier series of $\log\sin$ and $\log\cos$.

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