TITLE: How to prove $\int_0^1 \frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}d x=\pi^2/24$?
QUESTION [8 upvotes]: Since I'm stuck at this final step of the solution here. I wished to try contour integral, taking the contour a quadrant with centre ($0$) and two finite end points of arc at $(1),(i)$:
Then:
$$\operatorname{Res}\limits_{x=i}\frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}=-i\pi/8$$
But then:
$$2\pi i(-i\pi/8)=\pi^2/4??$$

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}{1 \over 1 + x^{2}}\,
     \arctan\pars{\root{1 - x^{2} \over 2}}\,\dd x={\pi^{2}  \over 24}:\
     {\large ?}}$.

In order to to 'remove' the ${\tt sqrt}$'s from the ${\tt arctan}$ function argument it's convenient to integrate by parts:
\begin{align}&\color{#66f}{\large
\int_{0}^{1}{1 \over 1 + x^{2}}\,\arctan\pars{\root{1 - x^{2} \over 2}}\,\dd x}
\\[5mm]&=\int_{x\ =\ 0}^{x\ =\ 1}
\arctan\pars{\root{1 - x^{2} \over 2}}\,\dd\arctan\pars{x}
=-\root{2}\int_{0}^{1}
{x\arctan\pars{x} \over \root{1 - x^{2}}\pars{x^{2} - 3}}\,\dd x
\end{align}

We 'avoid' the $\ds{\arctan\pars{z}}$ branch cuts by using an integral representation of it:
\begin{align}&\color{#66f}{\large
\int_{0}^{1}{1 \over 1 + x^{2}}\,\arctan\pars{\root{1 - x^{2} \over 2}}\,\dd x}
\\[5mm]&=-\root{2}\int_{0}^{1}
{x \over \root{1 - x^{2}}\pars{x^{2} - 3}}\int_{0}^{1}{x \over x^{2}t^{2} + 1}
\,\dd t\,\dd x
\\[5mm]&=-\root{2}\int_{0}^{1}\ \overbrace{\int_{0}^{1}
{x^{2} \over \root{1 - x^{2}}\pars{x^{2} - 3}\pars{x^{2} + t^{-2}}}
\,\dd x}^{\ds{\dsc{x}=\dsc{\cos\pars{\theta}}\ \imp\ \dsc{\theta}=\dsc{\arccos\pars{x}}}}\ \,{\dd t \over t^{2}}
\\[5mm]&=\root{2}\int_{0}^{1}\int_{0}^{\pi/2}{\cos^{2}\pars{\theta}\over
\pars{3 - \cos^{2}\pars{\theta}}\pars{t^{-2} + \cos^{2}\pars{\theta}}}
\,\dd\theta \,{\dd t \over t^{2}}
\\[5mm]&={\root{2} \over 3}\int_{0}^{1}\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\over
\pars{\sec^{2}\pars{\theta} - 1/3}\bracks{\sec^{2}\pars{\theta} + t^{2}}}
\,\dd\theta\,\dd t
\\[5mm]&={\root{2} \over 3}\int_{0}^{1}\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta \over
\pars{\tan^{2}\pars{\theta} + 2/3}\bracks{\tan^{2}\pars{\theta} + 1 + t^{2}}}
\,\dd t
\\[5mm]&={\root{2} \over 3}\int_{0}^{1}\bracks{%
\underbrace{\int_{0}^{\pi/2}
{\sec^{2}\pars{\theta}\,\dd\theta \over \tan^{2}\pars{\theta} + 2/3}}
_{\dsc{{\root{6} \over 4}\,\pi}}\ -\
\underbrace{\int_{0}^{\pi/2}
{\sec^{2}\pars{\theta}\,\dd\theta \over \tan^{2}\pars{\theta} + 1 + t^{2}}}
_{\dsc{\pi \over 2\root{1 + t^{2}}}}}
\,{\dd t \over t^{2} + 1/3}
\\[5mm]&={\root{3} \over 6}\,\pi\
\overbrace{\int_{0}^{1}{\dd t \over t^{2} + 1/3}}^{\dsc{\pi \over \root{3}}}\ -\
{\root{2} \over 6}\,\pi\ \overbrace{%
\dsc{\int_{0}^{1}{\dd t \over \root{t^{2} + 1}\pars{t^{2} + 1/3}}}}
^{\dsc{3\pi \over 4\root{2}}}\ =\ {\pi^{2} \over 6} - {\pi^{2} \over 8}
\\[5mm]&=\color{#66f}{\large{\pi^{2} \over 24}}
\approx {\tt 0.4112}
\end{align}

The last integral can be evaluated as follows:

\begin{align}&\overbrace{%
\dsc{\int_{0}^{1}{\dd t \over \root{t^{2} + 1}\pars{t^{2} + 1/3}}}}
^{\ds{t\ \mapsto\ {1 \over t}}}\ =\
3\int_{\infty}^{1}{-\dd t/t^{2} \over \root{1/t^{2} + 1}\pars{3/t^{2} + 1}}
=3\int_{1}^{\infty}{t\,\dd t \over \root{t^{2} + 1}\pars{t^{2} + 3}}
\\[5mm]&={3 \over 2}\int_{1}^{\infty}{\dd t \over \root{t + 1}\pars{t + 3}}\ =\
{3 \over 2}\ \overbrace{\int_{2}^{\infty}{\dd t \over \root{t}\pars{t + 2}}}
^{\ds{\dsc{t}\ \mapsto\ \dsc{t^{2}}}}
=3\ \overbrace{\int_{\root{2}}^{\infty}{\dd t \over t^{2} + 2}}
^{\dsc{\pi \over 4\root{2}}}\ = \ \dsc{3\pi \over 4\root{2}}
\end{align}