How to prove $\int_0^1 \frac1{1+x^2}\arctan\sqrt{\frac{1-x^2}2}d x=\pi^2/24$?

Question

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: $$2\pi i(-i\pi/8)=\pi^2/4??$$

Answer 1

Disclaimer: now I have a working real-analytic technique, but it is quite a tour-de-force.

We have: $$\begin{eqnarray*} I &=& \int_{0}^{1}\frac{1}{1+x^2}\arctan\sqrt{\frac{1-x^2}{2}}\,dx = \frac{1}{2}\int_{0}^{1}\frac{1}{(1+x)\sqrt{x}}\arctan\sqrt{\frac{1-x}{2}}\,dx\\&=&\int_{0}^{1/2}\frac{1}{(2-2x)\sqrt{1-2x}}\arctan\sqrt{x}\,dx=\int_{0}^{1/\sqrt{2}}\frac{x\arctan x}{(1-x^2)\sqrt{1-2x^2}}\,dx\end{eqnarray*}$$ and integrating by parts we get: $$ I = \sqrt{2}\int_{0}^{1}\frac{\arctan\sqrt{1-x^2}}{2+x^2}\,dx=\sqrt{2}\int_{0}^{\pi/2}\frac{\cos\theta\arctan\cos\theta}{3-\cos^2\theta}d\theta.\tag{1}$$ so: $$ I = \frac{1}{2\sqrt{2}}\int_{-\pi}^{\pi}\frac{\cos\theta\arctan\cos\theta}{3-\cos^2\theta}\,d\theta=\frac{1}{2\sqrt{2}}\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{-\pi}^{\pi}\frac{\cos^{2n+2}\theta}{3-\cos^2\theta}\,d\theta.\tag{2}$$ On the other hand, $$ I_n=\int_{-\pi}^{\pi}\frac{\cos^{2n+2}\theta}{3-\cos^2\theta}\,d\theta=-\int_{-\pi}^{\pi}\cos^{2n}\theta\,d\theta+3 I_{n-1}=-\frac{2\pi}{4^n}\binom{2n}{n}+3I_{n-1}\tag{3}$$ and $I_0=\pi\sqrt{\frac23}$. Now the plan is to solve recursion $(3)$ and compute the integral via $(2)$. $$ I = \frac{1}{2\sqrt{2}}\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\sum_{m\geq 1}\frac{2\pi}{4^{n+m} 3^m}\binom{2n+2m}{n+m}\tag{4}$$ leads to: $$ I = \frac{\pi}{\sqrt{2}}\sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^{m}}{3^m(2n+1)}\binom{-1/2}{n+m}=\frac{\pi}{\sqrt{2}}\int_{0}^{1}\sum_{n\geq 0}\sum_{m\geq 1}\frac{(-1)^{m}x^{2n}\,dx}{3^m}\binom{-1/2}{n+m}\tag{5}$$ but since: $$\sum_{n=0}^{s-1}(-1/3)^{s-n}(x^2)^n= \frac{1}{1+3x^2}\left((-1/3)^s-x^{2s}\right)$$ we have: $$ I = \frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{dx}{1+3x^2}\sum_{s=1}^{+\infty}\left((-1/3)^s-x^{2s}\right)\binom{-1/2}{s}\tag{6}$$ and finally: $$ I = \frac{\pi}{\sqrt{2}}\int_{0}^{1}\left(\sqrt{\frac{3}{2}}-\frac{1}{\sqrt{1+x^2}}\right)\frac{dx}{1+3x^2}\tag{7}$$ is easy to handle and leads to $\color{red}{\frac{\pi^2}{24}}$ as wanted, since: $$ \int\frac{dx}{(1+3x^2)\sqrt{1+x^2}}=\frac{1}{\sqrt{2}}\arctan\frac{x\sqrt{2}}{\sqrt{1+x^2}}.$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{1 \over 1 + x^{2}}\, \arctan\pars{\root{1 - x^{2} \over 2}}\,\dd x={\pi^{2} \over 24}:\ {\large ?}}$.

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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}

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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}

Answer 3

use the relation solved in this site $$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx=\frac{\pi^2}{24}$$ Put $$\tan{x}=u, dx=\frac{du}{1+u^2}$$ $$\sqrt{\frac{\cos 2x }{2 \cos^2 x}}=\sqrt\frac{1-u^2}{2}$$