On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$

Question

I'm having a difficult time evaluating the integral

$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$

This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \arctan \sqrt{\frac{1-\tan^2 x}{2}} \, \mathrm{d}x$;

\begin{align*} \require{cancel.js} \int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 t}{2}}\, \mathrm{d}t &\overset{1-\tan^2 t \mapsto 2t^2}{=\! =\! =\! =\! =\! =\!=\!=\!} \int_{0}^{\sqrt{2}/2} \frac{t \arctan t}{\sqrt{1-2t^2} \left ( 1-t^2 \right )} \, \mathrm{d}t \\ &=\cancelto{0}{\left [ - \arctan \sqrt{1-2t^2} \arctan t \right ]_0^{\sqrt{2}/2}} + \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t \end{align*}

My main guess is that differentiation under the integral sign is the way to go here. Any ideas?

Answer 1

\begin{align}J&=\int_0^{\frac{1}{\sqrt{2}}} \frac{\arctan\left(\sqrt{1-2x^2}\right)}{1+x^2}\,dx\\ &\overset{x=\frac{1}{\sqrt{2}}\sin u}=\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{1+\frac{1}{2}\sin^2 u}\,du\\ &=\sqrt{2}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{2+\sin^2 u}\,du\\ &=\left[\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\arctan(\cos u)\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\ &=\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\ &=\int_0^{\frac{\pi}{2}}\int_0^{\frac{1}{\sqrt{2}}}\left(\frac{\sin^2 u}{(1+\cos^2 u)(1+a^2\sin^2 u)}\,da\right)\,du\\ &=\int_0^{\frac{1}{\sqrt{2}}}\left[\frac{\sqrt{2}\arctan\left(\frac{1}{\sqrt{2}}\tan u\right)}{2a^2+1}-\frac{\arctan\left(\sqrt{1+a^2}\tan u\right)}{(2a^2+1)\sqrt{1+a^2}}\right]_{u=0}^{u=\frac{\pi}{2}}\,da\\ &=\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{\sqrt{2}}{2a^2+1}\,da-\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{1}{(2a^2+1)\sqrt{1+a^2}}\,da\\ &=\frac{\pi}{2}\Big[\arctan\left(\sqrt{2}a\right)\Big]_0^{\frac{1}{\sqrt{2}}}-\frac{\pi}{2}\left[\arctan\left(\frac{a}{\sqrt{1+a^2}}\right)\right]_0^{\frac{1}{\sqrt{2}}}\\ &=\frac{\pi}{2}\times \frac{\pi}{4}-\frac{\pi}{2}\times \frac{\pi}{6}\\ &=\boxed{\frac{\pi^2}{24}} \end{align}

Answer 2

This answer is based on Feynman's trick. Put \begin{equation*} I(a) = \int_{0}^{\pi/4}\arctan\left(a\sqrt{\dfrac{1-\tan^2 x}{2}}\right)\, dx . \end{equation*} Then \begin{gather*} I'(a) = \int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{1-\tan^2 x}{2}}\cdot \sqrt{\dfrac{1-\tan^2 x}{2}} \, dx = \\[2ex]\int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{\cos 2x}{1+\cos 2x}}\cdot \sqrt{\dfrac{\cos 2x}{1+\cos 2x}} \, dx = [y=\cos 2x]\\[2ex] = \dfrac{1}{2}\int_{0}^{1}\dfrac{1}{1+(a^2+1)y}\cdot\sqrt{\dfrac{y}{1-y}}\,dy= \left[z=\sqrt{\dfrac{y}{1-y}}\right] =\\[2ex] \dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{z^2}{(1+(a^2+2)z^2)(z^2+1)}\, dz = [\mbox{ residue calculus }]=\\[2ex] \dfrac{\pi}{2}\left(\dfrac{1}{a^2+1}-\dfrac{1}{(a^2+1)\sqrt{a^2+2}}\right) \end{gather*} Finally we get \begin{gather*} I(1)=I(1)-I(0)=\int_{0}^{1}I'(a)\, da =\dfrac{\pi}{2}\left[\arctan a -\arctan\dfrac{a}{\sqrt{a^2+2}}\right]_{0}^{1} =\dfrac{\pi^2}{24}. \end{gather*}

Answer 3

Following/Copying the answer over there we have that:

\begin{align*} \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t &= \int_{0}^{1} \frac{1}{1+x^2} \arctan \sqrt{\frac{1-x^2}{2}} \, \mathrm{d}x \\ &=-\sqrt{2} \int_{0}^{1} \frac{x \arctan x}{\sqrt{1-x^2} \left ( 3-x^2 \right )} \, \mathrm{d}x\\ &=-\sqrt{2} \int_{0}^{1}\frac{x}{\sqrt{1-x^2}\left ( 3-x^2 \right )} \int_{0}^{1} \frac{x}{1+x^2t^2} \, \mathrm{d}t \, \mathrm{d}x \\ &= -\sqrt{2} \int_{0}^{1} \int_{0}^{1} \frac{x^2}{\sqrt{1-x^2}\left ( 3-x^2 \right ) \left ( x^2+ \frac{1}{t^2} \right )} \frac{1}{t^2} \, \mathrm{d}x \, \mathrm{d}t\\ &\!\!\!\!\!\overset{x=\cos \theta}{=\! =\! =\! =\!} \sqrt{2} \int_{0}^{1} \int_{0}^{\pi/2} \frac{\cos^2 \theta}{\left ( 3 - \cos^2 \theta \right )\left ( \cos^2 \theta + \frac{1}{t^2} \right )} \, \mathrm{d}\theta \; \frac{\mathrm{d}t}{t^2} \\ &= \frac{\sqrt{2}}{3} \int_{0}^{1} \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\left ( \sec^2 \theta - \frac{1}{3} \right ) \left ( t^2 + \sec^2 \theta \right )} \, \mathrm{d} \theta \, \mathrm{d}t \\ &=\frac{\sqrt{2}}{3} \int_{0}^{1} \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\left ( \tan^2 \theta + \frac{2}{3} \right )\left ( \tan^2 \theta + 1 + t^2 \right )} \, \mathrm{d}\theta \, \mathrm{d}t \\ &=\frac{\sqrt{2}}{3} \int_{0}^{1} \left ( \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + \frac{2}{3}} \, \mathrm{d} \theta - \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + 1 + t^2} \, \mathrm{d}\theta \right ) \frac{\mathrm{d}t}{t^2+\frac{1}{3}} \end{align*}

For the remaining integrals we have:

\begin{align*} \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + \frac{2}{3}} \, \mathrm{d}\theta &\overset{u =\tan \theta}{=\! =\! =\! =\!} \int_{0}^{\infty} \frac{\mathrm{d}u}{u^2 + \frac{2}{3}} \\ &=\left [ \frac{\sqrt{3}}{2} \arctan \sqrt{\frac{3}{2}}u \right ]_0^\infty \\ &= \frac{\sqrt{3}\pi}{2\sqrt{2}} \\ &= \frac{\pi \sqrt{6}}{4} \end{align*}

and similarly

$$\int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + 1 + t^2} \, \mathrm{d}\theta = \frac{\pi}{2 \sqrt{1+t^2}}$$

Thus,

\begin{align*} \int_{0}^{1} \left ( \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + \frac{2}{3}} \, \mathrm{d} \theta - \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\tan^2 \theta + 1 + t^2} \, \mathrm{d}\theta \right ) \frac{\mathrm{d}t}{t^2+\frac{1}{3}} &= \frac{\pi \sqrt{6}}{4}\int_{0}^{1} \frac{\mathrm{d}t}{t^2 + \frac{1}{3}} - \frac{\pi}{2}\int_{0}^{1} \frac{\mathrm{d}t}{\sqrt{1+t^2} \left ( t^2 + \frac{1}{3} \right )} \\ &\!\!\!\!\!\overset{t \mapsto 1/t}{=\! =\! =\! =\! =\!}\frac{\pi \sqrt{6}}{4} \frac{\pi}{\sqrt{3}} - \frac{3 \pi}{2} \int_{1}^{\infty} \frac{t}{\sqrt{t^2+1} \left ( t^2+3 \right )} \, \mathrm{d}t \\ &\!\!\!\!\!\!\overset{t \mapsto t^2}{=\! =\! =\! =\!} \frac{\pi^2 \sqrt{2}}{4} - \frac{3\pi}{4} \int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t+1} (t+3)} \\ &=\frac{\pi^2 \sqrt{2}}{4} - \frac{3\pi}{4}\int_{2}^{\infty} \frac{\mathrm{d}t}{\sqrt{t} \left ( t+2 \right )} \\ &\!\!\!\!\!\overset{t \mapsto t^2}{=\! =\! =\! =\!} \frac{\pi^2 \sqrt{2}}{4} - \frac{3\pi}{2} \int_{\sqrt{2}}^{\infty} \frac{\mathrm{d}t}{t^2+2} \\ &= \frac{\pi^2 \sqrt{2}}{4} - \frac{3\pi^2}{8\sqrt{2}} \end{align*}

Collecting everything we get that

\begin{align*} \int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 \theta}{2}} \, \mathrm{d}\theta &= \frac{\sqrt{2}}{3} \left ( \frac{\pi^2 \sqrt{2}}{4} - \frac{3\sqrt{2} \pi^2}{16} \right ) \\ &= \frac{\sqrt{2}}{3} \cdot \frac{\sqrt{2}\pi^2}{16}\\ &= \frac{\pi^2}{24} \end{align*}

QED.

Thanks @Felix Martin.

Answer 4

Continue with\begin{align} & \int_{0}^{\pi/4 } \tan^{-1} \sqrt{\frac{1-\tan^2 x}{2}}dx \>\>\>\>\>\>\> \tan x =\sin t\\ =& \int_0^{\pi/2}\frac{\cos t}{2-\cos^2t}\ \tan^{-1}\frac{\cos t}{\sqrt2} \ dt\\ =& \int_0^{\pi/2}\int_0^{\pi/4} \frac{\sqrt2 \cos^2t}{(2-\cos^2t)(2\cos^2y+\sin^2 y\cos^2t)}dy\ dt\\ =& \ \frac\pi2\int_0^{\pi/4} \bigg(1-\frac{\cos y}{\sqrt{2-\sin^2y}}\bigg)dy=\frac\pi2\bigg(\frac\pi{4}-\frac\pi6\bigg)=\frac{\pi^2}{24} \end{align}