Coxeter integral

Evaluate:

$$\int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}\, {\rm d}\theta$$

Solution



We are invoking the following lemmas:

Lemma 1: $\displaystyle \int_{0}^{1}\frac{{\rm d}x}{\left ( x^2+1 \right )\sqrt{x^2+2}}= \frac{\pi}{6}$

Proof:
$$\begin{aligned}
\int_{0}^{1}\frac{{\rm d}x}{\sqrt{x^2+2}\left ( x^2+1 \right )} &\overset{x=\sqrt{2}\sinh t\Rightarrow a=\sinh^{-1}\left ( \frac{1}{\sqrt{2}} \right )}{=\! =\! =\! =\! =\! =\! =\!=\!=\! =\! =\! =\!=\! =\! =\! =\!=\!}\int_{0}^{a}\frac{{\rm d}t}{1+2\sinh^2 t} \\
 &= \int_{0}^{a}\frac{{\rm d}t}{\cosh 2t}\\
 &= \int_{0}^{a}\frac{\cosh 2t}{1+\sinh^2 2t}\, {\rm d}t\\
 &=\frac{1}{2}\tanh^{-1}\left ( \sinh 2a \right ) \\
 &= \frac{1}{2}\tanh^{-1}\left ( \sqrt{\left ( 1+2\sinh^2 a \right )^2}-1 \right )= \frac{1}{2}\tanh^{-1}\sqrt{3}\\
 &=\frac{\pi}{6}
\end{aligned}$$

Lemma 2: $\displaystyle \int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^2+a^2 \right )\left ( x^2+b^2 \right )}= \frac{\pi}{2ab \left ( a+b \right )}$

Proof:

$$\begin{aligned}
\int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^2+a^2 \right )\left ( x^2+b^2 \right )} &=\int_{0}^{\infty}\frac{1}{b^2-a^2}\left ( \frac{1}{x^2+a^2}- \frac{1}{x^2+b^2} \right )\, {\rm d}x \\
 &=\frac{1}{b^2-a^2}\left [ \frac{\pi}{2a}- \frac{\pi}{2b} \right ] \\
 &=\frac{\pi}{2ab(a+b)}
\end{aligned}$$

Lemma 3: $\displaystyle \tan^{-1}a = \int_{0}^{1}\frac{a}{1+a^2 x^2}\, {\rm d}x$

Using the three lemmas we are in position to evaluate the Coxeter's integral. Successively we have:

$$\begin{aligned}
\int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}\, {\rm d}\theta &=\int_{0}^{\pi/4}\int_{0}^{1}\frac{\sqrt{\frac{\cos 2\theta}{2\cos^2 \theta}}}{1+ \frac{\cos 2\theta}{2\cos^2 \theta}x^2}\, {\rm d}x \, {\rm d}\theta \\
 &= \int_{0}^{1}\int_{0}^{\pi/4}\frac{\sqrt{1-2\sin^2 \theta}}{2-2\sin^2 \theta+ \left ( 1-2\sin^2 \theta \right )x^2}\cdot \sqrt{2}\cos \theta \, {\rm d\theta}\, {\rm d}x\\
 &= \int_{0}^{1}\int_0^{\pi/4}\frac{\sqrt{1-\sin^2 \varphi }}{2-\sin^2 \varphi + \left ( 1-\sin^2 \varphi \right )x^2}\cos \varphi \, {\rm d}\varphi \, {\rm d}x\\
 &= \int_{0}^{1}\int_{0}^{\pi/4}\frac{\cos^2 \varphi}{\sin^2 \varphi + \left ( x^2 +2 \right )\cos^2 \varphi}\, {\rm d}\varphi \, {\rm d}x\\
 &= \int_{0}^{1}\int_{0}^{\pi/2}\frac{{\rm d}\varphi \, {\rm d}x}{\tan^2 \varphi + x^2 +2}\\
 &\overset{y=\tan \varphi}{=\! =\! =\! =\! =\!}\int_{0}^{1}\int_{0}^{\infty}\frac{{\rm d}y\, {\rm d}x}{\left ( y^2+x^2+2 \right )\left ( y^2+1 \right )} \\
 &=\frac{\pi}{2}\int_{0}^{1}\frac{{\rm d}y}{\left ( 1+\sqrt{2+y^2} \right )\sqrt{2+y^2}} \\
 &=\frac{\pi}{2}\left ( \frac{\pi}{4}-\frac{\pi}{6} \right ) \\
&=\frac{\pi^2}{24}
\end{aligned}$$