오늘의 계산 37 - Coxeter's Integrals (1)

Today's integral is a famous one.

$$ \int_{0}^{\frac{\pi}{4}} \arctan\sqrt{\frac{\cos2\theta}{2\cos^2\theta}} \, \mathrm{d}\theta = \frac{\pi^2}{24}. $$

To calculate this, we first note that

\begin{gather*} \arctan(\alpha) = \int_{0}^{1} \frac{\alpha}{1+\alpha^2 x^2} \, \mathrm{d}x, \tag{1} \\ \int_{0}^{\infty} \frac{\mathrm{d}x}{(x^2+\alpha^2)(x^2 + \beta^2)}=\frac{\pi}{2\alpha\beta(\alpha+\beta)}, \tag{2} \\ \int_{0}^{1} \frac{\mathrm{d}x}{(x^2+1)\sqrt{x^2+2}}=\frac{\pi}{6}. \tag{3} \end{gather*}

The identity $\text{(2)}$ is an easy consequence of the partial fraction decomposition:

\begin{align*}\int_{0}^{\infty} \frac{\mathrm{d}x}{(x^2+\alpha^2)(x^2 + \beta^2)} &= \int_{0}^{\infty}\frac{1}{\beta^2-\alpha^2}\left( \frac{1}{x^2 + \alpha^2} - \frac{1}{x^2+\beta^2}\right)\,\mathrm{d}x \\ &= \frac{1}{\beta^2-\alpha^2}\left(\frac{\pi}{2\alpha} - \frac{\pi}{2\beta}\right) \\ &=\frac{\pi}{2\alpha\beta(\alpha+\beta)} \end{align*}

For $\text{(3)}$, substituting $x=\sqrt{2}\sinh t$ and writing $a=\sinh^{-1}(1/\sqrt{2})$ recasts the integal as

\begin{align*} \int_{0}^{1} \frac{\mathrm{d}x}{(x^2+1)\sqrt{x^2+2}} &= \int_{0}^{a} \frac{\mathrm{d}t}{1+2\sinh^2 t} = \int_{0}^{a} \frac{\mathrm{d}t}{\cosh 2t} \\ &= \frac{1}{2}\arctan(\sinh 2a) = \frac{1}{2}\arctan\sqrt{3} = \frac{\pi}{6}. \end{align*}

Now we are ready to compute the Coxeter's integral.
\begin{align*} & \int_{0}^{\frac{\pi}{4}} \arctan\sqrt{\frac{\cos2\theta}{2\cos^2\theta}} \, \mathrm{d}\theta \\ &= \int_{0}^{\frac{\pi}{4}} \int_{0}^{1} \frac{\sqrt{\frac{\cos2\theta}{2\cos^2\theta}}}{1+\left(\frac{\cos2\theta}{2\cos^2\theta}\right)x^2} \, \mathrm{d}x\mathrm{d}\theta \tag{$\because (1)$} \\ &= \int_{0}^{1} \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{1-2\sin^2\theta}}{2-2\sin^2\theta+(1-2\sin^2\theta)x^2} \sqrt{2}\cos\theta \, \mathrm{d}\theta\mathrm{d}x \\ &= \int_{0}^{1}\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{1-\sin^2\phi}}{2-\sin^2\phi+(1-\sin^2\phi)x^2} \cos\phi \, \mathrm{d}\phi\mathrm{d}x \tag{$\sqrt{2}\sin\theta=\sin\phi$} \\ &= \int_{0}^{1} \int_{0}^{\frac{\pi}{2}} \frac{\cos^2\phi}{\sin^2\phi+(x^2+2)\cos^2\phi} \, \mathrm{d}\phi\mathrm{d}x \\ &= \int_{0}^{1}\int_{0}^{\frac{\pi}{2}} \frac{1}{\tan^2\phi + x^2 + 2} \, \mathrm{d}\phi\mathrm{d}x \\ &= \int_{0}^{1} \int_{0}^{\infty} \frac{1}{(y^2+x^2+2)(y^2+1)} \, \mathrm{d}y\mathrm{d}x \tag{$y=\tan\phi$} \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{d}y}{(1+\sqrt{2+y^2})\sqrt{2+y^2}} \tag{$\because (2)$} \\ &= \frac{\pi}{2} \int_{0}^{1} \left( \frac{1}{y^2+1} - \frac{1}{(y^2+)\sqrt{2+y^2}} \right) \, \mathrm{d}y \\ &= \frac{\pi}{2} \left( \frac{\pi}{4} - \frac{\pi}{6} \right) \tag{$\because (3)$} \\ &= \frac{\pi^2}{24}. \end{align*}