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Coxeter’s Integral

Prove that

    \[\int_{0}^{\pi/4}\arctan \sqrt{\frac{\cos 2\theta }{2\cos^2 \theta }} \, \mathrm{d}\theta = \frac{\pi^2}{24}\]

Solution

We state 3 lemmata:

Lemma 1: It holds that \displaystyle \arctan x = \int_{0}^{1} \frac{x}{1+x^2 t^2} \, \mathrm{d}t.

Lemma 2: It holds that \displaystyle \int_{0}^{1}\frac{\mathrm{d}x}{\left(x^2+1 \right)\sqrt{x^2+2}}=\frac{\pi }{6}.

Proof: We have successively:

    \begin{align*} \int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{x^2+2}(x^2+1)}&\overset{x=\sqrt{2} \sinh t}{=\! =\! =\! =\! =\! =\! =\!} \int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}}\frac{\mathrm{d}t}{1+2\sinh^2 t} \\ &=\int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}} \frac{\mathrm{d}t}{\cosh 2t} \\ &=\int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}}\frac{\cosh 2t}{1+\sinh^2 2t} \, \mathrm{d}t \\ &=\frac{1}{2} \operatorname{arctanh} \left ( \sinh 2\left ( \operatorname{arcsinh} \frac{1}{2} \right ) \right) \\ &= \frac{1}{2} \operatorname{arctanh} \sqrt{3} \\ &= \frac{\pi}{6} \end{align*}

Lemma 3: It holds that \displaystyle \int_{0}^{\infty}\frac{dx}{(x^2+a^2)(x^2+b^2)}=\frac{\pi }{2ab(a+b)} where a , b \neq 0.

Proof: We have successively:

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

We are ready to attack the initial monster. For that we have:

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

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