A logarithmic integral

Let $\alpha \in \mathbb{R}$ . Prove that:

$\int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2}\, \mathrm{d}x = \frac{\pi}{\sqrt{2}} \cdot \frac{\alpha}{\sqrt{1+ \sqrt{1+\alpha^2}}}$

Solution

Using the known Fourier series $\displaystyle \sum_{n=-\infty}^{\infty} \frac{1}{\left( x+2\pi n \right)^2}=\frac{1}{4} \csc^2 \frac{x}{2}$ as well as the known generating function of the Catalan numbers

$\sum_{n=0}^{\infty} \mathcal{C}_n x^n = \frac{1- \sqrt{1-4x}}{2x} \Rightarrow \sum_{n=0}^{\infty} \mathcal{C}_{2n} x^{2n} = \frac{\sqrt{1+4x} - \sqrt{1-4x}}{4x}$

Then, we have successively:

$\begin{align*} \int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2} \, \mathrm{d}x &=\frac{1}{2} \int_{-\infty}^{\infty}\frac{\arctan \left(\alpha \sin^{2} x\right)}{x^{2}}\, \mathrm{d}x\\ &=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan \left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}} \, \mathrm{d}x\\ &=\frac{1}{8}\int_{0}^{2\pi} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}\frac{x}{2} \, \mathrm{d}x\\ &=\frac{1}{4}\int_{0}^{\pi/2}\arctan \left(\alpha \sin^{2} x \right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right) \, \mathrm{d} x\\ &=\int_{0}^{\pi/2} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}x\, \mathrm{d} x\\ &=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{2n+1} \alpha^{2n+1} \int_{0}^{\pi/2}\sin^{4n}x \, \mathrm{d}x\\ &=\pi \alpha \sum_{n=0}^{\infty} \mathcal{C}_{2n}\left(-\frac{\alpha^{2}}{16}\right)^{n} \\ &=\frac{\pi}{i}\left(\sqrt{1+i \alpha}-\sqrt{1-i\alpha}\right)\\ &=\frac{\pi}{\sqrt{2}}\frac{\alpha}{\sqrt{1+\sqrt{1+\alpha^{2}}}} \end{align*}$