Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$ I know that for $n=1$ the integral is divergent and that for $n=2$ the integral has a closed form. However, I wonder if the general expression has a closed form.
My attempt: $$\int_0^\infty \left(\frac{\arctan(x)}{x}\right)^ndx=\frac{n}{1-n}\int_0^\infty\frac{\arctan^{n-1}(x)}{x^{n-1}(x^2+1)}dx=\frac{n}{1-n}\int_0^{(\frac{\pi}{2})^{n-1}} u^{n-1}\cot^{n-1}\left(u^{1/(n-1)}\right)du$$
I don't know if I'm on the right track here or not but I do not know through what methods to evaluate the last integral. Any help is appreciated.

$$\int_{0}^{+\infty}\left(\frac{\arctan x}{x}\right)^3\,dx \stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/2}\frac{\theta^3\,d\theta}{\tan^3\theta\cos^2\theta}\stackrel{\text{IBP}}{=}-\frac{\pi^3}{16}+\frac{3}{2}\int_{0}^{\pi/2}\frac{\theta^2\,d\theta}{\sin^2\theta}$$
and in general the computation of the wanted integrals boils down to the computation of $\int_{0}^{\pi/2}\left(\frac{\theta}{\sin\theta}\right)^m\,d\theta$ or the computation of $\oint_\gamma \frac{\log^m z}{z\left(z-\frac{1}{z}\right)^m}\,dz$ where $\gamma$ is the quarter circle joining $1$ and $i$. The integration of $\frac{\log^m z}{z\left(z-\frac{1}{z}\right)^m}$ along the segments joining $0$ and $1$ or $0$ and $i$ can be simply performed through Maclaurin series; in particular the wanted integrals can be always expressed in terms of standard or alternating Euler sums.
