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.

It wasn't requested, but instead of exact representations the OP might want an asymptotic expression for $n \to \infty.$  This one works well with the technique of Depoissonization.  Make an exponential power series and analyze it asymptotically:
$$ \sum_{n=0}^\infty \frac{y^n}{n!} C_n = \int_0^\infty \exp{\big(\frac{y}{x} \text{arctan}(x) \big)} dx , \quad C_n=\int_0^\infty \Big(\frac{\text{arctan}(x)}{x}\Big)^n dx$$ 
Now $\text{arctan}(x)/x = 1-x^2/3+x^4/5+...$ and with $y$ large and keeping on the first term in the asymptotic expansion
$$ e^{-y} \sum_{n=0}^\infty \frac{y^n}{n!} C_n \sim \int_0^\infty \exp{\big(-y\,\frac{x^2}{3}\big)} dx = \frac{1}{2} \sqrt{\frac{3\pi}{y}}.$$
By Depoissonization we can conclude that
$$ C_n \sim  \frac{1}{2} \sqrt{\frac{3\pi}{n}} .$$
For $n=50$ the asymptotic expression is within 2% of the value from a numerical integration.
