Closed form of $\int_0^1 (\arctan(x))^sdx$? I've recently been toying around with the integral:$$f(s)=\int_0^1 (\arctan(x))^sdx=\int_0^{\frac{\pi}{4}}u^{s}\sec^2(u)du$$
I know that $f(0)=1$ and $f(1)=-\frac{2\ln(2)-\pi}{4}$ but besides that, I do not know any specific values. I think that maybe higher values could be calculated in terms of the zeta function, but I don't know how to proceed. 
 A: You could proceed by repeated integration by parts analogously to how this question Closed-form of $\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx$ is answered, to see if you can find the general formula. Starting with
$$f(s)=\int_0^{\frac{\pi}{4}}u^{s}\sec^2(u)\,du$$
The first two integrations by parts give
$$f(s)= \left[u^s \tan u\right]_0^{\pi/4}-s \int_0^{\pi/4}u^{s-1}\tan u \,du$$
and then
$$f(s)= \left[u^s \tan u\right]_0^{\pi/4}-s\left[u^{s-1}(-\log \cos u)\right]_0^{\pi/4}+ s(s-1) \int_0^{\pi/4}u^{s-2}(-\log \cos u) \,du$$
For further integrations and integrations by parts note that
$$\log \cos u=-\log 2 +\sum_{k=1}^\infty (-1)^{k-1} \cos 2ku $$
A: From Maple,
$$
f(2) = -{\it Catalan}+(1/16)\,{\pi }^{2}+(1/4)\,\pi \,\ln  \left( 2 \right)
\\
f(3) = {\frac {9\,\zeta  \left( 3 \right) }{8}}+{\frac {{\pi }^{3}}{64}}+(3/16)
\,{\pi }^{2}\ln  \left( 1+i \right) -(3/4)\,\pi \,{\it Catalan}+(3/2)\,{
\rm Li}_3 \left(-i \right) 
$$
and so on.  Here
$$
Catalan = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}
$$
and $\mathrm{Li}_3$ is the trilogarithm.  Higher integer values involve these, together with $\zeta$ values and higher polylogarithms.  Actually, the $Catalan$ may be from $\mathrm{Li}_2(-i)$ and the $\log(1+i)$ may be from $\mathrm{Li}_1(-i)$.
