1

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.

James Arathoon
  • 3,659
aleden
  • 4,007
  • It appears that for positive integer $s$, the antiderivative can be expressed using polylog functions. For example, Maple says $$\int !{u}^{5} \left( \sec \left( u \right) \right) ^{2},{\rm d}u={ \frac {2,i{u}^{5}}{1+{{\rm e}^{2,iu}}}}-2,i{u}^{5}+5,{u}^{4}\ln \left( 1+{{\rm e}^{2,iu}} \right) -10,i{u}^{3}{\it polylog} \left( 2,-{{\rm e}^{2,iu}} \right) +15,{u}^{2}{\it polylog} \left( 3,-{ {\rm e}^{2,iu}} \right) +15,iu{\it polylog} \left( 4,-{{\rm e}^{2,i u}} \right) -15/2,{\it polylog} \left( 5,-{{\rm e}^{2,iu}} \right) $$ – Robert Israel Oct 21 '18 at 23:06

2 Answers2

1

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 $$\log \cos u=-\log 2 +\sum_{k=1}^\infty (-1)^{k-1} \cos 2ku $$

James Arathoon
  • 3,659
0

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 $\mathrm{Li}_1(-i)$.

GEdgar
  • 111,679