Just out of curiosity, I am trying to find a closed form for $$I(b)=\int_0^1\arctan^bt\ \mathrm{d}t$$ It converges for real $b>-1$, and behaves a lot like $\frac1{1+x}$.

I started out with noting that $$\arctan x=\frac i2\log\frac{1-ix}{1+ix}$$ $$\arctan^bx=\frac{i^b}{2^b}\log^b\frac{1-ix}{1+ix}$$ Which results in $$I(b)=\frac{i^b}{2^b}\int_0^1\log^b\frac{1-it}{1+it}\mathrm{d}t$$ Which I do not know how to find.

I did not want to give up, so I found $$I(1)=\int_0^1\arctan t\ \mathrm{d}t$$ Using $$\int f^{-1}(x)\mathrm{d}x=xf^{-1}(x)-(F\circ f^{-1})(x)$$ We have that $$I(1)=\frac\pi4-\frac12\log2$$

Also, rather trivially, it is easily shown that $$I(0)=1$$

But I still want to know if there is a closed form for $I(b)$. Can anyone help?


With the substitution $\tan u=t$, we the integral becomes $$I(b)=\int_0^{\frac\pi4}u^b\sec^2u\ \mathrm{d}u$$ Then we note that, for $|x|<\frac{\pi}2$, $$\tan x=\sum_{n\geq1}\frac{(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}$$ We differentiate both sides to obtain $$\sec^2x=\sum_{n\geq1}\frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-2}$$ And because $[0,\frac\pi4]\subset(\frac{-\pi}2,\frac{\pi}2)$, we have that $$I(b)=\sum_{n\geq1}\frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!}\int_0^{\frac\pi4}u^{2n+b-2}\mathrm{d}u$$ $$I(b)=\sum_{n\geq1}\frac{(2n-1)(-1)^{n-1}4^n(4^n-1)B_{2n}}{(2n)!(2n+b-1)}\bigg(\frac\pi4\bigg)^{2n+b-1}$$ $$I(b)=\bigg(\frac\pi4\bigg)^{b-1}\sum_{n\geq1}\frac{(-1)^{n-1}(2n-1)(4^n-1)B_{2n}}{(2n)!(2n+b-1)}\bigg(\frac\pi2\bigg)^{2n}$$ So, do any of you series-wizards out there have any suggestions?

  • 1
    $\begingroup$ Interesting problem, for sure ! I bet that $\zeta(.)$ and polygamma functions would appear somewhere. $\endgroup$ – Claude Leibovici Nov 20 '18 at 9:26
  • 1
    $\begingroup$ I'm betting on hypergeometric functions =)))) $\endgroup$ – Mikalai Parshutsich Nov 20 '18 at 10:00

Here is an attempt that is too long for a comment. Maybe you'll find it useful. \begin{align} \int_0^1 \arctan ^b x\,dx &= \frac{i^b}{2^b} \int_0^1 \log^b \left(\frac{1-ix}{1+ix} \right) \\ &= \frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0}\int_0^1 \left(\frac{1-ix}{1+ix} \right)^\alpha \\ &=\frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0}\int_0^1 \left(1-ix\right)^{2\alpha}(1+x^2)^{-\alpha} \,dx \\ &=\frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^1 \left(1-ix\right)^{2\alpha}\int_0^\infty \nu^{\alpha-1} e^{-\nu(1+x^2)} \,d\nu dx \\ &=\frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^\infty e^{-\nu} \nu^{\alpha-1} \int_0^1 \left(1-ix\right)^{2\alpha} e^{-\nu x^2} \,dxd\nu \\ &=\frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^\infty e^{-\nu} \nu^{\alpha-1} \int_0^1 \sum_{k=0}^{2\alpha} {2\alpha\choose k} (-1)^k i^k x^k e^{-\nu x^2} \,dxd\nu \\ &=\frac{i^b}{2^b} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^\infty e^{-\nu} \nu^{\alpha-1} \sum_{k=0}^{2\alpha} {2\alpha\choose k} (-1)^k i^k \int_0^1 x^k e^{-\nu x^2} \,dxd\nu \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^\infty e^{-\nu} \nu^{\alpha-3/2} \sum_{k=0}^{2\alpha} {2\alpha\choose k} (-1)^k i^k \nu^{-k/2} \left( \Gamma\left( \frac{k+1}{2}\right) - \Gamma\left( \frac{k+1}{2},\nu \right) \right) d\nu \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)}\int_0^\infty e^{-\nu} \nu^{\alpha-3/2} \sum_{k=0}^{2\alpha} {2\alpha\choose k} (-1)^k i^k \nu^{-k/2} \left( \Gamma\left( \frac{k+1}{2}\right) - \Gamma\left( \frac{k+1}{2}\right) e^{-\nu} \sum_{l=0}^{k/2-1/2} \frac{\nu^l}{\Gamma(l+1)}\right) d\nu \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)} \sum_{k=0}^{2\alpha} \Gamma\left( \frac{k+1}{2}\right){2\alpha\choose k} (-1)^k i^k \int_0^\infty e^{-\nu} \nu^{\alpha-k/2-3/2} \left( 1 - e^{-\nu} \sum_{l=0}^{k/2-1/2} \frac{\nu^l}{\Gamma(l+1)}\right) d\nu \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)} \sum_{k=0}^{2\alpha} \Gamma\left( \frac{k+1}{2}\right){2\alpha\choose k} (-1)^k i^k \int_0^\infty \left( e^{-\nu} \nu^{\alpha-k/2-3/2} - \sum_{l=0}^{k/2-1/2} \frac{e^{-2\nu} \nu^{\alpha-k/2-3/2+l}}{\Gamma(l+1)}\right) d\nu \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)} \sum_{k=0}^{2\alpha} \Gamma\left( \frac{k+1}{2}\right){2\alpha\choose k} (-1)^k i^k \left( \Gamma\left( \alpha-k/2-1/2 \right) - \sum_{l=0}^{k/2-1/2} \frac{\Gamma \left( \alpha-k/2-1/2+l\right) }{2^{\alpha-k/2-1/2+l}\Gamma(l+1)}\right) \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \frac{1}{\Gamma(\alpha)} \sum_{k=0}^{2\alpha} \Gamma\left( \frac{k+1}{2}\right){2\alpha\choose k} (-1)^k i^k \left( \frac{_2F_1\left(1,\alpha; k/2+3/2;1/2 \right)\Gamma(\alpha)}{2^\alpha \Gamma(k/2+3/2)} \right) \\ &=\frac{i^b}{2^{b+1}} \frac{\partial^b}{\partial \alpha^b}\Biggr|_{\alpha=0} \sum_{k=0}^{2\alpha} \Gamma\left( \frac{k+1}{2}\right){2\alpha\choose k} (-1)^k i^k \,\frac{_2F_1\left(1,\alpha; k/2+3/2;1/2 \right)}{2^\alpha \Gamma(k/2+3/2)} \\ \end{align}

  • $\begingroup$ Now split into even and odd k, reverse the order of summation, and you still have a mess. Of course, with this comment I had to upvote you. $\endgroup$ – marty cohen Nov 21 '18 at 4:42
  • $\begingroup$ How is it now haha? I did manage to integrate everything out. $\endgroup$ – Zachary Nov 21 '18 at 4:45
  • $\begingroup$ Wow that's pretty amazing. I'll have to examine it before I accept it. $\endgroup$ – clathratus Nov 21 '18 at 6:33
  • $\begingroup$ Looks legit. Does this only work for $b\in\Bbb N$? $\endgroup$ – clathratus Nov 21 '18 at 18:12
  • $\begingroup$ This method actually works for any $b\in(-1,\infty)$, which is the range for which the integral converges. However, you will just have to extend the usual definition of the derivative of non-integer order. $\endgroup$ – Zachary Nov 21 '18 at 21:08

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