I need help with calculating this integral: $$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx,$$ Where $_pF_q$ is a generalized hypergeometric function.

I was told it has a closed-form representation in terms of elementary functions and integers.

  • 1
    $\begingroup$ @Laila Maybe this hypergeometric function can be expressed in terms of elementary functions? Have you tried to find such representation? $\endgroup$ – Liu Jin Tsai May 11 '13 at 22:47
  • 7
    $\begingroup$ Look here, starting from "An infinite family of rational values"... $\endgroup$ – Myself May 11 '13 at 22:59

This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.

\begin{align} I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx \\ &=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\ &=0.7857194\dots \end{align} where $\alpha$ is the positive root of the polynomial $625\alpha^4-500\alpha^3-100\alpha^2-20\alpha-4$. It can be expressed in radicals as follows:

$$\alpha=\frac15+\sqrt\beta+\sqrt{\frac15-\beta +\frac1{25\sqrt\beta}},$$ where $$\beta=\frac1{30}\left(\frac\gamma5-\frac4\gamma+2\right),$$ where $$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$

| cite | improve this answer | |
  • 5
    $\begingroup$ Of course, you can express $\pi$ as $4 \arctan 1$ to get rid of non-integer constants. $\endgroup$ – Vladimir Reshetnikov May 11 '13 at 23:24
  • 5
    $\begingroup$ Commiserations to anyone who attempted this by hand. $\endgroup$ – L. F. May 11 '13 at 23:32
  • 5
    $\begingroup$ That's so obvious, I an kicking myself (very softly) for not thinking of it. $\endgroup$ – marty cohen May 12 '13 at 0:53
  • 13
    $\begingroup$ @L.F. I recalculated the integral completely by hand, and now the result looks much nicer. $\endgroup$ – Vladimir Reshetnikov May 12 '13 at 5:52
  • 7
    $\begingroup$ This is a true marvel...! $\endgroup$ – Sangchul Lee May 28 '13 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.