$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$

Question

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.

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

score 30 · Answer 1 · edited Jan 03 '17 at 05:40

30

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 $$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$

edited Jan 03 '17 at 05:40

Tito Piezas III

60,745

answered May 11 '13 at 23:19

Vladimir Reshetnikov

32,650

5

Of course, you can express $\pi$ as $4 \arctan 1$ to get rid of non-integer constants. – Vladimir Reshetnikov May 11 '13 at 23:24
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Commiserations to anyone who attempted this by hand. – L. F. May 11 '13 at 23:32
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That's so obvious, I an kicking myself (very softly) for not thinking of it. – marty cohen May 12 '13 at 00:53
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@L.F. I recalculated the integral completely by hand, and now the result looks much nicer. – Vladimir Reshetnikov May 12 '13 at 05:52
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This is a true marvel...! – Sangchul Lee May 28 '13 at 02:15
@VladimirReshetnikov: +1 I added an equivalent form. I hope its ok. The polynomial in $\alpha$ had a cube factor and $\alpha$ also had a shorter and "natural" defining polynomial as $\alpha^4(1-\alpha)=\frac4{3125}$. – Tito Piezas III Jan 02 '17 at 07:51
@VladimirReshetnikov: I reverted my edits as I asked a generalized question. – Tito Piezas III Jan 03 '17 at 05:48

$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$

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