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I found an interesting problem about generalized hypergeometric series in MO, stating that:

$$\, _4F_3\left(\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{2}{3};1,\frac{4}{3},\frac{4}{3};1\right)=\sum_{n=0}^\infty \left(\frac{(\frac13)_k (\frac23)_k}{(1)_k (\frac43)_k}\right)^2=\frac{\Gamma \left(\frac{1}{3}\right)^6}{36 \pi ^2}$$

This numerically agrees, but I found no proof using either elementary properties of hypergeometric functions (e.g. cyclic sum) or classical Gamma formulas (e.g. Dougall formula). I bet it has something to do with modular forms and elliptic $K$ integral, but the exact relation remain elusive.

How to prove this identity? What will be its motivation? Can we generate other Gamma evaluation of high order hypergeometric series using the method of proving it? Any help will be appreciated.

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3 Answers 3

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Let $S$ be the given $_4F_3$, then (first equality comes from termwise integration), $$\begin{aligned} S &= -\frac{1}{9}\int_0^1 t^{-2/3} (\log t) {_2F_1}(2/3,2/3;1;t)dt =-\frac{1}{9} \frac{d}{da} \left(\int_0^1 t^{-2/3+a} {_2F_1}(2/3,2/3;1;t)dt \right)_{a=0}\\ &= -\frac{1}{9}\frac{d}{da}\left(\frac{\, _3F_2\left(\frac{2}{3},\frac{2}{3},a+\frac{1}{3};1,a+\frac{4}{3};1\right)}{ a+1/3}\right)_{a=0} \end{aligned}$$

It is easily seen $A=\sqrt{\pi } \Gamma \left(\frac{7}{6}\right)/\Gamma \left(\frac{5}{6}\right)^2$ is the value of the $_3F_2$ at $a=0$ (Dixon). Set $$\begin{aligned} &{d_{2/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3} + a,\frac{2}{3},\frac{1}{3};1,\frac{4}{3};1)} \right)_{a = 0}} \qquad {d_1} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3};1 + a,\frac{4}{3};1)} \right)_{a = 0}} \\ &{d_{1/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3} + a;1,\frac{4}{3};1)} \right)_{a = 0}} \qquad {d_{4/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3};1,\frac{4}{3} + a;1)} \right)_{a = 0}}\end{aligned}$$

By multivariable chain rule, $$S = A -\frac{1}{3}(d_{1/3}+d_{4/3})\tag{*}$$


In general, derivative of $_pF_q$ with respect to a parameter is intractable. One can only handle them in an ad hoc manner. In our situation, it is well-known that $_3F_2$ at $1$ satisfies certain transformations: two generators are the 1st and 3rd entry here. Using these two entries, we obtain $$\begin{aligned} & \quad _3F_2\left(\frac{2}{3},\frac{2}{3},a+\frac{1}{3};1,a+\frac{4}{3};1\right) \\ &= \frac{\Gamma \left(\frac{2}{3}\right) \Gamma \left(a+\frac{4}{3}\right) \, _3F_2\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}-a;1,\frac{4}{3};1\right)}{\Gamma \left(\frac{4}{3}\right) \Gamma \left(a+\frac{2}{3}\right)} \\ &= \frac{\Gamma \left(\frac{2}{3}\right) \, _3F_2\left(a+\frac{1}{3},a+\frac{2}{3},a+\frac{2}{3};a+1,a+\frac{4}{3};1\right)}{\Gamma \left(\frac{2}{3}-a\right) \Gamma (a+1)} \\ &= \frac{\Gamma \left(-\frac{1}{3}\right) \Gamma \left(a+\frac{1}{3}\right) \Gamma \left(a+\frac{4}{3}\right) \, _3F_2\left(\frac{1}{3},\frac{2}{3},\frac{2}{3};\frac{4}{3},a+1;1\right)}{\Gamma \left(\frac{1}{3}\right)^2 \Gamma \left(a+\frac{2}{3}\right) \Gamma (a+1)}+\frac{\Gamma \left(\frac{1}{3}\right) \Gamma \left(a+\frac{1}{3}\right) \Gamma \left(a+\frac{4}{3}\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(a+\frac{2}{3}\right)^2} \end{aligned}$$

Observe that for all four $_3F_2$ above, their arguments are all like $(2/3,2/3,1/3;1,4/3)$, the only difference is $a$ appears at different places. This reveals why $(2/3,2/3,1/3;1,4/3)$ is special.

Introduce an operational definition: write $x\equiv y$ if $x-y$ is a "linear combination of gamma factors". For example, $x\equiv y$ if $x-y = A$. Now take derivative at $a=0$, we obtain $$\tag{**}d_{1/3}+d_{4/3} \equiv -d_{2/3} \equiv d_{1/3}+2d_{2/3}+d_1+d_{4/3} \equiv -d_1$$ Solving this system gives $$d_1 \equiv d_{2/3} \equiv d_{1/3}+d_{4/3} \equiv 0$$

Thus $d_{1/3}+d_{4/3}$ can be expressed into gamma function, so can $S$ according to $(*)$.

There is no difficulty in making $(**)$ explicit: $$d_{1/3}+d_{4/3}=\left(3-\frac{\pi }{\sqrt{3}}\right) A-d_{2/3}=d_1+d_{1/3}+2 d_{2/3}+d_{4/3}+\frac{1}{6} A \left(\sqrt{3} \pi -9 \log (3)\right)=-d_1+\frac{1}{2} A \left(\pi \sqrt{3}-6+3 \log (3)\right)+\frac{3 \left(3 \sqrt{3}-2 \pi \right) \Gamma \left(\frac{1}{3}\right)^2 \Gamma \left(\frac{7}{6}\right)^2}{\sqrt[3]{2} \pi ^2}$$

Solving gives $d_{1/3}+d_{4/3} = \dfrac{2 \sqrt{\pi } \left(27-4 \sqrt{3} \pi \right) \Gamma \left(\frac{13}{6}\right)}{21 \Gamma \left(\frac{5}{6}\right)^2}$. We also obtain values of $d_1, d_{2/3}$ as by-products.

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  • $\begingroup$ It's quite a nice solution. I actually think that if you can generalize it a bit you might be able to make a small publication out of it. For instance, if you get the case for general that I mention below, or generalize it in some other way. $\endgroup$
    – user387394
    Commented Sep 3, 2020 at 20:20
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Wow, amazing! Solved 9 years later! Thanks all for digging this up, and then for solving it. Can this give a general form for

$$_4F_3(\frac1m,\frac1m,\frac2m,\frac2m;\frac{m+1}m,\frac{m+1}m,1;1)$$

I should probably give some motivations for this. In the following paper, I looked at the expected exit time of a planar Brownian motion starting at 0 from a regular $m$-gon centered at 0:

https://projecteuclid.org/euclid.ecp/1465262013

It is (up to a constant which depends on the size of the polygon)

$$_4F_3(\frac1m,\frac1m,\frac2m,\frac2m;\frac{m+1}m,\frac{m+1}m,1;1)\times \frac{m^2}{\beta(1/m,(m-2)/m)^2},$$

which doesn't exactly roll off the tongue. However, for an equilateral triangle there is a different method for calculating this, and it gives $1/6$. So we get an identity by equating the two, and that is the identity. Now, the question is, can we use this method to get a nicer expression for the $_4F_3$ for larger $m$? This would then be a nicer expression for the expected exit time of Brownian motion from the regular $m$-gon.

A purely analytic (i.e. not probabilistic) version of this all can be found here, because the expected exit time is basically the Hardy H^2 norm of the domain, up to a constant.

https://arxiv.org/abs/1205.2458

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Step 1: Evaluating $$ \int_{0}^{\infty}x^{s-1}\,_2F_1\left ( \frac23,\frac23; 1;x \right )\text{d}x, $$ gives $$ \int_{0}^{1}\left ( x^{s-1}+ \frac{1-\sqrt{3}\cot(\pi s) }{2}x^{-s-\frac13} \right ) \,_2F_1\left ( \frac23,\frac23; 1;x \right )\text{d}x =\frac{1+\sqrt{3}\cot(\pi s) }{2} \frac{2\pi\sqrt{3} }{\Gamma\left ( \frac13 \right )^3 } \int_{0}^{1} x^{-s-\frac13}\,_2F_1\left ( \frac23,\frac23;\frac43;x\right)\text{d}x. $$ Step 2: Differentiating with respect to $s$, after some simplifications we arrive at $$ \,_4F_3\left ( \frac13,\frac13,\frac23,\frac23; 1,\frac43,\frac43;1 \right ) =\frac{\Gamma\left ( \frac13 \right )^6 }{12\pi^2} + \frac{2\pi}{3\sqrt{3}\,\Gamma\left ( \frac13 \right )^3} \underbrace{\int_{0}^{1} \frac{\ln(x)}{x^{2/3}} \,_2F_1\left ( \frac23,\frac23;\frac43;1-x \right )\text{d}x}_{I}. $$ Expanding the $\,_2F_1$, end up to evaluate $$ I=\sum_{n=0}^{\infty} \frac{\left ( \frac{2}{3} \right )_n^2 }{ \left ( \frac43 \right )_n} \frac{\Gamma\left ( \frac13 \right ) \left ( \psi^{(0)}\left ( \frac13 \right ) -\psi^{(0)} \left ( n+\frac43 \right ) \right ) }{\Gamma\left ( n+\frac43 \right ) }. $$ Step 3: From Entry 2, Entry 4, we have \begin{aligned} &\sum_{n=0}^{\infty} \frac{\left ( \frac23 \right )^2_n }{ \left ( \frac43 \right )_n(1)_n} \frac{\Gamma(n+a)}{\Gamma\left ( n+\frac{a+1}{2}+\frac13 \right ) } =\frac{\sqrt{\pi}\,\Gamma\left ( \frac76 \right )\Gamma(a)\Gamma\left ( \frac{1-a}{2}+\frac13 \right ) }{ \Gamma\left ( \frac56 \right )^2 \Gamma\left ( \frac{1+a}{2} \right ) \Gamma\left ( \frac{1-a}{2}+\frac23 \right )}, &\sum_{n=0}^{\infty} \frac{\left ( \frac23 \right )^2_n }{ \left ( 1 \right )_n} \frac{\Gamma(n+a)}{\Gamma\left ( n+a+\frac13 \right )^2 } =\frac{\Gamma\left ( \frac{a}2+1 \right )\Gamma\left ( \frac{a}2-\frac13 \right ) }{ a\,\Gamma\left ( a-\frac13 \right )\Gamma\left ( \frac{a}2+\frac13 \right )^2 }. \end{aligned} Therefore \begin{aligned} &\sum_{n=0}^{\infty} \frac{\left ( \frac23 \right )_n^2 }{ \left ( \frac43 \right )_n } \frac{\psi^{(0)}(n+1)-2\psi^{(0)} \left ( n+\frac43 \right ) }{\Gamma\left ( n+\frac43 \right ) } =\frac{\left ( 12\gamma-11\pi\sqrt{3}+27\ln(3) \right )\Gamma\left ( \frac13 \right )^8 }{128\pi^4},\\ &\sum_{n=0}^{\infty} \frac{\left ( \frac23 \right )_n^2 }{ \left ( \frac43 \right )_n } \frac{\psi^{(0)}(n+1)-\frac12\psi^{(0)} \left ( n+\frac43 \right ) }{\Gamma\left ( n+\frac43 \right ) } =\frac{\left ( \pi\sqrt{3}-3\gamma \right )\Gamma\left ( \frac13 \right )^8 }{64\pi^4}. \end{aligned} We immediately solve $$ \sum_{n=0}^{\infty} \frac{\left ( \frac23 \right )_n^2 }{ \left ( \frac43 \right )_n } \frac{\psi^{(0)} \left ( n+\frac43 \right ) }{\Gamma\left ( n+\frac43 \right ) } =\frac{\left ( -18\gamma+13\pi\sqrt{3}-27\ln(3) \right )\Gamma\left ( \frac13 \right )^8 }{192\pi^4}. $$ Eventually all these equations above allowed us to show $$ I=-\frac{\Gamma\left(\frac13\right)^9}{4\sqrt{3}\pi^3}.\square $$

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