I would like to sum over the index $h$ from 3 to $\infty$, the expression
(3 (-1)^h 4^(5 - h) (-3 + 2 h) Gamma[3/4] Gamma[7/2 + h] Gamma[
9/2 + h] Gamma[-4 +
2 h] (-8 (2 + h (3 + h + 2 h^2)) HypergeometricPFQRegularized[{3/
4, 1, -(1/2) + h}, {3/4 + h, 5/2 + h},
1] + (-1 +
2 h) (-8 (2 + h (5 + 4 h)) HypergeometricPFQRegularized[{7/4,
2, 1/2 + h}, {7/4 + h, 7/2 + h}, 1] -
56 h (1 + 2 h) HypergeometricPFQRegularized[{11/4, 3,
3/2 + h}, {11/4 + h, 9/2 + h}, 1] +
Gamma[1/2 +
h] (4 (2 +
h (3 + h + 2 h^2)) HypergeometricPFQRegularized[{3/4, 1,
1/2 + h, 1/2 + h}, {3/4 + h, 3/2 + h, 5/2 + h}, 1] +
2 (1 + 2 h)^2 (2 +
h (5 + 4 h)) HypergeometricPFQRegularized[{7/4, 2,
3/2 + h, 3/2 + h}, {7/4 + h, 5/2 + h, 7/2 + h}, 1] +
7 h (3 + 4 h (2 + h))^2 HypergeometricPFQRegularized[{11/4,
3, 5/2 + h, 5/2 + h}, {11/4 + h, 7/2 + h, 9/2 + h},
1]))))/((-1 + 2 h) (1 + 2 h)^2 (3 + 2 h)^2 (5 +
2 h)^2 (7 + 2 h) \[Pi]^(5/2) Gamma[1 + h]^2)
Seeing that the hypergeometic functions here are clearly related, some standard hypergeometric transformations might be appropriate.
A solution to this problem would provide PART of the answer to my attempt to find a possible exact value underlying the numerical integration result (the “two-rebit separability probability based on the operator monotone function \sqrt{x}”) of 0.26223 reported by Lovas and Andai on p. 16 of https://arxiv.org/abs/1610.01410 (see also sec. VII.C of https://arxiv.org/abs/1701.01973).
A higher precision calculation gives
0.26223001318248382218964168547671934611262918230711855809009398124851872043672`48.
It is my working hypothesis that if one subtracts from this numerical value the result of the desired calculation, then the new result might be found (perhaps with the help of WolframAlpha) to correspond to another exact value, thus providing an exact (but still unproven) value for the Lovas-Andai numerical integration result of 0.26223.
But perhaps not!
I arrived at this formulation by integrating the general term of a series expansion of the Lovas-Andai integrand. However, Mathematica asserted that the integration would only be permissible over a restricted set of two indices ($h>2,h<\frac{3}{4}+k$, rather than both from 0 to $\infty$). Thus, the result of the desired Mathematica calculation would only provide “part of the full answer”. (I obtained the summand above by summing over the index $k$ from $h$ to $\infty$.)
I was able to compute the value of the summand for $h=3,...,66$. Each result has the form \begin{equation} \label{one} a + \frac{b}{\pi^2} +\frac{c}{\pi} -\frac{2 c \log{2}}{\pi^2}, \end{equation} with $a,b,c$ being rational. Application of the FindSequenceFunction command gave for the general form of the parameter $a$, the expression
(150 (2 + h) (1 + 3 h) Pochhammer[7/2, -3 + h]^2)/(77 (-2 + h) (-1 + h) h
Gamma[1 + h] Pochhammer[15/4, -3 + h])
Summing over $h$ from 3 to $\infty$ one obtains
25/462 (27 HypergeometricPFQ[{1, 1, 7/2, 7/2}, {3, 3, 15/4}, 1] +
21 HypergeometricPFQ[{1, 1, 7/2, 7/2}, {3, 15/4, 4}, 1] +
2 HypergeometricPFQ[{1, 1, 7/2, 7/2}, {15/4, 4, 4}, 1])
which, remarkably, evaluates to $\frac{75}{14}$. (In line with my general approach here, this should be subtracted from the 0.26223.....result.)
A further FindSequenceFunction application yields a difference function expression for $c$, which after dividing by the formula for $a$, and then adjusting, gives the result for $c$,
75 (8 h (30 + h (9 + h (-43 + 12 (-3 + h) h))) -
h (1 + h) (-17 + 38 h + 24 h^2) \[Pi] +
2 (2 + h) (-1 + 2 h) (1 + 3 h) (3 + 4 h) HarmonicNumber[-(1/4) +
h] + 6 (\[Pi] - 6 Log[2]) +
6 h (1 + h) (-17 + 38 h + 24 h^2) Log[2]) Pochhammer[7/
2, -3 + h]^2)/(77 (-2 + h) (-1 + h) h (-1 + 2 h) (3 + 4 h) Gamma[
1 + h] Pochhammer[15/4, -3 + h])
which I have been unable so far to (fully) sum over $h$ from 3 to $\infty$. (A part of the sum for $c$, now, appears to be $\frac{225 \log{2}}{14}$--which should also be appropriately subtracted in accordance with the $a + \frac{b}{\pi^2} +\frac{c}{\pi} -\frac{2 c \log{2}}{\pi^2}$ expression.)
I have not succeeded in finding a formula for the parameter $b$, which seems to be considerably more complicated (larger numerators, denominators) than $a,c$.
