# Perform an integration involving the product of two hypergeometric functions

I've encountered the following product, $$\, _2F_1\left(3 d+2,3 d+2;6 d+4;1-\frac{1}{t^2}\right) \, _3F_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;t^2\right)$$ of rather highly-structured hypergeometric functions and speculate that it might have some still further simplified form. This product is a factor in an integrand, $$w(t,d)= t^{-3 (d+1)} \left(t^2-1\right)^d \, _2F_1\left(3 d+2,3 d+2;6 d+4;1-\frac{1}{t^2}\right) \, _3F_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;t^2\right),$$ I am trying to integrate over $t \in [0,1]$, with $d$ being a free parameter, the positive integer values of which are the ones of (quantum-information-theoretic, https://arxiv.org/abs/1610.01410) interest.

I investigated using the transformation 15.8.3 in http://dlmf.nist.gov/15.8#E1 , which would convert the 2F1 argument $1-\frac{1}{t^2}$ to $t^2$, as in the 3F2 function, and which then might lead to some simplification, but this seemed to produce singularities due to the equality between the first two entries of the Gauss hypergeometric function. I seem to be able to transform the argument to $1-t^2$, but not $t^2$ itself.

After changing variables from $t$ to $t=\sqrt{T}$, for $d=2$, if one performs (in Mathematica) an indefinite integration of $$\tilde{w}(T,d) =\frac{1}{2} (T-1)^d T^{-\frac{3 d}{2}-2} \, _2\tilde{F}_1\left(3 d+2,3 d+2;6 d+4;\frac{T-1}{T}\right) \, _3\tilde{F}_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;T\right),$$ over $T$ for $d=2$, one obtains

(37 - 444 T + 2442 T^2 - 8140 T^3 + 7950 T^4 - 233982 T^5 - 572866 T^6 + 19458 T^7 + 729165 T^8 + 127450 T^9 - 62280 T^10 - 8790 T^11 + 2100 T^4 Log[1/T] + 78540 T^5 Log[1/T] + 508620 T^6 Log[1/T] + 901740 T^7 Log[1/T] + 401100 T^8 Log[1/T] - 52500 T^9 Log[1/T] - 35700 T^10 Log[1/T] - 2100 T^11 Log[1/T])/(2560481280000 (-1 + T)^12)

But for $d=4,6,8,\ldots$ (odd $d$'s are intractable), one gets sums of ($1 +\frac{3d}{2}$) MeijerG functions divided by $T^{1 +\frac{3d}{2}}$, descending from $$T^{-\frac{3 d}{2}-1} G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} -3 d-1,-3 d-1,-\frac{3 d}{2} \\ 0,0,-\frac{3 d}{2}-1 \\ \end{array} \right)$$ to $$G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} 1,-3 d,-3 d \\ 1,1,0 \\ \end{array} \right),$$ such as, for $d=4$,

-(MeijerG[{{-13, -13, -6}, {}}, {{0, 0}, {-7}}, 1/T]/(

4330257806605605116960008147304448000000000 T^7)) + MeijerG[{{-13, -13, -5}, {}}, {{0, 0}, {-6}}, 1/T]/( 909354139387177074561601710933934080000000 T^6) - ( 31 MeijerG[{{-13, -13, -4}, {}}, {{0, 0}, {-5}}, 1/T])/( 14549666230194833192985627374942945280000000 T^5) + MeijerG[{{-13, -13, -3}, {}}, {{0, 0}, {-4}}, 1/T]/( 466335456095988243364923954325094400000000 T^4) - ( 43 MeijerG[{{-13, -13, -2}, {}}, {{0, 0}, {-3}}, 1/T])/( 36374165575487082982464068437357363200000000 T^3) + ( 31 MeijerG[{{-13, -13, -1}, {}}, {{0, 0}, {-2}}, 1/T])/( 90935413938717707456160171093393408000000000 T^2) - MeijerG[{{-12, -12, 1}, {}}, {{1, 1}, {0}}, 1/ T]/24249443716991388654976045624904908800000000

I am studying the structure of such results ($d=4, 6, 8,\ldots,150$) and have found (using the Mathematica FindSequenceFunction command) that the coefficients of the highest power ($3 d-5$) of $T$ occurring in the denominators of the MeijerG terms follow the rule $$\frac{2^{8 d+1} 3^{\frac{9 d}{2}+3} \Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3 \Gamma \left(\frac{d+1}{2}\right)^2 \Gamma (d+1) \Gamma \left(\frac{3 d}{2}+1\right)^4 \Gamma (3 d+2)}{\pi ^4},$$ and the coefficients of the second-highest power ($3 d-6$) of $T$, the rule $$\frac{3^{9 d+\frac{13}{2}} \Gamma \left(\frac{d}{2}+2\right) \Gamma \left(\frac{3 d}{2}\right) \Gamma \left(d+\frac{2}{3}\right)^2 \Gamma (d+1)^3 \Gamma \left(d+\frac{4}{3}\right)^3 \Gamma \left(d+\frac{5}{3}\right) \Gamma (3 d+2)}{16 \pi ^3 (d+1)^2}.$$ The ratio of this expression to the previous one is $$\frac{2^{-d-1} 3^{\frac{3 d}{2}+\frac{1}{2}} \Gamma \left(\frac{d}{2}+\frac{2}{3}\right) \Gamma \left(\frac{d}{2}+\frac{4}{3}\right) \Gamma (d+3)}{\sqrt{\pi } d^2 (d+1)^3 \Gamma \left(\frac{3 d}{2}\right) \Gamma \left(\frac{d+1}{2}\right)}.$$

In fact, the numerators of the coefficients of the highest ($3 d-5$) power are all -1, and of the second highest ($3 d-6$) power of $T$ occurring in the denominator are +1. The alternation in sign continues as the coefficients of the third, fourth,...highest powers are examined, but the numerators cease to be just -1 or 1.

The reciprocal of the coefficients of the third highest ($3 d-7$) power of $T$ occurring in the denominator follow the rule $$-\frac{2^{8 d} 27^{3 \left(\frac{d}{2}-1\right)+4} \left(6 \left(\frac{d}{2}-1\right)+7\right) \left(\frac{d}{2}+2\right) \Gamma \left(2 \left(\frac{d}{2}-1\right)+3\right) \Gamma \left(3 \left(\frac{d}{2}-1\right)+4\right)^3 \Gamma \left(3 \left(\frac{d}{2}-1\right)+7\right) \Gamma \left(6 \left(\frac{d}{2}-1\right)+6\right) \Gamma \left(\frac{d}{2}+\frac{1}{2}\right)^2 \Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3}{\pi ^4 \left(2 \left(\frac{d}{2}-1\right)+3\right)^2 \left(2 \left(\left(\frac{d}{2}-1\right) \left(\frac{d}{2}+4\right)+7\right) \left(\frac{d}{2}-1\right)+5\right)}.$$ And similarly, for the fourth highest ($3 d-8$) power, we have, more succinctly, $$\frac{144 \Gamma \left(\frac{d}{2}+4\right) \Gamma \left(\frac{3 d}{2}+3\right) \Gamma (3 d+2)^4}{d (d+1)^2 (d (d (d (8 d (d+8)+49)-322)-96)+288)}.$$ We are examining the coefficients of still further lower powers.

As noted for even $d$ (including for $d=2$, as we have found), one gets sums of ($1 +\frac{3d}{2}$) MeijerG functions divided by $T^{1 +\frac{3d}{2}}$. We have observed that for each of $d=2,4,6,...222$, the sum of the coefficients of these ($1 +\frac{3d}{2}$) terms is zero.

• Related to your question mathoverflow.net/questions/279442/…. – user64494 Aug 29 '17 at 17:00
• Thanks, user 64494. In the original posting of the prior question you noted, I hadn't yet realized that the function $f(s,t)$ could, in fact, be integrated over $s \in [0,\infty]$. I've now amended that question to note this observation. The present question was posted as a direct result of that observation. Maybe I should have just amended the original question, and not posted this one. But the idea of simplifying the product of two hypergeometric functions (the central question put here) was not even implicitly posed in that one. – Paul B. Slater Aug 31 '17 at 12:55
• Also, in this question, as opposed to the earlier one, I'm now focused on the "heart" of the integration problem, eliminating (at this stage, irrelevant) t-free terms, that can be inserted back after the integration is performed. – Paul B. Slater Aug 31 '17 at 15:29

We have found that for any specific even value of $d$, the indefinite integration of $\tilde{w}(T,d)$ over $T$ yields a weighted sum of $1 +\frac{3 d}{2}$ MeijerG functions $$T^{-\frac{3 d}{2}-1} G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} -3 d-1,-3 d-1,-\frac{3 d}{2} \\ 0,0,-\frac{3 d}{2}-1 \\ \end{array} \right) +$$ $$\sum _{i=1}^{\frac{3 d}{2}-1} T^{-\frac{3 d}{2}+i-1} f(d,i) G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} -3 d-1,-3 d-1,i-\frac{3 d}{2} \\ 0,0,-\frac{3 d}{2}+i-1 \\ \end{array} \right)+$$ $$f\left(d,\frac{3 d}{2}\right) G_{3,3}^{2,3}\left(\frac{1}{T}| \begin{array}{c} 1,-3 d,-3 d \\ 1,1,0 \\ \end{array} \right)$$ times a factor (its reciprocal was given in the statement of the question) of $$-\frac{\pi ^4 2^{-8 d-1} 3^{-\frac{9 d}{2}-3}}{\Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3 \Gamma \left(\frac{d+1}{2}\right)^2 \Gamma (d+1) \Gamma \left(\frac{3 d}{2}+1\right)^4 \Gamma (3 d+2)}.$$

Let us note that $$f\left(d,\frac{3 d}{2}\right) = \frac{3 i^d \Gamma \left(\frac{3 d}{2}\right)^2}{8 \Gamma (d) \Gamma (2 d)},$$ and $$\frac{f\left(d,\frac{3 d}{2}-1\right)}{f(d,1)}=\frac{(27 i)^d 4^{-2 d-3} (d+2) (5 (d-1) d+2) \Gamma \left(\frac{d}{2}-\frac{1}{3}\right) \Gamma \left(\frac{d}{2}+\frac{1}{3}\right) \Gamma \left(\frac{d}{2}+\frac{2}{3}\right) \Gamma \left(\frac{d}{2}+\frac{4}{3}\right)}{\sqrt{\pi } (d-1) \Gamma \left(d+\frac{1}{2}\right) \Gamma \left(\frac{d+3}{2}\right)^2} .$$

Also, $$\sum _{i=1}^{\frac{3 d}{2}} f(d,i)=-1.$$

Further, giving us the weights $f(d,i)$ to be employed, we now have the linear difference equation (constructed based on multiple applications of the Mathematica FindSequenceFunction command),

f (d, i) = DifferenceRoot[ Function[{y, n}, {-n (-2 n + d) (-2 n + 3 d) y[n] + (12 + 36 n + 12 n^3 + 4 n^2 (9 - 2 d) - n d (28 + 11 d) + d (-16 + d (-3 + 4 d))) y[ 1 + n] + (-12 n^3 - 8 n^2 (9 + d) + 2 (2 + d) (-24 + d (11 + 2 d)) + n (-144 + d (-20 + 11 d))) y[ 2 + n] + (3 + n) (6 + 2 n + d) (6 + 2 n + 3 d) y[3 + n] == 0, y[1] == -((4 d (1 + d)^2)/( 4 + 8 d + 3 d^2)), y[2] == ( 8 d (1 + d)^2 (-4 + d^2 (4 + d)))/((2 + d) (4 + d) (2 + 3 d) (4 + 3 d)), y[3] == -(( 4 d (1 + d)^2 (288 + d (-96 + d (-322 + d (49 + 8 d (8 + d))))))/( 9 (2 + d) (4 + d) (6 + d) (2 + 3 d) (4 + 3 d)))}]][i]

Now, one must evaluate (without fixing $d$) the weighted sum given above, at the specific end points $T=1$ and $T=0$, taking the difference to arrive at the desired (definite integration) result.

If, on the basis of certain subject matter considerations we avoided earlier in order to concentrate on the core integration problem, we replace the factor given above $$-\frac{\pi ^4 2^{-8 d-1} 3^{-\frac{9 d}{2}-3}}{\Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^3 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^3 \Gamma \left(\frac{d+1}{2}\right)^2 \Gamma (d+1) \Gamma \left(\frac{3 d}{2}+1\right)^4 \Gamma (3 d+2)}.$$ by $$\frac{\pi ^{3/2} (-1)^{d+1} 2^{2-d} 3^{-3 d-2} \Gamma \left(\frac{3 (d+1)}{2}\right) \Gamma \left(\frac{5 d}{2}+2\right)}{\Gamma \left(\frac{d}{2}+\frac{5}{6}\right)^2 \Gamma \left(\frac{d}{2}+\frac{7}{6}\right)^2 \Gamma (d+1)^2 \Gamma \left(\frac{3 d}{2}+1\right)^4},$$ then (see eq. (4) of https://arxiv.org/abs/1301.6617 for the cited results) for the "two-qubit" case $d=2$, we found these two ($T=1$ and $T=0$) values to be $148 +\frac{8}{33}$ and 148, respectively, giving us the correct result for the difference of $\frac{8}{33}$. For the "two-quaterbit" $d=4$ instance, we have $-333631 -\frac{297}{323}$ and -333632, giving us the $\frac{26}{323}$ oucome. Now for $d=6$, we have $1008871862+\frac{2999}{103385}$ and $10088871862$, consistent with the cited paper, and similarly for $d=8$, with the upper value being $-3543784402375-\frac{4046867}{4091349}$ and the lower value being -3543784402376. So, for odd values of $\frac{d}{2}$, we appear to have pairs of positive limits, and for even values, negative limits. (These four lower [$T=0$] integer limits we computed exactly, while the upper [$T=1$] set was computed numerically to high accuracy.)

Actually, the article cited in the first answer, https://arxiv.org/abs/1301.6617, in its Fig. 3, provides an apparent (quite different appearing) answer to the integration problem (with the same replacement of leading factors, as noted in the first answer), as high-precision numerical integrations indicate for $d=1,2,\ldots,8$. (For $d =\frac{3}{2}$, the absolute values obtained from the formula and the numerical integration agree.)

Substituting $d = \frac{\alpha}{2}$ in the result in the cited Fig. 3, we have the Mathematica expression

(4^(-3 - d)
Gamma[5/2 + (3 d)/2] Gamma[
2 + (5 d)/
2] ((-54 +
1/2 d (39 +
5/2 d (628 +
25/2 d (161 +
d (-581 + 370 d))))) HypergeometricPFQ[{1,
2/5 + d/2, 3/5 + d/2, 4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/
64] + (347274 +
5/2 d (-312019 +
25/2 d (22255 +
4 d (-2431 + (925 d)/2)))) HypergeometricPFQ[{2,
2/5 + d/2, 3/5 + d/2, 4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/64] +
10 ((-769797 +
25/2 d (66227 +
2 d (-12843 + 1850 d))) HypergeometricPFQ[{3,
2/5 + d/2, 3/5 + d/2, 4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/64] +
75 ((44133 + 4 d (-6131 + 925 d)) HypergeometricPFQ[{4,
2/5 + d/2, 3/5 + d/2, 4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/64] +
8 ((-7981 + 1850 d) HypergeometricPFQ[{5, 2/5 + d/2,
3/5 + d/2, 4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/64] +
3700 HypergeometricPFQ[{6, 2/5 + d/2, 3/5 + d/2,
4/5 + d/2, 5/6 + d/2, 7/6 + d/2,
6/5 + d/2}, {13/10 + d/2, 3/2 + d/2, 17/10 + d/2,
19/10 + d/2, 2 + d/2, 21/10 + d/2}, 27/
64])))))/(3 Gamma[1 + d/2] Gamma[3 + d] Gamma[
13/2 + (5 d)/2])

(Qing-Hu Hou was able, as reported in the cited article, to apply "Zeilberger's algorithm" to this expression to obtain a concise reformulation of it, involving an infinite summation.) But as noted, being the heart of the posed question, we do not seem to be able to obtain this large Mathematica expression by direct integration (leaving $d$ as a free parameter). It had been derived instead through use of the "moment-based density approximants" procedure of S. B. Provost http://www.mathematica-journal.com/issue/v9i4/contents/DensityApproximants/DensityApproximants_7.html applied to certain "determinantal moments".

The large Mathematica hypergeometric-based expression in the previous answer can be expressed (p. 26 of https://arxiv.org/pdf/1609.08561.pdf, taking $k=0$ and $\alpha =\frac{d}{2}$ in $2 Q(k,\alpha)$), as $$1-\frac{\sqrt{\pi } 2^{-\frac{9 d}{2}-\frac{5}{2}} \Gamma \left(\frac{3 (d+1)}{2}\right) \Gamma \left(\frac{5 d}{4}+\frac{19}{8}\right) \Gamma (2 d+2) \Gamma \left(\frac{5 d}{2}+2\right) \, _6\tilde{F}_5\left(1,d+\frac{3}{2},\frac{5 d}{4}+1,\frac{1}{4} (5 d+6),\frac{5 d}{4}+\frac{19}{8},\frac{3 (d+1)}{2};\frac{d+4}{2},\frac{5 d}{4}+\frac{11}{8},\frac{1}{4} (5 d+7),\frac{1}{4} (5 d+9),2 (d+1);1\right)}{\Gamma (d)} .$$