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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 $a,c$.

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  • $\begingroup$ After numerical summation, I think the result is close to 0.00427272, not 0.26223. $\endgroup$ – Roman Jan 16 at 10:52
  • $\begingroup$ Thanks, Roman! I should make things clearer--I am not trying here to reproduce the 0.26223 result, but only the part of that result which presently seems amenable to exact calculation. (Then, subtracting this from the 0.26223....., possbly, say, WolframAlpha might suggest an underlying exact result.) I took series expansions within the original Lovas-Andai integrand (yielding the 0.26223 outcome). Mathematica was able to integrate this, but with the restriction that my two indices (k and h), running both from 0 to $\infty$ would be restricted to $h>2, h<3/4+k$. That's my attempted calculation. $\endgroup$ – Paul B. Slater Jan 16 at 15:12
  • $\begingroup$ An intriguing observation! Subtracting the Roman result of 0.00427272 from 0.26223......, I get 0.257957, which WolframAlpha suggests might equal $\frac{4 b_2(2)}{5}+2$, where $b_2(2)$ is the "Madelung constant" (mathworld.wolfram.com/MadelungConstants.html) $-\pi \log{2}$. However, for the 0.00427272, WolframAlpha doesn't suggest anything immedately very "elegant"--at best, it would seem is $\frac{x^2_{{min}}}{500}$, where $x_{min}$ "is the value at which $\Gamma$ function is minimal for positive argument". More digits in Roman's calculation might be helpful. $\endgroup$ – Paul B. Slater Jan 16 at 15:58
  • $\begingroup$ I can offer 0.0042727174678 as an improvement. $\endgroup$ – Roman Jan 16 at 16:34
  • $\begingroup$ Well, based on my exact summing over $h$ from 3 to 64, rather than to $\infty$, I get 0.0042727174581--obviously agreeing very strongly with Roman's calculation. His increased precision, though, seems to undercut the relevance of the Madelung constant. Also, I am somewhat surprised by the smallness of the sum--since I have exact calculations that indicate that this (0.004...) result should be the sum of $\frac{75}{14}+\frac{225}{14 \pi }-\frac{225 \log (2)}{7 \pi ^2} \approx 8.2154287602604004$ and some yet undetermined terms on the order of -8.21115604. ($14, \pi,\log{2}$ seem relevant here.) $\endgroup$ – Paul B. Slater Jan 16 at 22:11
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Here are some more accurate results. We have seen that the infinite sum can be expressed in the form $$a + \frac{b}{\pi^2} + \frac{c}{\pi} -\frac{2c\log2}{\pi^2}$$ Using the asymptotic solutions of the fourth-order recurrence for the truncated sum, I obtain high-precision approximations to the constants $a$, $b$, and $c$. For this purpose I computed the exact value of the sum of the first 100000 summands. As it was shown by Paul, we have $a=\frac{75}{14}$, which in my calculation is confirmed up to 200 decimal places. For $b$ I get $$\tiny -21.9925102620586535247078385978969147938281514892602670680311339663952051483715365838587761188481762314729119244723836227340793118276275037392946039511371501323268243912386427544836042858242 $$ and for $c$ $$\tiny -17.56863808233113893803620645282291474290576304500611202270609786346160868771274443094724380146230555853504152542638799919971888457249682140926560502481227332897826339344548117642261443082869 $$ Putting these together we obtain a better approximation for the infinite sum: $$\tiny 0.004272717467855486550885164553229384966927537681571131827105267094105211667361629419844797930085821788087215928355998198692754575756851901698699049941694679686972783950784736797482855586737 $$ Unfortunately, I was not able to identify a closed form for any of these numbers (using both the inverse calculator and the AskConstants package). The other bad news is that it falsifies the previous guess by WolframAlpha (involving Catalan's constant) whose decimal expansion starts as $0.00427271746785548655070264$, since it already differs in the 22nd place. I'm pretty sure that all my digits are correct.

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Actually, if one feeds the calculation of Roman of 0.0042727174678554865508851 into WolframAlpha, it supplies a possible corresponding exact value of $\frac{1}{175} \left(315 C-227-337 \pi +25 \pi ^2+215 \pi \log (2)+82 \pi \log (3)\right) \approx 0.00427271746785548655070264$, where $C$ is Catalan's constant (http://mathworld.wolfram.com/CatalansConstant.html) $0.915965594177219015054603514932384110774$.

This does not seem wildly implausible, as $\pi$, $\pi^2$ and $\log{2}$ occur in the individual values of the summand $h=3,\ldots,66$ noted in the original question itself. (A detail WolframAlpha is NOT given.)

Christoph Koutschan, in his very detailed answer, which I will post below, also similarly finds these three constants ($\pi$, $\pi^2$, $\log{2}$) present--and he also adds two digits (64) to the value of Roman's, that is he obtains 0.004272717467855486550885164.

But now if one subtracts 0.004272717467855486550885164 from the high-precision value using the Lovas-Andai formula for the "two-rebit operator monotone $\sqrt{x} separability probability", 0.26223001318248382218964168547671934611262918230711855809009398124851872043672`48, one gets 0.25795729571462833563875652--for which WolframAlpha (nor the InverseSymbolicCalculator) does not seem to supply an appealing/plausible exact value (the ultimate goal, of course).

To make any possible inroads into finding an exact underlying value for the indicated difference 0.25795729571462833563875652, it seems that one would have to be able to sum

(32 (1/(1 - w))^k w^(-3 + 

h) ((8 - 3 w) w + 2 h (-4 + w) (-4 + 3 w)) Gamma[-(1/2) + h] Gamma[ 1/2 + h])/((1 + 2 k)^2 (-3 + 4 k (1 + k)) [Pi]^3 (1 - w)^(3/4) Pochhammer[1, h]^2)

over both $h$ and $k$ from 0 to $\infty$, but OMIT precisely the range for which $h > 2$ and $h < 3/4 + k$. Then, the numerical integral of the result over $w \in [-\infty,0]$ should give the indicated 0.25795729571462833563875652. However, we would hope to compute the integral exactly. (Perhaps I shall post this restricted summation problem as an independent question.)


Here is the detailed analysis of Koutschan, which he supplied in an email to me:

I have had a look at your problem. I'm not able to come up with a final answer, but here is what I found: I'm looking at the truncated sum s[n] = Sum[c3, {h, 3, n}] The first few values can be computed exactly without problems: s[3] = 625/231 - 5189120/(160083*Pi^2) + 175040/(53361*Pi) - (350080*Log[2])/(53361*Pi^2), s[4] = 4395/1232 - 45725270/(1013859*Pi^2) + 1912448/(337953*Pi) - (3824896*Log[2])/(337953*Pi^2), s[5] = 42447/10640 - 194118306352/(3692136525*Pi^2) + 2532700832/(335648775*Pi) - (5065401664*Log[2])/(335648775*Pi^2), s[6] = 1037727/244720 - 574475243800424/(9935539388775*Pi^2) + 70415723512/(7719921825*Pi) - (140831447024*Log[2])/(7719921825*Pi^2), etc. With my HolonomicFunctions program, I can compute a recurrence for s[n]; it is of order 4 and reads as follows:

(4882907610000*n + 45409645769940*n^2 + 182645407267146*n^3 + 415999962341523*n^4 + 577982802817866*n^5 + 454895806317386*n^6 + 59225519859538*n^7 - 343166189289461*n^8 - 511988360267166*n^9 - 437806177932076*n^10 - 266108651837064*n^11 - 122546379647520*n^12 - 43866716329280*n^13 - 12318858825600*n^14 - 2711291561728*n^15 - 463107569920*n^16 - 60175779328*n^17 - 5747031040*n^18 - 380086272*n^19 - 15532032*n^20 - 294912*n^21)*s[n] + (-103803348096000*n - 784602554561820*n^2 - 2768105704869738*n^3 - 6026823217096953*n^4 - 9020605003947467*n^5 - 9775342297829008*n^6 - 7842619884164520*n^7 - 4642095266086329*n^8 - 1932925496383987*n^9 - 457611245448706*n^10 + 41209636133904*n^11 + 97240054251968*n^12 + 51453003285792*n^13 + 17442939655872*n^14 + 4286099946496*n^15 + 786501567232*n^16 + 107466914048*n^17 + 10655647232*n^18 + 725864448*n^19 + 30400512*n^20 + 589824*n^21)*s[1 + n] + (416937889589760 + 3554863148398464*n + 14328974200154544*n^2 + 36268879481293680*n^3 + 64595995906430976*n^4 + 85992224535253612*n^5 + 88741243539351209*n^6 + 72676116321097082*n^7 + 47967434809619206*n^8 + 25766469377698904*n^9 + 11327025146369045*n^10 + 4082525146254538*n^11 + 1204382850934684*n^12 + 289244537708488*n^13 + 55994945590928*n^14 + 8602481665120*n^15 + 1023967361344*n^16 + 90980820352*n^17 + 5673752064*n^18 + 221276160*n^19 + 4055040*n^20)*s[2 + n] + (-991927338716160 - 7959959312866704*n - 30438759074620104*n^2 - 73625416232650920*n^3 - 126182888327402802*n^4 - 162759219972139479*n^5 - 163886906596739718*n^6 - 131915392870032037*n^7 - 86229362512951236*n^8 - 46250572451287905*n^9 - 20482816805074942*n^10 - 7510999432218307*n^11 - 2279867079855182*n^12 - 570856754453560*n^13 - 117095161342864*n^14 - 19457792537456*n^15 - 2575687266912*n^16 - 264896044288*n^17 - 20371899392*n^18 - 1100574720*n^19 - 37183488*n^20 - 589824*n^21)*s[3 + n] + (574989449126400 + 4504016604954240*n + 16848977783257440*n^2 + 39941997048959832*n^3 + 67197715675727256*n^4 + 85209617638015468*n^5 + 84466109548900131*n^6 + 67022670913239937*n^7 + 43247189158707820*n^8 + 22929016930240154*n^9 + 10051209082086679*n^10 + 3653373301666929*n^11 + 1100790554316050*n^12 + 274025929788560*n^13 + 55976134921664*n^14 + 9280502487568*n^15 + 1228325908256*n^16 + 126624089216*n^17 + 9789531136*n^18 + 533520384*n^19 + 18259968*n^20 + 294912*n^21)*s[4 + n] == 0

The recurrence allows to compute s[n] exactly for quite high n. For example, it took me about 30s to obtain s[50000] = 0.004272717467855486550885164... (for space reasons I give the rounded value here, the exact expression involves integers with several 10000 digits).

From the initial values and the recurrence it becomes also apparent that each s[n] is a linear combination of {1/Pi^2, Log[2]/Pi^2, 1/Pi, 1} with rational number coefficients. So we could study the four coefficient sequences separately, i.e., c1[n] = coeff of 1/Pi^2 in s[n], and similarly c2[n], c3[n], c4[n]. Clearly, these four sequences satisfy the same recurrence as s[n]. Numerical experiments suggest that c4[n] approaches a constant (which could be 75/14 ?), while c1, c2, c3 seem to diverge. However, their (weighted) sum s[n] converges to a constant, which is close to the above value for s[50000] (I believe that the first 20 decimal places are correct).

We can do all kinds of asymptotics studies with this recurrence, for example, one finds that the solutions to the above recurrence behave like one out of {(-1)^n/n^(23/4), 1/n^(3/4), 1, n^(1/4)} from which we conclude that the sequences c1, c2, c3 grow like n^(1/4). So I expect the following asymptotic picture: c1[n] = a1*n^(1/4) + b1 + l.o.t. (lower-order terms) c2[n] = a2*n^(1/4) + b2 + l.o.t. c3[n] = a3*n^(1/4) + b3 + l.o.t. c4[n] = b4 + l.o.t. We have s[n] = c1[n]/Pi^2 + c2[n]*Log[2]/Pi^2 + c3[n]/Pi + c4[n] so for convergence we have to assume that a1/Pi^2 + a2*Log[2]/Pi + a3/Pi = 0. Then the desired value s[Infinity] would be given by b1/Pi^2 + b2*Log[2]/Pi + b3/Pi + b4

One could use the asymptotic information to get more precise estimates on the involved constants. I don't know if this is a direction that you would like to follow. So please have a look at what I wrote and then let me know what you think.

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