I have this product of two hypergeometric functions

Hypergeometric2F1[2 + 3 d5 + 2 k5, 2 + 3 d5 + 2 k5, 4 + 6 d5 + 4 k5, 1 - 1/t^2] 
  HypergeometricPFQ[{d5/2, d5, -(d5/2) - k5}, {1 + d5/2, 1 + (3 d5)/2 + k5}, t^2]

to which I want to apply the conversion rule posted here to a single hypergeometric function, using (as suggested by a comment of Johannes Trost) the values $z=1,c=t^2,d=1-\frac{1}{t^2}$. (Feel free to suggest other values.) I've been trying to do so, and am not sure I have been fully successful--in particular, I've not been certain as to the import/consequence of the Condition operator ( /; ) at the end of the summation. Also, my attempts to perform the indicated infinite summation have not produced any significant results--which may be unavoidable, due to the difficulty of the underlying problem.

In any case, I hope to perform an integration over $t \in [0,1]$ incorporating this hypergeometric product or its equivalent as discussed in this post.

(I changed the $d,k$ symbols in that posting to $d5,k5$ to avoid conflict of symbols.)

Here are the variable/parameter settings I have been using.

r = 2; s = 1; p = 3; q = 2; z = 1; c = t^2; d = 1 - 1/t^2; 
{Subscript[a, 1], Subscript[a, 2], Subscript[a, 3]} = {d5/2, d5, -(d5/2) - k5}; 
{Subscript[b, 1], Subscript[b, 2]} = {1 + d5/2, 1 + (3 d5)/2 + k5}; 
Subscript[α, 1] = 2 + 3 d5 + 4 k5; 
Subscript[α, 2] = 2 + 3 d5 + 4 k5; 
Subscript[β, 1] = 4 + 6 d5 + 4 k5;

closed as unclear what you're asking by Sektor, halirutan, MarcoB, Coolwater, Henrik Schumacher May 28 '18 at 7:38

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  • $\begingroup$ Please, post a self-contained snippet of code for others to play with. If I have to piece together something that supposedly works from 4 different places I would be discouraged to say the least. $\endgroup$ – Sektor May 25 '18 at 20:40
  • $\begingroup$ Variable/parameter settings for what? Pretend that you know nothing about hypergeometric functions, but you are a Mathematica power user. Now, read your Q. Would you be able to piece the information sufficient to solve the problem? Doubt it! Describe succinctly what you have as a given, your assumptions and what you are trying to prove/compute. $\endgroup$ – Sektor May 25 '18 at 22:28
  • $\begingroup$ As a suggestion -- Avoid Subscript and Superscript when using Mathematica. Please, refer to the help centre on how to properly format your Qs. $\endgroup$ – Sektor May 25 '18 at 22:30

The /; is not at the end, but in the middle. The stuff to the right is a (very complicated) condition that must be satisfied for the transformation to proceed.

Note that the code at http://functions.wolfram.com/ isn't real Mathematica code, but Mathematica-like pseudocode.

  • $\begingroup$ But can't one employ the Input Form and/or Standard Form posted on the page to get "real Mathematica code"? I'll add some code to my question above, setting the values of the various parameters/variables, to hopefully make implementation more accessible.. $\endgroup$ – Paul B. Slater May 25 '18 at 22:09

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