MMA does not calculate the limit, however the limit exists as seen in the approximative plot.

Limit[HypergeometricPFQ[{-1/2}, {1/2, m/2}, x]/Gamma[m], m -> 0]

How to achieve the limit?

MMA 13.0

  • $\begingroup$ Do you know of a transformation that would show the limit exists? Otherwise this looks like a math question, rather than a MMA one. $\endgroup$
    – MarcoB
    Sep 16, 2022 at 14:14

2 Answers 2


Use this identity from the Wolfram functions website. This reads:

equation = HypergeometricPFQ[{a1}, {b, b2}, z] == 
((2 + 2 b - b2) HypergeometricPFQ[{a1}, {1 + b, b2}, z])/ b + ((-2 - b + b2)/b
 + z/( b (1 + b))) HypergeometricPFQ[{a1}, {2 + b, b2}, z] 
 + ((-2 + a1 - b) z HypergeometricPFQ[{a1}, {3 + b, b2}, z])/( b (1 + b) (2 + b));

To match your case, use the substitution rule (note that the function is symmetric in the two lower parameters):

substitute = {a1 -> -1/2, b -> m/2, b2 -> 1/2};

This means you want the limit of the following expression:

expression = (equation[[2]] /. substitute)/Gamma[m]

enter image description here

Expand the expression around m==0 to leading order, and truncate the result to the leading term:

limit = Series[expression, {m, 0, 0}] // Normal

enter image description here

You can check that the plot is correct:

Plot[limit, {z, -2, 2}]

enter image description here

  • $\begingroup$ @granularbastard Note that the whole purpose of using the identity is to unfold the b parameter into b+1, b+2, b+3 to resolve the degenerate point b->0. This means you should be using the symmetry of the function under interchange of the two b parameters and do the substitution as: a1->m-1/2; b2->1/2+m/2; b->m/2. Then it should go through as you expect. $\endgroup$
    – Kagaratsch
    Sep 16, 2022 at 14:46

Here is a solution with a bit of manual work. Note that HypergeometricPFQRegularized is close to what OP wants. So define


Using FunctionExpand[g] it is easy to see that

(* True *)

Therefore the limit OP is looking for can be obtained as follows:

(* -2*x*HypergeometricPFQ[{1/2},{3/2,2},x] *)
  • 1
    $\begingroup$ The conversion can be obtained by Solve[ HypergeometricPFQRegularized[{a}, {b, c}, x] == (HypergeometricPFQRegularized[{a}, {b, c}, x] // FunctionExpand), HypergeometricPFQ[{a}, {b, c}, x]] $\endgroup$
    – Bob Hanlon
    Sep 16, 2022 at 14:28

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