# Limit of hypergeometric series and gamma function

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

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

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]


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

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


You can check that the plot is correct:

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


• @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. Sep 16 at 14:46

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

f=HypergeometricPFQ[{-1/2},{1/2,m/2},x];
g=HypergeometricPFQRegularized[{-1/2},{1/2,m/2},x];


Using FunctionExpand[g] it is easy to see that

FullSimplify[f/Gamma[m]==g*Sqrt[Pi]*(Gamma[m/2]/Gamma[m])]
(* True *)


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

Limit[g*Sqrt[Pi],m->0]*Limit[Gamma[m/2]/Gamma[m],m->0]
(* -2*x*HypergeometricPFQ[{1/2},{3/2,2},x] *)

• The conversion can be obtained by Solve[ HypergeometricPFQRegularized[{a}, {b, c}, x] == (HypergeometricPFQRegularized[{a}, {b, c}, x] // FunctionExpand), HypergeometricPFQ[{a}, {b, c}, x]] Sep 16 at 14:28