Re: Evaluating a Meijer G-function

*To*: mathgroup at smc.vnet.net*Subject*: [mg69468] Re: Evaluating a Meijer G-function*From*: "sashap" <pavlyk at gmail.com>*Date*: Wed, 13 Sep 2006 04:02:37 -0400 (EDT)*References*: <ee64o4$7pj$1@smc.vnet.net>

Hi Raul, In the current development version of Mathematica Limit can prove this. Series expansion about t==0 is In[2]:= Series[g[t], {t, 0, 1}] Out[2]= (-1)^Floor[(Pi + 2*Arg[t])/(2*Pi)]* (Floor[(Pi + 2*Arg[t])/(2*Pi)]*SeriesData[t, 0, {I*Pi}, 1, 2, 1] + SeriesData[t, 0, {(-2*EulerGamma + Log[4] - 2*Log[t] - 3*PolyGamma[0, 1/2] + PolyGamma[0, 3/2])/2}, 1, 2, 1]) Oleksandr Pavlyk Special Functions Developer Wolfram Research Raul Martinez wrote: > I have the following special case of Meijer's G-function: > > g = MeijerG[{{1/ 2, 1/ 2}, {1}}, {{0, 0, 0}, { }}, 4 / t^2] / (2 Pi), > where t is real. > > When I evaluate it numerically for a sequence of decreasing small t, > 0 < t < 1, it is clear that the value approaches 0 as t -> 0. > > But neither > > N[g /. t -> 0] > > nor > > Limit[g, t -> 0] > > yields the result that g = 0. > > Can anyone show that g -> 0 as t -> 0? > > I've consulted functions.wolfram.com, mathworld.wolfram.com, and many > other web sites and reference works, to no avail. > > Thanks in advance. > > Raul