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Re: Evaluating a Meijer G-function


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


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