Re: MeijerG
- To: mathgroup at smc.vnet.net
- Subject: [mg95188] Re: [mg95159] MeijerG
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 9 Jan 2009 06:23:54 -0500 (EST)
- Reply-to: hanlonr at cox.net
Seems consistent with its series and plots
f[c_, x_] := -Log[x^2] - Sqrt[Pi]*
MeijerG[{{0}, {}}, {{0, 0}, {1/2}}, x^2/(4*c)];
Limit[f[c, x], x -> 0]
Log[1/c] + 2*EulerGamma
Series[f[c, x], {x, 0, 1}] // Normal //
FullSimplify[#, Element[{c, x}, Reals]] &
Piecewise[{{Log[1/c] + 2*EulerGamma, c != 0 && x >= 0}},
Log[1/c] - 2*I*Pi + 2*EulerGamma]
func = Table[f[c, x], {c, 1/2, 3/2, 1/2}];
Plot[Tooltip[func], {x, 0, 1},
Epilog -> {Red, AbsolutePointSize[4],
Point[Table[{0, Log[1/c] + 2*EulerGamma},
{c, 1/2, 3/2, 1/2}]]},
Frame -> True, Axes -> False]
Bob Hanlon
On Wed, Jan 7, 2009 at 6:56 AM , dimitris wrote:
> Can I trust the following result?
>
> In[29]:= Limit[-Log[x^2] -
> Sqrt[Pi]*MeijerG[{{0}, {}}, {{0, 0}, {1/2}}, x^2/(4*c)], x -> 0]
> (*c>0*)
>
> Out[29]= 2*EulerGamma + Log[1/c]
>
> Another well known CAS gave
>
>> limit(-ln(x^2)-sqrt(Pi)*MeijerG([[0], []],[[0, 0],
>> [1/2]],1/4*x^2/c),x = 0);
>
>> infinity
>
>
> Regards
> Dimitris