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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


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