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Re: Integrate[1/2+1/2 Erf[z],{z,-inf,0}]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg21088] Re: Integrate[1/2+1/2 Erf[z],{z,-inf,0}]
  • From: "Stephen P Luttrell" <luttrell at signal.dra.hmg.gb>
  • Date: Sun, 12 Dec 1999 23:51:55 -0500 (EST)
  • Organization: Defence Evaluation and Research Agency
  • References: <5jf25g$g6k@smc.vnet.net> <7h7sn4$kmv@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hendrik van Hees <h.vanhees at gsi.de> wrote in message
news:7h7sn4$kmv at smc.vnet.net...
> This is really a problem Mathematica seems not to be able to solve. The
> problem is that it fails to calculate the limit
>
> inf(1-erf(inf)) for inf->Infinity.
>
> Doing this from hand with help of de L'Hospital's rule gives clearly 0.
> Although I used Analytic->True within the Limit-command it didn't apply
> this rule.


You need to use the Calculus`Limit` package.

In[1]:=Needs["Calculus`Limit`"];

In[2]:=Limit[x(1-Erf[x]),x->Infinity ]

Out[2]=0



--
Stephen P Luttrell
DERA Malvern




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