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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg17531] Re: Integrate[1/2+1/2 Erf[z],{z,-inf,0}]
  • From: "Atul Sharma" <mdsa at musica.mcgill.ca>
  • Date: Fri, 14 May 1999 01:13:06 -0400
  • References: <5jf25g$g6k@smc.vnet.net> <7h7sn4$kmv@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Try first loading the package Calculus`Limit` and then evaluating the
integrals, which I think should solve the problem :

In[1]:= <<Calculus`Limit`
In[2]:= Limit[inf(1-Erf[inf]),inf->Infinity]
Out[2]= 0

In[3]:=Integrate[1/2+1/2 Erf[z],{z,-inf,0}]

Out[3]=inf/2 + 1/(2*Sqrt[Pi]) - 1/(E^inf^2*2*Sqrt[Pi]) -
  1/2*inf*Erf[inf]

Atul
----------------------------------------------------------------------------
-------
A. Sharma MD,FRCP(C)
Pediatric Nephrology,
McGill University, Montreal Children's Hospital

Hendrik van Hees wrote in message <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.
>
>--
>Hendrik van Hees Phone:  ++49 06159 71-2755
>c/o GSI-Darmstadt SB 3.162 Fax:    ++49 06159 71-2990
>Planckstr. 1 mailto:h.vanhees at gsi.de
>D-64291 Darmstadt http://theory.gsi.de/~vanhees/vanhees.html
>




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