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 >