Re: definite integration of 1/a

*To*: mathgroup at smc.vnet.net*Subject*: [mg86411] Re: definite integration of 1/a*From*: David Bailey <dave at Remove_Thisdbailey.co.uk>*Date*: Tue, 11 Mar 2008 02:54:50 -0500 (EST)*References*: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <fr2mik$o6g$1@smc.vnet.net>

Chris Chiasson wrote: > I meant that I was expecting Log@a > > On Sun, Mar 9, 2008 at 9:30 PM, Chris Chiasson <chris at chiasson.name> wrote: > >> In[1]:= >> Integrate[1/a,{a,1,ablah},Assumptions\[RuleDelayed]ablah>0] >> >> Out[1]= >> \!\(If[ablah > 1, Log[ablah], >> Integrate[1\/a, {a, 1, ablah}, >> Assumptions \[Rule] ablah \[LessEqual] 1]]\) >> >> I was expecting 1/a. Is there something I can do to get that? Thank you. >> >> -- >> http://chris.chiasson.name/ > > > > Actually, I think you meant Log@ablag :) Try: Integrate[1/a,{a,1,ablah},Assumptions:>ablah>0,GenerateConditions->False] David Bailey http://www.dbaileyconsultancy.co.uk