Re: definite integration of 1/a

*To*: mathgroup at smc.vnet.net*Subject*: [mg86410] Re: definite integration of 1/a*From*: dh <dh at metrohm.ch>*Date*: Tue, 11 Mar 2008 02:54:39 -0500 (EST)*References*: <fr2mi9$o63$1@smc.vnet.net>

Hi Chris, a is only a dummy variable, therefore, it can not appear in the result. Log[ablah] is o.k. However, it is not clear to me why the second part is not simply: -Log[ablah], but instead left unevaluated. hope this helps, Daniel Chris Chiasson 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. >