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.
>