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Re: definite integration of 1/a


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/




-- 
http://chris.chiasson.name/



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