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Re: definite integration of 1/a
*To*: mathgroup at smc.vnet.net
*Subject*: [mg86443] Re: definite integration of 1/a
*From*: Bill Rowe <readnewsciv at sbcglobal.net>
*Date*: Tue, 11 Mar 2008 03:00:52 -0500 (EST)
On 3/10/08 at 2:04 AM, chris at chiasson.name (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.
If you were expecting a function of a, then you should be doing
an indefinite integral rather than a definite integral, i.e.,
Integrate[1/a, a, Assumptions :> a > 0]
log(a)
Alternatively, if you did want the definite integral then you
upper limit of integration needs to be larger than the lower
limit. So,
Integrate[1/a, {a, 1, ablah}, Assumptions :> ablah > 1]
log(ablah)
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