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Re: definite integration of 1/a

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  • Subject: [mg86443] Re: definite integration of 1/a
  • From: Bill Rowe <readnewsciv at>
  • Date: Tue, 11 Mar 2008 03:00:52 -0500 (EST)

On 3/10/08 at 2:04 AM, chris at (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

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]


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]

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