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Re: Re: definite integration of 1/a

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86407] Re: [mg86398] Re: definite integration of 1/a
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 11 Mar 2008 02:54:06 -0500 (EST)
  • References: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <200803100705.CAA24798@smc.vnet.net>

Surely, you meant you were expecting Log[ablah]. You can get it by  
giving Integrate one of the two mutually exclusive assumptions, which  
together amount to the one that you want to give:

Integrate[1/a, {a, 1, ablah}, Assumptions :> 0 < ablah < 1]
Log[ablah]
Integrate[1/a, {a, 1, ablah}, Assumptions :>
      1 <= ablah]
Log[ablah]

The only way I know of getting this answer without giving one of these  
partial assumptions is by not using assumptions at all:

Integrate[1/a, {a, 1, ablah}, GenerateConditions -> False]
log(ablah)

Andrzej Kozlowski




On 10 Mar 2008, at 08:05, Chris Chiasson wrote:

> 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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