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Re: definite integration of 1/a

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86411] Re: definite integration of 1/a
  • From: David Bailey <dave at Remove_Thisdbailey.co.uk>
  • Date: Tue, 11 Mar 2008 02:54:50 -0500 (EST)
  • References: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <fr2mik$o6g$1@smc.vnet.net>

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/
> 
> 
> 
> 
Actually, I think you meant Log@ablag :)

Try:

Integrate[1/a,{a,1,ablah},Assumptions:>ablah>0,GenerateConditions->False]

David Bailey
http://www.dbaileyconsultancy.co.uk


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