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Re: definite integration of 1/a

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86410] Re: definite integration of 1/a
  • From: dh <dh at metrohm.ch>
  • Date: Tue, 11 Mar 2008 02:54:39 -0500 (EST)
  • References: <fr2mi9$o63$1@smc.vnet.net>


Hi Chris,

a is only a dummy variable, therefore, it can not appear in the result. 

Log[ablah] is o.k. However, it is not clear to me why the second part is 

not simply: -Log[ablah], but instead left unevaluated.

hope this helps, Daniel





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.

> 




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