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) -- To reply via email subtract one hundred and four

**Follow-Ups**:**Re: Re: definite integration of 1/a***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>