Re: Re: definite integration of 1/a

*To*: mathgroup at smc.vnet.net*Subject*: [mg86453] Re: [mg86443] Re: definite integration of 1/a*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 11 Mar 2008 05:33:22 -0500 (EST)*References*: <200803110800.DAA24976@smc.vnet.net>

On 11 Mar 2008, at 09:00, Bill Rowe wrote: > 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) > It seems to me that this is only an illusion that Assumptions do anything in the case of indefinite integration. I can't see how they could really have any relevance since the Risch algorithm cannot make any use of them. Indeed note that In[79]:= Integrate[1/a, a] Out[79]= Log[a] In[80]:= Integrate[1/a, a, Assumptions :> a < 0] Out[80]= Log[a] and even Integrate[1/a, a, Assumptions :> Re[a] == 0 && Im[a] == 0] Log[a] So one would assume that they are being completely ignored. However, this is not quite so: In[85]:= Integrate[1/a, a, Assumptions :> a == 0] Out[85]= ComplexInfinity However, I am sure that this is not really intended to mean anything and is just a side-effect of something else. Assumptions are not meant to work with indefinite integration and it probably should be considered if not exactly a bug then slghly careless design that they actually work instead of producing some syntax error message. Andrzej Kozlowski

**References**:**Re: definite integration of 1/a***From:*Bill Rowe <readnewsciv@sbcglobal.net>