Re: Re: definite integration of 1/a

*To*: mathgroup at smc.vnet.net*Subject*: [mg86457] Re: [mg86416] Re: definite integration of 1/a*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 11 Mar 2008 05:34:08 -0500 (EST)*References*: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <fr2mik$o6g$1@smc.vnet.net> <200803110755.CAA24701@smc.vnet.net>

It seems ot me that your post contains quite an interesting observation that has not been made explicit. Note the following, (to me) surprising fact: In[90]:= Simplify[Integrate[1/a, {a, 1, ablah}], ablah > 0] Out[90]= log(ablah) but, (here is what I find surprising): Simplify[Integrate[1/a, {a, 1, ablah}, Assumptions -> ablah > 0], ablah > 0] If[ablah > 1, Log[ablah], Integrate[1/a, {a, 1, ablah}, Assumptions -> Inequality[0, Less, ablah, LessEqual, 1]]] In other words, in this particular case, providing the Assumption ablah>0 within Integrate actually prevents Simplify from getting the more general result. We can see how this comes about. Without the Assumption ablah>0 Integrate gives a general complex result: Integrate[1/a, {a, 1, ablah}] If[Re[ablah] >= 0 || Im[ablah] != 0, Log[ablah], Integrate[1/a, {a, 1, ablah}, Assumptions -> ! (Re[ablah] >= 0 || Im[ablah] != 0)]] Note tha condition Re[ablah]>=0 which, of course is satisfied whenver ablah >0. But, with the Assumption ablah>0 inIntegrate we get a less general real result: Integrate[1/a, {a, 1, ablah}, Assumptions :> ablah > 0] If[ablah > 1, Log[ablah], Integrate[1/a, {a, 1, ablah}, Assumptions -> Inequality[0, Less, ablah, LessEqual, 1]]] so this is one of the (I think rather rare) cases when Integrate does better by working in the complex plane and then interpreting the result on the real line than by working on the real line form the start. Andrzej Kozlowski On 11 Mar 2008, at 08:55, David W.Cantrell wrote: > "Chris Chiasson" <chris at chiasson.name> wrote: >> I meant that I was expecting Log@a > > Surely you really meant that you were expecting Log[ablah]. > > How about the following?: > > In[2]:= Simplify[Integrate[1/a, {a, 1, ablah}], ablah > 0] > > Out[2]= Log[ablah] > > But I must say that I'm slightly surprised that (using version 6.0.2), > > In[1]:= Integrate[1/a, {a, 1, ablah}, Assumptions -> ablah > 0] > > Out[1]= If[ablah > 1, Log[ablah], > Integrate[1/a, {a, 1, ablah}, Assumptions -> 0 < ablah <= 1]] > > did not give simply Log[ablah]. > > David > >> 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/ >

**References**:**Re: definite integration of 1/a***From:*"David W.Cantrell" <DWCantrell@sigmaxi.net>