Re: Question about Integrate and If

*To*: mathgroup at smc.vnet.net*Subject*: [mg46645] Re: Question about Integrate and If*From*: "John Jowett" <John.Jowett at cern.ch>*Date*: Thu, 26 Feb 2004 17:53:43 -0500 (EST)*Organization*: CERN*References*: <c1iqub$69a$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hello, You can get the behaviour you want with: Integrate[1/x, {x, a, 2a}, Assumptions -> a > 0] If you look at the section on "Definite Integrals" in the Mathematica Book or the Further Examples for the built-in function Integrate in the Help Browser you will find some explanation. John Jowett "Antonio González" <SPAMgonferNO at esi.us.es> wrote in message news:c1iqub$69a$1 at smc.vnet.net... > Setting the trivial integral > > Integrate[1/x,{x,a,2a}] > > I obtain, in Mathematica 4.1 the simple result > > -Log[a] + Log[2a] > > that upon simplification gives Log[2]. > > However, If I ask the same integral to Mathematica 5.0, > I obtain > > If[a == 0, Log[2], Integrate[1/x, {x, a, 2*a}, Assumptions -> a != 0]] > > which is really stupid, as it is saying that if a==0, > the integral (that would be singular in that case) is Log[2], > but if a!=0 (which is the logical assumption) instead of > giving the correct answer lets the question unanswered. > In this case, the answer is trivial, but I have found the > same behavior in more complicated integrals. > > The situation is still more complicated if I try to use > assumptions. Making a copy and paste and asking > > Integrate[1/x, {x, a, 2*a}, Assumptions -> a != 0] > > the result is > > If[Im[a] != 0 || Re[a] != 0, Log[2], Integrate[x^(-1), {x, a, 2*a}, > Assumptions -> a != 0 && Im[a] == 0 && Re[a] == 0]] > > that is even worse, as if I assume that a!=0, it is > obvious that Im[a] != 0 || Re[a] != 0 (how could it be > that a != 0 && Im[a] == 0 && Re[a] == 0 ?) > > Can anybody explain this strange behavior? Why does not give > the answer > > If[a!= 0, Log[2],Indeterminate] > > or something like that? > > Antonio > >