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MathGroup Archive 2004

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Question about Integrate and If

  • To: mathgroup at smc.vnet.net
  • Subject: [mg46615] Question about Integrate and If
  • From: "Antonio González" <SPAMgonferNO at esi.us.es>
  • Date: Wed, 25 Feb 2004 13:07:08 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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



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