       • 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.

However, If I ask the same integral to Mathematica 5.0,
I obtain

If[a == 0, Log, 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,
but if a!=0 (which is the logical assumption) instead of
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, 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

If[a!= 0, Log,Indeterminate]

or something like that?

Antonio

```

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