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MathGroup Archive 2002

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Integration yields different results... Why?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35397] Integration yields different results... Why?
  • From: Binesh Bannerjee <binesh-dated-1026891966.8faddf at hex21.com>
  • Date: Thu, 11 Jul 2002 05:24:01 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hi.
	I'm wondering if I haven't understood something here...

Let me get the mathematica script that's giving me trouble out of the way.
F[a_, b_, c_, d_, e_, X_] := (a + b X + c X^2)Exp[-(d + e X)^2]
I0 = Simplify[
      Integrate[F[a, b, c, d, e, X]*(Exp[X]*f - g), {X, h, Infinity}, 
        Assumptions -> {Im[e] == 0}]];
F2[Params_] := 
  Simplify[
	Integrate[
		F[a,b,c,Apply[d,Params],Apply[e,Params],X]*(Exp[X]*f-g),
		{X, h, Infinity}, 
		Assumptions -> {Im[Apply[e, Params]] == 0, Im[e] == 0}
	] /. {
		Apply[e, Params] -> e,
		Apply[d, Params] -> d
	}
  ]
Arguments = {{z}, {y, z}, {x, y, z}, {w, x, y, z}}
Map[F2[#] - I0 &, Arguments]

I've cut and pasted that, and it works, so, you can try that.

Basically, I have the original function, then I make the parameters to the
function be functions whose parameters do _NOT_ include X _or_ a,b,c,d,e,f,g,h.
and, then, I substitute out the functional parameters fromthe output
via the /. operator.

For some reason, it works fine up to a 3 parametered function.
(You see the final output is { 0,0,0,If[h<0,(* Ridiculous expression *),
Integrate[...]] }
If I make it a 4 parametered function, then it gives me a _different_
answer, EVEN if I strip out the conditional in the If[(h<0)]...
Why is this happening? Can I just assume that this is a bug in Mathematica?
Would it be ok for me to just do this:
	FourParams = Integrate[F[a,b,c,d,e,X]*(Exp[X]*f-g),{X,h,Infinity}] /. {
		d->d[w,x,y,z],
		e->e[w,x,y,z]
	}

Is that functionally identical?

I'm using Mathematica 4.0.0.0. Unfortunately, I can't as of now afford
to upgrade to 4.2.0.0 just now...

Any help appreciated,
Thanks,
Binesh Bannerjee


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