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I am new to mathematica and I hope this is a simple problem. I have two identical functions that only differ by the variable names:

PolarInverseFT[f_, p_, u_, r_, t_] := 
 Integrate[
  p f Exp[-I 2 π r p Cos[u - t]], {p , 0, ∞}, {u, 0, 2 π} ]

PolarInverseFT2[f_, r_, t_, p_, u_] := 
 Integrate[
  r f Exp[-I 2 π  p r Cos[t - u]], {r, 0, ∞}, {t, 0, 2 π} ]

I would expect them to behave identically. However,

PolarInverseFT[λ^2/(λ^2 + 4 π^2 p^2), p, u, r, t]

returns the error

-I Sin[2 p π r #1] is not a valid variable

But

PolarInverseFT2[λ^2/(λ^2 + 4 π^2 r^2), r, t, p, u]

gives the correct answer. For other f_ both can return an answer without failing, but the answers differ!

Any help would be greatly appreciated.

Michael E2
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JarrodHadfield
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  • Welcome to Mathematica.SE! I suggest the following:
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     –  Mar 06 '15 at 11:20
  • You can format inline code and code blocks by selecting it and clicking the {} button above the edit window. The edit window help button ? is also useful for learning how to format your questions and answers. – Michael E2 Mar 06 '15 at 11:34
  • In general, why including your integration variables in the arguments ? – b.gates.you.know.what Mar 06 '15 at 13:02
  • @b.gatessucks that was my first thought as well. The integration variables appear in the first argument, so this makes sense ( the second and third arguments in each case must be passed in as undefined symbols ). That said, I obtain different results for the two cases, starting with a clean kernel each time. (not the same error though, the first case returns an unevaluated integral w/v9) – george2079 Mar 06 '15 at 15:05
  • @b.gatessucks True, but besides the point. I note that the integrals themselves (executed directly and not through the above definitions) exhibit the same strange behavior. – Sjoerd C. de Vries Mar 06 '15 at 15:22
  • @SjoerdC.deVries There was a question about something very similar, but I can't find it right now. The problem was caused by the lexicographical ordering of the symbol names, as strange as it may seem. Do you remember that question? – Dr. belisarius Mar 06 '15 at 16:28
  • @belisarius I remember two questions that were related, Daniel lichtblau commented there. – Sjoerd C. de Vries Mar 06 '15 at 17:33
  • @SjoerdC.deVries Related 1 and Related 2 – Dr. belisarius Mar 06 '15 at 17:59
  • confirmed the lexicographical ordering of the symbol names being subtracted for the Cos argument is the distinction here. Interestingly changing to Cos[u+t] fixes it. – george2079 Mar 06 '15 at 18:49
  • @george2079 It's more interesting if you also consider that FullForm@Cos[u - t] == FullForm@Cos[t - u] – Dr. belisarius Mar 06 '15 at 19:31
  • @belisarius: And gets even more interesting, if you reverse the integration parameters: All of a sudden, FT2 yields 0, while FT1 runs on and on and on (and then stops without any result, but some (other) General::ivar errors). – Jinxed Mar 07 '15 at 00:02
  • A simpler way to trigger the General::ivar messages: Integrate[Exp[-I Cos[u - t]], {u, 0, 2*Pi}]. The result from this Integrate is correct, but the messages shouldn't be there. – Szabolcs Apr 05 '15 at 19:03
  • I believe that you should remove p_ and u_ from the arguments to PolarInverseFT as they are used as limit variables for the integration. Similarly remove r_ and t_ from the arguments to PolarInverseFT2. I don't think this will fix your problem but I think the code is better. – Jack LaVigne Oct 02 '15 at 21:47

1 Answers1

1

not an answer, just a simpler demonstration of the issue ( version 9.01 )

Integrate[r (1/(1 + 4 Pi^2 r^2)) Exp[-I 2 Pi  r Cos[t - u]],
       {r, 0, Infinity}, {t, 0, 2 Pi}]

enter image description here

In the second case I've just copied the expression and changed t to z .. puzzling. It seems in the first case u has been assumed real and the second not.

george2079
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