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Re: Problem with change of variables in an integral

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  • Subject: [mg131605] Re: Problem with change of variables in an integral
  • From: Alexei Boulbitch <Alexei.Boulbitch at>
  • Date: Tue, 10 Sep 2013 03:47:36 -0400 (EDT)
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Dear Alexei,

Integrate[...] in Map[...] will be evaluated first before Map can apply ReplaceAll[...], so nothing is done by Map.

There is a typo in
Integrate[E^(i \[Phi])/(1 + (E^(i \[Phi]) r)^3), {r, 0, \[Infinity]}]
i needs to be changed to I. Then it will produce the same result.

Transforming integrals by change of variables is not straightforward in Mathematica, especially if the integral can be evaluated in closed form. One method involves wrapping the integral with Hold or HoldForm so that evaluation is not done. In addition, simple replacement of the integration variables will not work and the integrand must be divided by the Jacobian corresponding to the variable transformation.

Another method is to write a customized code based on the box language of Mathematica to interpret the 2-D form of integrals, which does not evaluate the integrals but does so only when asked. This is what is done in the SymbolicComputing package (, WTC 2011). The commands are:


SCTransInt[\[Integral]f[z] \[DifferentialD]z,
  TransVar -> {z, r, z == r E^(I \[Phi])},
  ReplVar -> {r, 0, \[Infinity]}] /. \[Phi] -> 2 \[Pi]/3

It is to be noted that once the package is loaded, integrals, products and sums in 2-D form are not evaluated. Evaluation can be done using the functions SCEvalInt, SCEvalProd, and SCEvalSum, respectively. Kernel functions Integrate, Product, and Sum are the same and can be used for immediate evaluation.


Youngjoo Chung

Dear Youngjoo,

Thank you for explanations. It is very helpful. I would like to ask you to give few more details.

First of all, you write that the Integrate statement evaluates before the other ones. Since it is a very important information please give me kindly a reference to the place in the documentation, where this is written.

Second, it would be very helpful, if in the light of the above statement you explain the result of the following:

Clear[f, \[Phi]];
Map[ReplaceAll[#, {z -> r*Exp[I \[Phi]], \[Phi] -> 2 \[Pi]/3}] &,
 Integrate[f[z], {z, 0, \[Infinity]}]]

\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(f[
\*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r] \[DifferentialD]\((
\*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r)\)\)\)

Thank you for the pointing out my stupid mistake with i instead of I, this, indeed, explains the result I got. The reference to your package is also very helpful. Thank you.

Best regards, Alexei

Alexei BOULBITCH, Dr., habil.
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