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Topic: Problem with change of variables in an integral
Replies: 3   Last Post: Sep 12, 2013 11:11 PM

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Murray Eisenberg

Posts: 16
Registered: 4/14/10
Re: Problem with change of variables in an integral
Posted: Sep 5, 2013 8:11 AM
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First, I don't see how you can get the stated "correct result" unless your original function f[z] is specified -- presumably, to be 1/(1+z^3).

Second, is Phi a fixed number 2 Pi/3 or is it a new variable (so that you're attempting to write the original single integral as a double integral)? At one place you seem to be setting Phi -> 2 Pi/3, yet in your "correct result" you show what seems to be a variable Phi.

On Sep 3, 2013, at 11:34 PM, Dr. Robert Kragler <> wrote:

> Hello,
> Although I know how to make a change of variables in an integral I can only do
> it manually by applying a substitution rule to the integrand and the
> differential e.g
> {f[z],\[DifferentialD]z}//. {z-> r E^(I
> \[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi])/3}
> But it cannot applied this substitution rule directly to the integral, e.g.
> Integrate[f[z],{z,0,\[Infinity]}] //. {z-> r E^(I
> \[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi])/3}
> Comparing with the correct result, the exponential factor E^((2 I \[Pi])/3) =
> (-1)^(2/3) is missing in the evaluation of the integral. The correct appearance
> of the
> integral is : Integrate[1/(1+r^3) E^((2 I \[Pi])/3),{r,0,\[Infinity]}]
> How can I force Mathematica (V8) to perform the correct transformation of
> variables as regards to the integral (and not to its separate parts of it as
> {f[z],\[DifferentialD]z} ?
> Any suggestions are appreciated.
> Robert Kragler
> --
> Robert Kragler
> Email :
> URL :

Murray Eisenberg
Mathematics & Statistics Dept.
Lederle Graduate Research Tower
University of Massachusetts
710 North Pleasant Street
Amherst, MA 01003-9305

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