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Re: Problem with change of variables in an integral
Posted:
Sep 5, 2013 8:11 AM


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 <kragler@hsweingarten.de> 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 : kragler@hsweingarten.de > URL : http://portal.hsweingarten.de/web/kragler
 Murray Eisenberg murray@math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower University of Massachusetts 710 North Pleasant Street Amherst, MA 010039305



