Re: Problem with change of variables in an integral
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- Subject: [mg131572] Re: Problem with change of variables in an integral
- From: "Alexander Elkins" <alexander_elkins at hotmail.com>
- Date: Thu, 5 Sep 2013 08:12:58 -0400 (EDT)
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Dear Dr. Robert Kragler,
Basicly the problem is two-fold. First and foremost Mathematica
performs the Integrate before doing the replacement.
Secondly \[DifferentialD]z is for display only and cannot be accessed.
In order to make it work for you we must use Hold combined with Release.
I have inferred f [z] as follows to get the integrand shown:
In[1]:= With[{f = Function[z, 1/(1 + z^3)]},
f[z] Hold[ D[z, r]] /. z -> E^((2 I \[Pi])/3) r] // Release
Out[1]= E^((2 I \[Pi])/3)/(1 + r^3)
So Integrate[1/(1 + z^3), {z, 0, \[Infinity]}] is the integral in which you
wish to replace z. Here is how to do that:
In[2]:= Hold[Integrate[1/(1 + z^3), {z, 0, \[Infinity]}]] /.
Hold[Integrate[fv_, v_]] :>
Hold[Integrate[Hold[fv D[z, r]] /. z -> E^((2 I \[Pi])/3) r // Release,
v /. z -> r]] // Release
Out[2]= (2 (-1)^(2/3) \[Pi])/(3 Sqrt[3])
Best regards,
Alexander Elkins
"Dr. Robert Kragler" <kragler at hs-weingarten.de> wrote in message
news:l069n5$atc$1 at smc.vnet.net...
>
> 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 at hs-weingarten.de
> URL : http://portal.hs-weingarten.de/web/kragler