Re: Problem with change of variables in an integral

*To*: mathgroup at smc.vnet.net*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)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <l069n5$atc$1@smc.vnet.net>

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