Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2013

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Problem with change of variables in an integral =>

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131606] Re: Problem with change of variables in an integral =>
  • From: "Dr. Robert Kragler" <kragler at hs-weingarten.de>
  • Date: Wed, 11 Sep 2013 03:49:31 -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: <6618272.26257.1378266502354.JavaMail.root@m06> <000501cea929$4f3aa600$edaff200$@comcast.net>
  • Reply-to: kragler at hs-weingarten.de

To David Park

first of all thank you for your reply as regards to the problem of change of 
variables in an integral. I tried to reply already on Sept. 4 but my email 
returned as undeliverable although the email address is correct. Somehow, my 
email address must be on a "black mailing list" on Park's mail server?!

Thus, it may be a little bit unusual to reply to this private communication 
through MathGroup but the problem is (and it occurred not the first time) that 
all email replies to David Park directly return with a "Delivery Status 
Notification (Failure). I hope that through MathGroup this email will be read by 
David Park.

Regards
Robert Kragler


Am 04.09.2013 06:43, schrieb djmpark:
> Hi Robert,
>
> You could try the Presentations Student's Integral section. (Do you know
> what f is? It won't fully evaluate without knowing that.) In any case try:
>
> integrate[f[z], z]
> % // ChangeIntegralVariable[r -> z/ E^(I (2 \[Pi])/3), z,
>    DirectSubstitution -> z -> r E^(I (2 \[Pi])/3)]
>
>
> David Park
> djmpark at comcast.net
> http://home.comcast.net/~djmpark/index.html
>
>
>
> From: Dr. Robert Kragler [mailto:kragler at hs-weingarten.de]
>
>
> 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/mathematica
>
>
web/kragler




  • Prev by Date: Re: Problem with change of variables in an integral
  • Next by Date: Re: Integrating special functions
  • Previous by thread: Re: question about ndsolve
  • Next by thread: Incorrectly Cropped ListPlot