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MathGroup Archive 1997

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Re: Integrate product of Gaussian*Sin

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6973] Re: [mg6855] Integrate product of Gaussian*Sin
  • From: Richard Finley <trfin at fiona.umsmed.edu>
  • Date: Wed, 30 Apr 1997 22:26:01 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

>Date: Thu, 24 Apr 1997 10:46:08 -0500
>To: James Perry <perryj at rpi.edu>
>From: Richard Finley <trfin at fiona.umsmed.edu>
>Subject: Re: [mg6855] Integrate product of Gaussian*Sin
>
Oops!   Sorry!... when I wrote down this integral on paper I neglected to
include the translation part of the argument in the Gaussian (a), and as
several people pointed out, this Gaussian is not even except for a = 0.
Thanks for keeping me honest.
RF

James,
>
>I am a little confused because there is an end bracket missing in your
equation.  I presume that you mean the integral:
>
>Integrate[Exp[-alpha*(x-a)^2]*Sin[n Pi x/L],{x,-Infinity,Infinity}]
>
>If this is the integral you are interested in there is no need to change
variables because it is the integral of a product of odd and even functions
over the real line and is therefore identically zero for all values of the
parameters. 
>
>hope that helps.
>
>regards,   RF 
>
>
>
>At 02:44 AM 4/24/97 -0400, you wrote:
>>Hi,
>>
>>	Can anyone suggest a good change of variables to carry out the
>>integration
>>
>>	Integrate[Exp[-alpha*(x-a)^2*Sin[n Pi x/L],{x,-Infinity,Infinity}]
>>
>>	I can't find this form in my integral tables (I'm going to check
>>the library today for a more comprehensive list, I might find a form that I
>>can convert my expression to), and Mathematica can't find a solution unless
>>I take the limits of the integral {x,-c,c}. However, treated as an improper
>>integral
>>
>>	Integrate[Exp[-alpha*(x-a)^2*Sin[n Pi x/L],{x,-c,c}]
>>	Limit[%,c->Infinity]
>>or	Limit[%,c->-Infinity]
>>
>>	Still does not give a solution, since the answer to the integral
>>(with limits {x,-c,c}) is a combination of Erf[x] and Erfi[x], and the
>>Limit[Erfi[x],x->+/- Infinity]->+/- Infinity. The Erf[x] has a limit of +/-
>>1 as x->+/- Infinity.
>>	I'm not sure if there is a solution to this, anyone with
>>Gaussian-type function experience?
>>
>>Thank you
>>Jim
>>
>>
>>
>>
>



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