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MathGroup Archive 1997

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Re: Integrate product of Gaussian*Sin

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6983] Re: [mg6855] Integrate product of Gaussian*Sin
  • From: schnizer at itp.tu-graz.ac.at (B. Schnizer)
  • Date: Thu, 1 May 1997 14:48:32 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In article <5jrmmf$em3$3 at dragonfly.wolfram.com>, Richard Finley
<trfin at fiona.umsmed.edu> wrote:

> 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
> >
> >
> >
> >

RF's  remark is not applicable. In fact, the integral is different from zero !

You can solve it in MMA 3.0  by the following substitutions:

f = E^(- alpha (x - a) ^2) Sin[n Pi] x/L]

h = f /. x -> y + a

g  =  E^(-y^2 alpha) TrigExpand[Sin[Apart[n Pi (a + y)/L]]]

Integrate[g,{y,-Infinity,Infinity}]


This result agrees with that from Gradsthein Ryshik, Nr. 3.896.4

Yours sincerely

B. Schnizer


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