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