Re: Integrate product of Gaussian*Sin

*To*: mathgroup at smc.vnet.net*Subject*: [mg6982] Re: [mg6855] Integrate product of Gaussian*Sin*From*: Richard Finley <trfin at fiona.umsmed.edu>*Date*: Thu, 1 May 1997 14:48:31 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Jim, In my previous note, I carelessly looked only at a=0 where the integral is identically zero. If you look at the form of the integral, it appears that the answer should depend on a in a periodic fashion (you can see this by graphing numerically for a range of values of a). If you assume that the real part of alpha is greater than zero, and that n/L is real, then you can make the transformation x -> x + a and solve the integral: Integrate[Exp[-alpha*(x)^2]*Sin[n Pi (x+a)/L],{x,-Infinity,Infinity}] which gives for a solution: Sqrt[Pi/alpha] Exp[-n^2 Pi^2/(4 alpha L^2)] Sin[n Pi a/L] which has the expected dependence on "a". I think this is correct this time...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 > > > >