       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
>
>
>
>

```

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