Infinite integrals of Exp[-A x^2 + 2 B x]

• To: mathgroup at smc.vnet.net
• Subject: [mg68603] Infinite integrals of Exp[-A x^2 + 2 B x]
• From: AES <siegman at stanford.edu>
• Date: Fri, 11 Aug 2006 04:40:31 -0400 (EDT)
• Organization: Stanford University
• Sender: owner-wri-mathgroup at wolfram.com

```Infinite integrals of the general form  Exp[-A x^2 + 2 B x]  with
Re[A]>0 are common in many problems, can be integrated analytically, and
also evaluate pretty rapidly in Mathematica, at least when expressed in
that simple form.

But if you try to evaluate this same integral with even slightly more
complicated (but still constant) expressions in place of the parameters
A and B, the evaluation seems to get remarkably slow.

Consider for example the following inputs:

1)  An offset 2D (but separable) gaussian function:

f[x_,y_,x1_,y1_] = (1/(Pi w^2)) Exp[-((x-x1)/w)^2-((y-y1)/w)^2]

2)  Its 2D (but of course still separable) Fourier transform:

g[xt_,yt_,x1_,y1_] = Exp[-Pi^2w^2(xt^2+yt^2)] Exp[-I 2Pi(x1 xt+y1 yt)]

3)  The overlap integral of this transform with another offset gaussian:

c[x1_, y1_, x2_, y2_] =
Integrate[g[xt, yt, x1, y1] f[xt, yt, x2, y2],
{xt, -Infinity, Infinity}, {yt, -Infinity , Infinity},
Assumptions -> Re[(1/w^2) + Pi^2 w^2] > 0]

This final integral, even with the Re[ ] assumption specifically set
forth, takes *over a minute and half* to evaluate in Mathematica 5.1 on
a Mac iBook G4 running OS 10.3.9 -- and it took substantially longer to
evaluate it the first time through, when I was trying to determine the
necessary assumption needed to get the convergent solution.

anyway) on this fairly commonplace calculation.

I could of course simplify the constants in the integrand "by hand", for
example by converting to normalized variables, and also break the
separable 2D integration into two 1D integrations "by hand".

But, hey!  Am I supposed to be making life easier for Mathematica?  Or
is Mathematica supposed to make life easier for me?!?

```

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