Using $ to designate "integral from -infinity to + infinity," the
Gaussian integral:

$dx e^(-.5Ax^2) (1)

is of course central to Quantum Field Theory, Fourier transforms, etc.

Can someone please point me to the definite integral:

$dx e^[-.5(Ax^2+By^2)] (2)

Even better, the integral:

$dx e^(-.5Ax^2 +Jx) (3)

is also important in QFT.

Can someone please point me to:

$dx e^[-.5(Ax^2+By^2)+Jx+Ky] (4)

I refer to these as "elliptical" Gaussians because Ax^2+By^2 is, of
course, the heart of the equation for an ellipse.

These would have application in QFT when there are two fields, x and y,
in the same action.

Thanks,

Jay.
_____________________________
Jay R. Yablon
Email:

In article ,
Jay R. Yablon wrote:
Using $ to designate "integral from -infinity to + infinity," the
Gaussian integral:

$dx e^(-.5Ax^2) (1)

is of course central to Quantum Field Theory, Fourier transforms, etc.

Can someone please point me to the definite integral:

$dx e^[-.5(Ax^2+By^2)] (2)

I think I must be misunderstanding what you're looking for here.
You can just pull the exp(-By^2 / 2) outside of the integral, and
you're left with (1).


Even better, the integral:

$dx e^(-.5Ax^2 +Jx) (3)

For this one, you complete the square -- that is, you find a
substitution x=X+c such that the argument of the exponential
is -A X^2 / 2 + B for some constant B. then pull exp(B) out
of the integral, and you're done.

is also important in QFT.

Can someone please point me to:

$dx e^[-.5(Ax^2+By^2)+Jx+Ky] (4)

This is like (2): you just pull the y-dependent bits out of the integral,
and the remaining integral is (3).

In (2) and (4), do you perhaps mean integrals with respect to both
x and y, rather than just integrals with respect to x (as you've written)?
If so, what I've said still applies; you just do the integral
with respect to x first, as I've described, and then the remaining
expression will be a similar-looking Gaussian integral over y.

The still more general way to think about these cases is to look for a
linear or affine transformation (x,y) - (X,Y) that turns the
elliptical contours of the integrand into circular countours centered
on the origin.

-Ted

--
[E-mail me at , as opposed to .]

wrote in message
...

In article ,
Jay R. Yablon wrote:

.. . .

In (2) and (4), do you perhaps mean integrals with respect to both
x and y, rather than just integrals with respect to x (as you've
written)?

Yes, that was a typo.

If so, what I've said still applies; you just do the integral
with respect to x first, as I've described, and then the remaining
expression will be a similar-looking Gaussian integral over y.


This integral is arrived at to begin with, for one variable Ax^2, by
squaring the integral and using a dummy variable y, then passing into
polar coordinates, and taking the square root of the result. Jx comes
in when one completes the square. What I suspected, and what has now
been confirmed to me by replies here and elsewhere, is that one can
simply multiply one result times another. That is, derive for Ax^2 +
Jx, derive for By^2 + Ky, then multiply the results together for the
Ax^2 + Jx + By^2 + Ky result. Very simple in the end.

Thanks,

Jay.

On Jun 21, 8:32 am, "Jay R. Yablon" wrote:
Using $ to designate "integral from -infinity to + infinity," the
Gaussian integral:

Can someone please point me to:

$dx e^[-.5(Ax^2+By^2)+Jx+Ky] (4)

You may want to check out Appendix A of Zee's QFT book or the Appendix
to Chapter 9 in Weinberg's QFT vol.1. Similarly, many other QFT books
will have a section on evaluating Gaussian integrals in general.

This particular integral is 2pi/sqrt(AB)*exp(-J^2/2a-K^2/2b).

Igor