2D Integral, Gaussian and 2 Sinc Functions

[beautiful1]

I am looking for help with the following integral

A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy)

where sinc(x) =\sin(x) / x for x \neq 0 and sinc(0) = 1

(pls forgive my poor latex)

Either in the indefinite form or with the upper/lower limits at +/-\infty

The real-valued constants a, b, c, and d are positive.

My original idea was to switch to coordinates w = x+y and u=x-y but I can not get pass the sinc functions...any help would be appreciated.

 

[beautiful1]

I am looking for help with the following integral

A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy)

I have now transformed coordinates using

w = x + y

u = x - y

dw du = -2 dx dy

and I get

<br /> A = -\frac{1}{2} \int dw e^{-a w^2} \int du e^{ibu} {\rm sinc}(\alpha w + \beta u) {\rm sinc}(\alpha w - \beta u)<br />

where I define

\alpha = (c+d)/2 and \beta = (c-d)/2

Noting that

<br /> {\rm sinc}[\alpha w + \beta u] {\rm sinc}[\alpha w - \beta u] <br /> = -\frac{1}{2} \frac{(\cos[2\alpha w]-\cos[2\beta u])}{(\alpha w)^2 - (\beta u)^2}<br />

I find

<br /> 4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} <br /> -\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2} <br />

I do not know what to do from here.

I have put each of the integrals over u into "the integrator" at http://integrals.wolfram.com/, which returns an answer in terms of the exponential integral Ei(x), but I am unsure how to get to that point. I have not tried to perform the subsequent integral over w.

Any help would be appreciated.

 

[beautiful1]

In case anyone cares

I find

<br /> 4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} <br /> -\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2} <br />

For the first integral over u

<br /> \begin{equation*}<br /> \begin{split}<br /> I_{1}(x) &amp;= \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} \\<br /> \\<br /> &amp; = e^{-i \alpha w} \int du \frac{e^{i \beta u}}{(\beta u)(\alpha w - \beta u)} \\<br /> \\<br /> &amp; = \frac{e^{-i \alpha w} }{2\alpha w} \int du e^{i \beta u} <br /> \left( \frac{1}{\beta u} + \frac{1}{\alpha w + \beta u} \right)<br /> \end{split}<br /> \end{equation*}<br />

My interest is in the case the upper and lower limits of integration are + and - \infty, respectively.

Then

<br /> \begin{equation*}<br /> \begin{split}<br /> \int_{-\infty}^{\infty} du \frac{e^{iu}}{u} &amp;= \int _{0}^{\infty} du \frac{e^{iu}}{u} + \int _{-\infty}^{0} du \frac{e^{iu}}{u} \\<br /> \\<br /> &amp; = \int _{0}^{\infty} du \frac{e^{iu}}{u} - \int _{0}^{\infty} du \frac{e^{-iu}}{u} \\<br /> \\<br /> &amp;= \frac{1}{2i} \int _{0}^{\infty} du \frac{\sin u}{u}<br /> \end{split}<br /> \end{equation*}<br />

The last integral is the sine integral. Note the integrand is the sinc function. And (somehow, not sure yet, may be another post) for the given limits, this evaluates to

<br /> \int _{0}^{\infty} du \frac{\sin u}{u} = \frac{\pi}{2}<br />

I believe the remaining integrals over u will all evaluate in a similar manner.
As for w...