2D Integral, Gaussian and 2 Sinc Functions

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  • #1
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Main Question or Discussion Point

I am looking for help with the following integral

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

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

(pls forgive my poor latex)

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

The real-valued constants [tex] a, b, c, [/tex] and [tex] d [/tex] are positive.

My original idea was to switch to coordinates [tex] w = x+y [/tex] and [tex] u=x-y [/tex] but I can not get pass the sinc functions...any help would be appreciated.
 
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  • #2
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beautiful1 said:
I am looking for help with the following integral

[tex] A = \int dx \int dy \exp(-a (x+y)^2 +ib(x-y)) sinc(cx+dy) sinc(dx+cy) [/tex]
I have now transformed coordinates using

[tex] w = x + y [/tex]

[tex] u = x - y [/tex]

[tex] dw du = -2 dx dy [/tex]

and I get

[tex]
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)
[/tex]

where I define

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

Noting that

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

I find

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

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 [tex] Ei(x) [/tex], 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.
 
  • #3
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In case anyone cares

beautiful1 said:
I find

[tex]
4A= \int dw e^{-a w^2} \cos[2\alpha x] \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2}
-\int dw e^{-a w^2} \int du e^{i \beta u} \frac{\cos[2\beta u]}{(\alpha w)^2 - (\beta u)^2}
[/tex]
For the first integral over u

[tex]
\begin{equation*}
\begin{split}
I_{1}(x) &= \int du \frac{e^{i \beta u}}{(\alpha w)^2 - (\beta u)^2} \\
\\
& = e^{-i \alpha w} \int du \frac{e^{i \beta u}}{(\beta u)(\alpha w - \beta u)} \\
\\
& = \frac{e^{-i \alpha w} }{2\alpha w} \int du e^{i \beta u}
\left( \frac{1}{\beta u} + \frac{1}{\alpha w + \beta u} \right)
\end{split}
\end{equation*}
[/tex]

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

Then

[tex]
\begin{equation*}
\begin{split}
\int_{-\infty}^{\infty} du \frac{e^{iu}}{u} &= \int _{0}^{\infty} du \frac{e^{iu}}{u} + \int _{-\infty}^{0} du \frac{e^{iu}}{u} \\
\\
& = \int _{0}^{\infty} du \frac{e^{iu}}{u} - \int _{0}^{\infty} du \frac{e^{-iu}}{u} \\
\\
&= \frac{1}{2i} \int _{0}^{\infty} du \frac{\sin u}{u}
\end{split}
\end{equation*}
[/tex]

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

[tex]
\int _{0}^{\infty} du \frac{\sin u}{u} = \frac{\pi}{2}
[/tex]

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

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