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What do I have to do, or what technique do I have to use, to perform the following integral?

$$ \frac{1}{4\pi^{2}}\iint_{-\infty}^{\infty}\mathrm{d}x\mathrm{d}y (x^{2}+y^{2}) \left| E(x,y) \right|^{2} = \dfrac{1}{4\pi^{2}}\iint_{-\infty}^{\infty}\mathrm{d}k_{x}\mathrm{d}k_{y} \left( \left| \frac{\partial A}{\partial x} \right|^{2} + \left| \dfrac{\partial A}{\partial y} \right|^{2} \right)$$

where:

$$ E(x,y) = \iint_{-\infty}^{\infty} A(k_{x},k_{y}) \exp\left[i\left(k_{x}x+k_{y}y+ z \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right)\right] \mathrm{d}k_{x}\mathrm{d}k_{y} $$

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    Dear Rodrigo, if by "performing', you mean just calculating it analytically, well, I am afraid that you will have to say what $A(k_x,k_y)$ is. For a general $A$, the formula above is clearly the simplest explicit way to write the same expression.2011-05-29
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    Maybe if you put us in context, where this horrible integral might have a reason to be, we could be more of use. Usually, shooting 4 integrals in the same expression isn't something nice to see.2011-05-29
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    @LubošMotl $A$ is a general function, as you pointed. Could you please show me the "simplest explicit way" that you said above? I just can't see this step.2011-05-29
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    @PatrickDaSilva well, the l.h.s of the first equation its the spread of the field intensity in a plane normal to the direction of the propagation ($\vec{z}$) of electric field $E(x,y)$. This field could be regarded as a infinitesimal superposition of plane waves.2011-05-29
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    Dear @Rodrigo, what I wrote was that the expressions you wrote are the simplest ways to write the same result if $A$ is a general function - so what you wrote is already a solution. One may write it both in terms of $E$, as an $xy$ integral, or in terms of $A$, as an $k_xk_y$ integral, just like you did. Do you want us to prove the identity on the first line assuming the definition of $E$ on the second line? It's simple - just substitute it and use the usual Fourier integral identities. At any rate, it's not clear what you want.2011-05-29
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    @PatrickDaSilva and, as you could note, the exponential term does not represent a plane wave... Its because I used the paraxial approximation, which says that: $ k_{x}^{2}+k_{y}^{2} \ll k^{2} $.2011-05-29
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    @Luboš: Simple is in the eye of the beholder. I think if Rodrigo thought it was simple, he wouldn't be asking us.2011-05-29
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    @Rodrigo: It would be much clearer if you wrote $\partial A/\partial k_x$ instead of $\partial A/\partial x$.2011-05-29
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    @Rodrigo: The exponential term does represent a plane wave -- only in $x$ and $y$ and not in $z$, but since your integrals are only over $x$ and $y$, $z$ and $k_z$ are effectively just parameters that occur in the Fourier transform of $E(x,y)$.2011-05-29
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    @Rodrigo: Unless my answer is mistaken, the identity in the first line is just an approximation. This being a math site and not a physics site, it would seem preferable to write it as such.2011-05-29
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    @joriki sorry about this misunderstood... I just thought that here would be more constructive.2011-05-29
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    @Rodrigo: Sorry, I don't understand your above comment. Could you please rephrase it?2011-05-29
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    @joriki I asked here because my biggest problem was mathematical, not physical. I didnt understand what property had been used to manipulate the integral. So, I thought that here was the best place to show my doubt.2011-05-29
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    @joriki Otherwise, is there any way to move this question to physics.se?2011-05-29
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    @Rodrigo: Sorry, that was a misunderstanding. I didn't mean that you should have posed the question at a physics site; merely that, this being a math site, you should be more rigorous in distinguishing between equations and approximations. Since the first line is an approximation (if I'm not mistaken), it seems preferable to write it as such, not as an equality. Sorry about the misunderstanding.2011-05-29
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    @Rodrigo: About migrating to physics.se, I think the community moderators might be able to do that; us "normal" moderators have only two options for suggesting migration, stats.se and meta.math.se -- I'm not sure why that is; perhaps someone who knows might comment? However, again, I wasn't suggesting such a migration.2011-05-29

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Assuming that what you're asking is how to get the identity in the first line, given the Fourier decomposition in the second line: Note that multiplying by $x$ in real space corresponds to applying $\mathrm i\partial/\partial k_x$ in Fourier space. You can write $x^2|E(x,y)|$ as $|xE(x,y)|^2$, and then use Parseval's theorem to replace this by the integral over the squared magnitude of the Fourier transform, where the Fourier transform is given by

$$\mathrm i\frac{\partial}{\partial k_x}\left(A(k_x,k_y)\exp\left[\mathrm iz \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right]\right)\;.$$

Differentiating $A$ gives you the term you want in the result, and differentiating the other factor gives a factor

$$\frac{\partial}{\partial k_x}\left(k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)=\frac{k_x}{k}-\frac{k_x}{k}+\frac{(k_x^2+k_y^2)k_x}{2k^3}=\frac{k_x}{2k}\frac{k_x^2+k_y^2}{k^2}\;.$$

I guess the last term is dropped because of your approximation $k_{x}^{2}+k_{y}^{2} \ll k^{2}$?

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    don't you forget a piece of the argument of the exponential? The complete argument is: $i\left(k_{x}x+k_{y}y+z\left(k−\dfrac{k^{2}_{x}+k^{2}_{y}}{2k}\right)\right)$. Then the last term will not vanish.2011-05-29
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    @Rodrigo: No. I was treating the entire double integral in the second line of your question as the Fourier decomposition of $E(x,y)$. The factor $\exp[\mathrm i(k_xx+k_yy)]$ is part of the Fourier decomposition itself, so it doesn't appear in the Fourier transform; the Fourier transform is just $A(k_x,k_y)\exp\left[\mathrm iz \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right]$.2011-05-29
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    Ok, now I understand what do you do.2011-05-29