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

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


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

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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. – Luboš Motl May 29 '11 at 5:37
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. – Patrick Da Silva May 29 '11 at 5:47
@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. – Rodrigo Thomas May 29 '11 at 6:05
@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. – Rodrigo Thomas May 29 '11 at 6:22
@Luboš: Simple is in the eye of the beholder. I think if Rodrigo thought it was simple, he wouldn't be asking us. – joriki May 29 '11 at 6:43
up vote 3 down vote accepted

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. – Rodrigo Thomas May 29 '11 at 14:29
@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 $A(k_x,k_y)\exp\left[\mathrm iz \left( k - \frac{k_{x}^{2}+k_{y}^{2}}{2k}\right)\right]$. – joriki May 29 '11 at 14:58
Ok, now I understand what do you do. – Rodrigo Thomas May 29 '11 at 15:14

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