Electric field Fourier decomposition

I have the following decomposition for the electric component of light:

$$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{E}(\vec r)=\frac1{4\pi^2} \iint_\Omega \vec A(k_x, k_y) \mathrm{e}^{i \vec{k} \cdot \vec{r}} \, \mathrm{d}k_x \mathrm{d}k_y.$$

Similarly, the magnetic field is: $$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{H}(\vec r)=\frac1{4\pi^2} \iint_\Omega \frac{\vec k}{\omega \mu_0} \times \vec A(k_x, k_y) \mathrm{e}^{i \vec{k} \cdot \vec{r}} \, \mathrm{d}k_x \mathrm{d}k_y.$$

Fine, this is the setting. Now I wish to compute the average Poynting vector $\langle \vec S \rangle$:

$$\langle \vec S \rangle = \frac12 \operatorname{Re} [\vec E(\vec r) \times \vec H(\vec r)^*].$$ Is there a way to express $\langle \vec S \rangle$ in a nice form? Integrals of functions of $\vec A$ for example. I get horrendous expressions with convolutions that don't give me a nice compact formula.

I know that this might actually be more mathematics than physics, but it might be possible that the physical boundary conditions give a better solution.

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I'll check Jackson's electrodynamics book and see if it says anything relevant, though if you are set on expressing $\langle S\rangle$ in terms of the momentum-space vector potential, it's possible you might not be able to do any better than a convolution. –  David Z Mar 2 '12 at 6:34
@David: I want an expression that is the least computationally expensive. If that has a convolution, then so be it :-). –  Jonas Teuwen Mar 2 '12 at 12:01
Convolution sucks, computationally; but there are always fourier transform tricks to save time. –  Colin K Mar 17 '12 at 19:43

You could potentially simplify this slightly further, were there any relations between $\bb{A}$ and $\bb{k}$. Going back to the coordinate expression for $\bb{S}$ will require another integration and will be no simpler than the original form you had given.