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.
 A: A way to simplify the expressions would be to calculate $\bf S$ in momenta space:
$$
\newcommand{\dd}{\mathrm{d}}
\newcommand{\bb}[1]{{\bf #1}}
\bb{S}_p({\bf k}'') = \int e^{-i\bb{k}''\bb{r}}\dd \bb{r}\left(\dfrac{1}{2}\mathrm{Re}(\bb{E}(\bb{r})\times\bb{H}(\bb{r})^*)\right)\\
=\dfrac{1}{4}\int e^{-i\bb{k}''\bb{r}}\dd \bb{r}\left(\bb{E}(\bb{r})\times\bb{H}(\bb{r})^*+\bb{E}(\bb{r})^*\times\bb{H}(\bb{r})\right)\\
=\dfrac{1}{4}\dfrac{1}{16\pi^4 \omega \mu_0}\int \dd \bb{r} \int \dd \bb{k}\int \dd \bb{k}'e^{-i\bb{k}''\bb{r}}[\left((\bb{A}(\bb{k})e^{i\bb{k}\bb{r}})\times(\bb{k}'\times \bb{A}(\bb{k'})^*e^{-i\bb{k'}\bb{r}})+\\
+(\bb{A}(\bb{k})^*e^{-i\bb{k}\bb{r}})\times(\bb{k}'\times \bb{A}(\bb{k'})e^{i\bb{k'}\bb{r}})\right)]
$$
Then we notice that the integration over $\bb{r}$ kills the exponents and provides $\delta$-function terms:
$$
\bb{S}_p({\bf k}'')=\dfrac{1}{4}\dfrac{1}{4\pi^2 \omega \mu_0} \int \dd \bb{k}\int \dd \bb{k}'[\left(\bb{A}(\bb{k})\times(\bb{k}'\times \bb{A}(\bb{k'})^*)\delta(\bb{k}-\bb{k}'-\bb{k}'')+\\
+\bb{A}(\bb{k})^*\times(\bb{k}'\times \bb{A}(\bb{k'}))\delta(\bb{k}'-\bb{k}-\bb{k}'')\right)]\\
=\dfrac{1}{4}\dfrac{1}{4\pi^2 \omega \mu_0} \int \dd \bb{k}'\left[\bb{A}(\bb{k}'+\bb{k}'')\times(\bb{k}'\times \bb{A}(\bb{k'})^*)+\bb{A}(\bb{k}'-\bb{k}'')^*\times(\bb{k}'\times \bb{A}(\bb{k'}))\right]
$$
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.
