The spectral representation and isotropic covariance functions

Caveat: My apologies if this question is poorly phrased. I am an engineer/computer scientist teaching myself mathematics.

The spectral representation of the covariance function of a second order stochastic process at a spatial location $\mathbf{h}$ (in $d$-dimensions) is given as

$$Cov(\mathbf{h}) \; = \; \int_{\mathbb{R}^d} e^{i \: \overline{\omega} . \mathbf{h}} f(\omega) d\omega$$

where $\overline{\omega}$ is a frequency vector i.e. $\overline{\omega} = (\omega_1 , \cdots , \omega_d)$, $f(\omega)$ the spectral density function and $\mathbf{h}$ is (commonly) the separation vector between two arbitrary locations $\mathbf{x}$ and $\mathbf{y}$ i.e. $\mathbf{x} - \mathbf{y}$ or $\mathbf{y} - \mathbf{x}$, and herein lies the problem; this fact implies that covariance is directional and no longer entirely a function of the separation vector, i.e.

$$Cov(\mathbf{x} - \mathbf{y}) \; \ne \; Cov(\mathbf{y} - \mathbf{x})$$

This appears to rule out the existence of isotropic covariance functions; and worse still, there is no basis for assuming that any two points separated by the same distance $r = \| \mathbf{x} - \mathbf{y} \|$ have the same covarience.

Yet I'm aware that isotropic covariance functions exists and are valid but I cannot explain how the above spectral representation admits them. This is what I need help with.

Edit

I think I've found the symmetry part of the answer. Because

$$e^{i \: \overline{\omega} . \mathbf{h}} \: = \: cos( \overline{\omega} . \mathbf{h} ) + i \: sin(\overline{\omega} . \mathbf{h})$$

$C(\mathbf{h}) = C(-\mathbf{h}) \:\:$ if $\:\:sin(\overline{\omega} . \mathbf{h}) = 0$.

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Thanks Steve. I've already come across your reply to that post. Unfortunately, it does not address my question. – Olumide Dec 31 '11 at 20:07