Fourier transform of n-th power of autocorrelation of a random process

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I'm having troubles in understanding how Fourier transform of the n-th power of a time function is obtained. In particular I came across to a particular result with respect to the calculation of the power spectral density of a log-normal process, that is calculated as

$$ S_{\lambda \lambda} (f) = \int_{-\infty}^{+\infty} e^{2m_3+\sigma_3^2(1+r_{\nu_3\nu_3}(\tau))}e^{-j2\pi f \tau} d\tau $$

where $r_{\nu_3\nu_3}(\tau)$ is a Guassian. The exponential in the previous expression is thus expressed as series, and the following result is obtained: $$ S_{\lambda \lambda} (f) = e^{2m_3+\sigma_3^2} \cdot \left[\delta(f) + \sum_{n=1}^{\infty} \frac{\sigma_3^{2n}}{n!} \frac{S_{\nu_3 \nu_3}\left(\frac{f}{\sqrt{n}}\right)}{\sqrt{n}} \right] $$

I was wondering if the following is always true, or if it is a specific property of Gaussians:

$$\mathfrak{F}\{(r_{\nu_3 \nu_3}(t))^n\} = \frac{S_{\nu_3 \nu_3}\left(\frac{f}{\sqrt{n}}\right)}{\sqrt{n}} $$

Thanks.

M.

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Note that the Fourier transform of the $n^{th}$ power of a function is equivalent to the function's Fourier transform convolved $n$ times with itself. Since the Fourier transform of a Gaussian is Gaussian, and since a Gaussian convolved with itself remains Gaussian, the result is true for Gaussians but not in general.