Autocorrelation function and power spectral density

Question

I want to get the autocorrelation function of the power spectral density of the wind.

This function is defined by: $$s(\omega)=\frac{c_1}{(1+1.5 \times c_2\omega)^{5/3}}$$ $c_1$ and $c_2$ are constants. Any help please.

I specify that the autocorrelation function is done by the integral of : $$\int_{-\infty}^{+\infty} \! s(\omega) \times \exp(I \times \tau \times \omega) \, \mathrm{d}\omega$$

with $-\infty < \tau <+\infty $ and $I^2=−1$

Answer 1

Assumig that $*$ means usual multiplication and $c_2>0$, the following plan should work:

1.) Rescale $c_2\omega =\tilde\omega$

2.) Substitute $\tilde\omega = q-1$

3.) Integrate by parts, choosing $f'(q)=q^{-5/3}, g(q)=e^{i q \tilde{\tau} }$ with $\tilde{\tau}=\tau/c_2$ to lower the degree of the singulartiy at $0$.

4.) Use formula 318 from this link . It can be proven by contour integration methods, using a half circle contour in the (upper/lower) halfplane excluding the branch cut on the positive/negative imaginary axis

Can you get it from here?

Answer 2

I have done this development: assume that : $\omega=\frac{q-1}{c2}$, so the integral becomes: $$I_1 = \int_{-\infty}^{+\infty}\! q^\left( -\frac{5}{3}\right)\times exp\left(i\omega\tau\right) \, \mathrm{d}\omega = \int_{-\infty}^{+\infty}\! q^\left( -\frac{5}{3}\right)\times exp\left(i\tau\frac{q-1}{c2}\right) \, \mathrm{d}\omega.$$ $$=\frac{1}{c2}\times \int_{-\infty}^{+\infty}\! q^\left( -\frac{5}{3}\right)\times exp\left(i\tau\frac{q-1}{c2}\right) \, \mathrm{d}q$$ After substitution of $\bar{\tau}=\frac{\tau}{c2}$, we find: $$I_1 = \frac{1}{c2\times exp(i\bar{\tau})} \int_{-\infty}^{+\infty} \! q^{-\frac{5}{3}}\times exp\left(iq\bar{\tau}\right) \, \mathrm{d}q$$ Can you help me for the rest.