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I am reposting this question from Cross Validated as I have not received any responses.

While reading this book, I got stuck on page 266 where the authors found the spectral measure $F(du)$ of the generalized covariance function $K(h) = \Gamma(-\alpha/2) |h|^{\alpha}, ~0<\alpha<2.$ $F(du)$ and $K(h)$ are related through the identity $$K(h) = \int\limits_{-\infty}^{\infty} \{\cos(2\pi uh)-1\}F(du).$$ The spectral measure had been derived to be $$F(du) = \pi^{-\alpha-1/2}\Gamma({\frac{\alpha+1}{2}}) |u|^{-\alpha-1}.$$

Any help towards how the authors derived the spectral measure will be highly appreciated. I tried (futilely) differentiating both sides with respect to $h$ and then taking sine Fourier transform.

Edit: So, I have to solve the equation, for $h>0$, $$2\int\limits_{0}^{\infty}\{\cos(2\pi uh)-1\} f(u)du = \Gamma(-\alpha/2) h^{\alpha}~~~~~~~~~~~~~~~(1)$$ of $f(u)$. Therefore, if I put the expression of $f(u)$ in (1), the RHS should be produced. I do not know how to integrate the LHS after putting the expression of $f$, so I checked with specific choices of $\alpha$ (say, $\alpha=1, 1/2$) that indeed the RHS is coming.

Next, I differentiate with respect to $h$ to get $$\int\limits_{0}^{\infty}\sin(2\pi uh)2\pi uf(u)du = c h^{\alpha-1}.~~~~~~~~~~~~~~~(2)$$ In this step, again I put the expression of $f$ in LHS and check for specific choices of $\alpha$ (say, $\alpha=1, 1/2$) that the RHS is still being produced.

Next, I take inverse Fourier sine transform to have $$2\pi uf(u) = c_1\int\limits_{0}^{\infty}\sin(2\pi uh)h^{\alpha-1}dh.~~~~~~~~~~~~~~~(3)$$ Now, in this step, if I put $\alpha=1$ then the RHS diverges. So, it seems that step (3) is wrong. Please advise where am I making a mistake and what should I do instead?

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    $\begingroup$ One can guess the form of $F(u) = c |u|^{-1-\alpha}$ by simple scaling: since $K(\lambda h) = \lambda^\alpha K(h)$, we necessarily have $F(\lambda^{-1} u) = \lambda^{-1 - \alpha} F(u)$. Then, represent $F(u)$ as in formula (1) in this answer and use Fubini's theorem together with the formula for the Fourier transform of a Gaussian to get the constant $c$ right. $\endgroup$
    – Mateusz Kwaśnicki
    Nov 30, 2019 at 7:59
  • $\begingroup$ Thank you very much, @MateuszKwaśnicki! What you have said does make sense to me. But, it would be great if the proof is more methodological without the guess at the starting. Because, otherwise, I have to ask for different functions differently which do not satisfy the $K(\lambda h)=\lambda^{\alpha}K(h)$ relation; for example, in the case of generalized power covariance $K(h) = c\{(1+|h|^{\alpha})^{\beta/\alpha} -1\}, 0<\alpha\leq2, -\infty<\beta<2$. Thanks again! $\endgroup$
    – Shanks
    Nov 30, 2019 at 9:47
  • $\begingroup$ Guessing the answer is one of the most productive methods in mathematics! Anyway, if you do not feel like guessing, you can evaluate the inverse Fourier transform (or the inverse cosine transform, since your $K(h)$ is symmetric) to $K(h)$. As you noticed, this causes some trouble, because $K(h)$ is not integrable. One way around is to use distributional Fourier transform; I do not know a good reference here, though. Another is related to what I wrote before: express various objects as mixtures of Gaussians, and use well-known properties of these. (1/2) $\endgroup$
    – Mateusz Kwaśnicki
    Nov 30, 2019 at 10:56
  • $\begingroup$ In particular, the examples of $K(h)$ that you are interested in, are of the form $\psi(h^2)$, where $\psi$ is a complete Bernstein function. Then $F(u)$ is the Lévy measure of the Lévy process (a subordinate Brownian motion) with characteristic exponent $\psi(h^2)$. This Lévy measure is a mixture of Gaussians, given by the Lévy measure of the subordinator with Laplace exponent $\psi(h)$. One can often find this Lévy measure by analytic extensions. This is discussed in detail, with a lot of examples, in the book Bernstein Functions: Theory and Appl. by Schilling, Song and Vondraček. (2/2) $\endgroup$
    – Mateusz Kwaśnicki
    Nov 30, 2019 at 11:01

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