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?

Bernstein Functions: Theory and Appl.by Schilling, Song and Vondraček. (2/2) $\endgroup$