# Deriving spectral measure

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?