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How to solve, or at least how to proceed to solve, the following equation for $g(u)$

$$\int_0^{\infty} \{1-\cos(2\pi uh)\} g(u)du = (1+h^{\alpha})^{\beta/\alpha} -1?$$

Here $0<\alpha\leq2$ and $-\infty<\beta\leq2$. Any help will be greatly appreciated!

Edit: By following the suggestion in the comment, first I differentiate with respect to $h$ to get $$\int_0^{\infty} \sin(2\pi uh)2\pi u g(u) du = \beta (1+h^{\alpha})^{\beta/\alpha-1} h^{\alpha-1}.$$ Now taking inverse sine Fourier transform gives $$g(u) = \frac{c}{u}\int\limits_0^{\infty}\sin(2\pi uh)(1+h^{\alpha})^{\beta/\alpha-1}h^{\alpha-1}dh,$$ where $c$ is a suitable constant. But, if I am not wrong, $\sin(2\pi uh)(1+h^{\alpha})^{\beta/\alpha-1}h^{\alpha-1}$ is not integrable for all $\alpha$ and $\beta$. Am I making some mistake?

Why do I expect a solution for $g$ of the above equation?: I am trying to find the spectral measure (which exists) of a generalized covaiance function of the form $c\{1-(1+h^{\alpha})^{\beta/\alpha}\}$ using the relation $K(h) = \int\limits_0^{\infty} \{\cos(2\pi uh)-1\}F(du)$ from this paper, equation 3.

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    $\begingroup$ Differentiate wrt to $h$, then take inverse sine Fourier transform? $\endgroup$
    – Nemo
    Commented Nov 7, 2019 at 14:10
  • $\begingroup$ Thanks for the comment @Nemo! But, it seems I got stuck again as I mentioned in the edited version. $\endgroup$
    – Shanks
    Commented Nov 23, 2019 at 23:23
  • $\begingroup$ the function $g(u)$ will contain singular (delta function) contributions at $u=0$; for example, for $\alpha=1$, $\beta=2$ the solution is $$g(u)=(2/\pi^2)u^{-2}[1-\tfrac{1}{2}u\delta'(u)]$$. $\endgroup$
    – Carlo Beenakker
    Commented Nov 24, 2019 at 16:15

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