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?