Assume (for definiteness) $g:\mathbb{R} \to \mathbb{R}$ is continuous and that $f$ is defined by
$$
f(E) = \int _0^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g(t) \, dt.
$$
I'm interested in whether $g$ can be recovered assuming we know $f$.
Does anyone know if this type of integrals have been studied before?
For instance I am familiar to the fact that (Riemann-Liouville) integrals of the form
$$
(J^\alpha g)(E) = \frac{1}{\Gamma (\alpha )}\int _0 ^E(E - t)^{\alpha -1}g(t) \, dt
$$
can be inverted when $\alpha $ is a half-integer by using identities of the form $J^\alpha \circ J^\beta = J^{\alpha + \beta }$ and then differentiate.
EDIT: I would just like to point out that I'm not necessarily looking for an explicit inversion formula. If the above equation fits into some general theory which concludes that $g$ can be recovered I'm happy.
EDIT II: I have narrowed the problem down into finding $g_0$ (only depending on $t$) with
$$
\int _1 ^{E-1} \Big ( (E - t)^2 - 1\Big )^{3/2} g_0(t) \, dt = 1, \qquad E>1.
$$
Not sure whether that helps though.
EDIT III: If it helps I actually do know the solution in my particular case is
$$
g(t) = \int _{\{h^{-1}(t)\}} \frac{1}{|\nabla h|}\,dS
$$
for some $h$ for which the gradient never vanishes on $\{h^{-1}(t)\}$. Here $dS$ is surface measure. (The reason I still want to solve the equation is that I know $f$ is a certain invariant and I need to show $g$ is also invariant.)
