## Integral equation

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.)

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An immediate problem with this is that your integral diverges at the upper limit. – Michael Renardy Oct 8 at 15:01
I'm only interested in $E$ belonging to a bounded interval and also in recovering $g$ on this interval. If it makes any difference I know $g\ge 0$ in my particular case. – Alex A Oct 9 at 6:38
We had a similar problem in arxiv.org/pdf/1201.0393.pdf and had to build a small "theory" of some junk like that because we couldn't find anything of value in any textbook. I guess our equation (26) plus the preceding differentiation tricks plus the considerations in the Appendix will give you an idea of what should be done here (the answer is "Yes" in any decent function class; we did it for $C^\infty$ only, but that's merely because we were lazy). – fedja Oct 23 at 23:05
By the way, Alex, it looks like we have at least 12 users with the same username as yours on MO. You may consider modifying your name a bit to be recognizable at the first glance... – fedja Oct 23 at 23:19
Thank you very much, Fedja! I will take a look in that paper of yours immediately. I will also consider changing my user name as I was not aware of this problem. Thanks. – Alex A Oct 24 at 9:28
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