As the title says: I'm interested in the following integro-differential equation. Let $g:(1,\infty) \to [0,1]$ be given, and assume $g$ is smooth. I want to find functions $F:[1,\infty) \to [0,1]$ that satisfy:
i) For every $x \in (1, \infty)$, $\displaystyle\int_1^x \frac{F(u)}{\sqrt{x^2-u^2}} \, du = g(x)$
ii) $F$ is non-decreasing, right continuous, $F(1) = 0$, and $\lim_{y \to \infty} F(y) = 1$, i.e. $F$ is a CDF supported on $(1,\infty)$.
The first thing to try is differentiating, but this doesn't seem to go anywhere: since that the integrand $\frac{F(u)}{\sqrt{x^2-u^2}}$ blows up at $u = x$, we can't (immediately) pass the derivative through the integral.
The next thing is integrating by parts, which yields
$\displaystyle g(x) = \frac{\pi}{2} F(x) - \int_1^x F'(u) \arctan\Big(\frac{u}{\sqrt{x^2-u^2}}\Big) \,du.$
(Some additional assumptions are needed here: for example, that $F$ is differentiable.) Now using the Leibniz integral rule,
$\displaystyle g'(x) = \int_1^x \frac{F'(u)}{x\sqrt{x^2-u^2}} \, du.$
But this doesn't seem to help much.
Are there standard methods for this kind of thing? Is there some way to make sense of passing the derivative inside the integral?
I would be happy to see an example of some $g$ and $F$ that satisfy this: I have no such examples right now.