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.
The integral equation \begin{equation} \int_1^x \frac{F(u)}{\sqrt{x^2-u^2}} \, du = g(x) \end{equation} is a classical variant of the Abel equation: \begin{equation} f(s)=\int_1^s\frac{y(t)\,dt}{\sqrt{s-t}} \end{equation} which has a solution \begin{equation} y(t)=\frac{1}{\pi}\frac{d}{dt}\int_1^t\frac{f(s)\,ds}{\sqrt{t-s}} \end{equation} This result can be checked by pluging this solution in the integral equation.
Changing $t=u^2$ in the Abel integral equation, $2\sqrt{t}y(t)=F(u)$, $s=x^2$ and $f(s)=g(x)$, the problem is then \begin{equation} g(x)=\int_1^x \frac{F(u)}{\sqrt{x^2-u^2}} \, du \end{equation} its solution is \begin{align} \frac{F(u)}{2u}&=\frac{1}{\pi}\frac{d}{dt}\int_1^t\frac{g(x)\,ds}{\sqrt{t-s}}\\ &=\frac{1}{\pi}\frac{1}{2u}\frac{d}{du}\int_1^u\frac{2xg(x)\,dx}{\sqrt{u^2-x^2}} \end{align} and thus \begin{equation} F(u)=\frac{2}{\pi}\frac{d}{du}\int_1^u \frac{xg(x)}{\sqrt{u^2-x^2}}\,dx \end{equation} Concerning now the requirement $F(1)=0$, one can show by expressing the derivative, that (see below), for $u\to 1$, $$ F(u)\sim \frac{\sqrt{2}}{\pi}g(u)(u-1)^{-1/2}$$ then the condition imposes $g(u)=o\left(\sqrt{u-1}\right)$.
As an example, taking $g(u)=\tfrac\pi2(u-a)$, we found $$ F(u)=\frac{1}{u}\frac{u^2-a}{\sqrt{u^2-1}}$$ and thus $F(1)=0$ if $a=1$, i.e. when $g(1)=0$. Additionally, in this example, $\lim_{u\to\infty}F(u)=1$.
To evaluate the derivative \begin{equation} \frac{\pi}{2}F(u)=\frac{d}{du}\int_1^u\frac{xg(x)\,dx}{\sqrt{u^2-x^2}} \end{equation} we first change the variable $x=uv$ to expressed \begin{align} \frac{\pi}{2}F(u)&=\frac{d}{du}u\int_{\tfrac{1}{u}}^1\frac{tg(ut)}{\sqrt{1-t^2}}\,dt\\ &=\frac{1}{u}\int_1^u\frac{xg(x)\,dx}{\sqrt{u^2-x^2}}+\frac{1}{u}\int_1^u\frac{x^2g'(x)\,dx}{\sqrt{u^2-x^2}}+\frac{1}{u}\frac{g(1)}{\sqrt{u^2-1}} \end{align} Both integrals vanish when $u=1$, the last term $\sim \frac{g(1)}{\sqrt{2(u-1)}}$ when $u\sim 1$.