Prove that there is a unique continuous solution to the following integral equation. I am trying to prove that there is a unique continuous solution to the integral equation $$F(\alpha) = \int_{0}^{\alpha}F\left(\frac{t}{1-t}\right)\frac{dt}{t}; \qquad F(\alpha)=1 \text{ for } \alpha\geq 1; \qquad F(\alpha)=0 \text{ for } \alpha\leq 0. $$
I believe this will involve using the contraction mapping theorem, but I am unsure how to go about this. I have tried to approach it by considering the cases where $\alpha\geq\frac{1}{2}$ and $\alpha<\frac{1}{2}$ separately but have had no luck so far.
Thanks
 A: Set $t = \frac{1}\alpha$ (as suggested in the wolfram link from comments), and define $F(1/t) = \rho(t)$. To define $F$ on the interval $[0,1]$, we need to define $\rho$ on the interval $[1,\infty)$. We see that the equation to solve is now
$$ \rho(t) = F(1/t) = \int_0^{1/t} \rho\left(\frac1s -1\right) \ \frac{ds}s$$
Making the change of variables $S = 1/s$ so that $dS = -ds/s^2$ i.e. $dS/S = ds/s$,
$$\rho(t) = \int_t^\infty \rho(S-1) \frac{dS}S $$
Since we formally have $\rho(1) = F(1) = 1$, we can write the above as
$$ \rho(t) = \left(\int_1^\infty - \int_1^t\right) \rho(S-1) -\frac{dS}S = 1 - \int_1^t \rho(S-1) \frac{dS}S. $$
We have reached a reasonable integral formulation for $\rho$(The usual delay ODE for Dickman,  $\rho'(t) = -\rho(t-1)/t$), written in integral form
$$ \rho(t) = \rho(n) - \int_{n}^t \frac{\rho(S-1)}S dS, \quad t\in[n,n+1],\quad n=1,2,3,\dots$$
that can be iteratively used to define $\rho$, since we have the initial condition $$\rho\big|_{[0,1]} = F\big|_{[1,\infty]} = 1.$$
Writing out the induction explicitly: Define $\rho_n : [n,n+1] \to\mathbb R$ inductively by $\rho_0(t) = 1$ for $t\in[0,1]$, and inductively solve for $t\in[n,n+1],$
\begin{align} 
 \rho_n'(t) &= -\rho_{n-1}(t-1)/t, \\
 \rho_n(n) & = \rho_{n-1}(n).
\end{align}
Each $\rho_n : [n,n+1]\to\mathbb R$ is uniquely (by the Fundamental Theorem of Calculus) given by $$\rho_n(t) =\rho_{n-1}(n) + \int_{n}^t \rho_{n-1}(S-1)\frac{dS}S.$$
Now define $\rho(t) := \rho_n(t)$ for each $t\in[n,n+1]$. Since $\rho_{n}(n) = \rho_{n-1}(n)$, we have defined a function $\rho$ on all of $\mathbb R$ that is continuous (even at the integers $n$):
$$
\rho(t) = 
\begin{cases}
1 & t\in[0,1]\\
1 - \log t & t\in[1,2]\\
1-\log(2) -\int_2^t (1- \log(S-1))\frac{dS}S & t \in [2,3]\\
\qquad \vdots & \vdots
\end{cases}$$
To finish, we need to verify that $F(0)= \rho(\infty)=0$. One can  inductively show that $\rho$ is decreasing from the ODE $\rho' = -(\dots)$, so that $$\lim_{t\to\infty} \rho(t) = \lim_{\substack{ n \to \infty \\ n\in\mathbb N}} \rho(n).$$ One also inductively obtains the inequalities coming from $\rho(n+1) = \rho(n) - \int_n^{n+1} \frac{\rho(S-1)}{S}dS$,
$$ \rho(n+1) \le \rho(n) - \frac{\rho(n)}{n+1} = \rho(n)\left( 1 - \frac1{n+1}\right) , \quad n\ge 1$$
$$\rho(n+1)\ge \rho(n) - \rho(n-1) \frac1n \ge \rho(n-1)\left(1-\frac1n\right),\quad n\ge 2$$
The second inequality gives $\rho\ge 0$ on $[0,\infty)$. The first gives
$$\rho(n+1) \le \rho(1) \prod_{m=1}^n \left(1 - \frac1{m+1}\right) =  \prod_{m=1}^n \left(1 - \frac1{m+1}\right) =: a_n$$
So we are done if we can show $a_n\to 0$. This holds since $a_n = \exp\left(\sum_{m=1}^n \log \left(1 - \frac1{m+1}\right) \right)$ and
$$ \sum_{m=1}^n \log \left(1 - \frac1{m+1}\right) \le \sum_{m=1}^n \frac{-1}{m+1} \xrightarrow[n\to\infty]{} -\infty.$$
Therefore, we finally have $\rho(\infty) = 0$, which completes the construction.
