Recurrence relation for a function with an integral of the function? Pardon my lack of tex skills, but what is the recommended procedure in the following scenario:
$$g(f) = 1+\int_0^{1-f} g\left(\dfrac{f}{1-x}\right)\,dx$$
I am not sure how to proceed in such a scenario. My expression is more complicated, but that is the gist of the concept I'm struggling with.
also, we know that g(1) = 1
I'm thinking some sort of Leibniz approach, but I'm an engineer by training so I'm out of my depth.
edit: If the above simplification does not have a solution/doesn't lend itself well to an example, here is the actual thing:
$$g(f) = [1-(1-f)^{2}] + 2\int_0^{1-f} (1+g\left(\dfrac{f}{1-x}\right))x\,dx$$
 A: Here's a solution to the simplified version.  There is a large amount of miraculous cancellation so maybe someone can find a more elegant way to attack it.
The derivative of the RHS (with respect to $f$) has two parts (you could view this as an application of the multivariable chain rule):
$$\frac{d}{df} \int_0^{1-f} g\left(\dfrac{f}{1-x}\right)\,dx = -g\left(\frac{f}{1-(1-f)}\right) + \int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{1}{(1-x)}\,dx$$
$$= -1 + \int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{1}{(1-x)}\,dx. $$
One can then apply integration by parts to this new integral using the fact that $$ \frac{d}{dx} g\left(\frac{f}{1-x}\right) = g'\left(\frac{f}{1-x}\right) \frac{f}{(1-x)^2}.$$
This gives
$$\int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{f(1-x)}{f(1-x)^2}\,dx = \frac{1-x}{f} g\left(\frac{f}{1-x}\right) \Bigg|_{x=0}^{1-f} + \int_0^{1-f} \frac{1}{f}  g\left(\frac{f}{1-x}\right)\, dx $$
$$ = 1 - \frac{g(f)}{f} + \frac{1}{f}\left(g(f)-1\right) = 1 - \frac1f.$$
This means that the derivative of the RHS (which is also equal to $g'(f)$) is just $-1/f$, so $g(f) = 1 - \ln f$.  This does seem to satisfy the original (simplified) equation.
