# 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:

$$g(f) = [1-(1-f)^{2}] + 2\int_0^{1-f} (1+g\left(\dfrac{f}{1-x}\right))x\,dx$$

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I think I've $\LaTeX{}$'d this up correctly - if not then feel free to revert the edit! – Clive Newstead Sep 20 '12 at 15:33
that is perfect. thank you! – RodericDay Sep 20 '12 at 15:33
If you take the derivative on both sides with respect to $x$ you get an intresting property of $g$. – mick Sep 20 '12 at 15:45
the fact that there is an initial condition suggested this is the approach, but I just cannot make it work – RodericDay Sep 20 '12 at 15:51
I see now how wrong i was. Feel silly :) – mick Sep 21 '12 at 20:01

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.

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this looks like the right direction I think the miraculous cancellations are a feature, not a bug haha – RodericDay Sep 20 '12 at 16:25
Im a bit confused. – mick Sep 20 '12 at 16:41