I have a function $f(z) \in [0,\infty]$ that satisfy $-1 < f(z) < 0$.

I would like to find the minimal $\gamma$ that satisfy:

$$ \int_0^{\gamma} f(z)dz = \log(1+f(0))$$

Clearly, I cannot bound such $\gamma$ since $f$ can go to 0 so I can further relax the requirment:

find minimal $\gamma$ such that there exist some $\xi$ and:

$$ \int_{\xi}^{\gamma} f(z)dz = \log(1+f(\xi))$$

That is, if $f$ increases, I can update the goal.

So the worse function $f$ that maximize $\gamma$ should satisfy:

$$ \int_{\xi}^{\gamma} f(z)dz = \log(1+f(\xi))$$

for all $\xi \in [0,\gamma]$.

Using $\int_{\xi}^{\gamma} = \int_{0}^{\gamma}-\int_0^{\xi}$ and: $\int_{0}^{\gamma} = \log(1+f(0))$

I need to find $f$ such that:

$$ \int_0^{\xi} f(z)dz = \log(1+f(0))-\log(1+f(\xi))$$

So if I would find such $f$ where $f(0)$ is the boundery condition, it would be the worse function and the $\gamma$ for that function is the greater such $\gamma$ for all function.

So, finaly, the quastion is: how can I solve this integral equation? or, is it possible to bound the function in some sence?