Following the color intensity problem I'm trying to solve.
I came up with the conditional probability (given the ratio of intensities $x_0$ what is the probability that the next ratio $X_1$ will be, in case two balls are still of different colors) $$ \mathbb{P}(X_1 < x | X_0 = x_0) = \begin{cases} 2 \cdot \int\limits_0^1 \int\limits_0^{xy}\frac{1}{x_0+1} \cdot \frac{1}{x_0+1} \mathop{\mathrm{d}x}\mathop{\mathrm{d}y},\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\, x_0 >x \\ 2 \cdot \left[\int\limits_0^{x_0 / x}\int\limits_0^{xy} \frac{1}{x_0+1} \cdot \frac{1}{x_0+1} \mathop{\mathrm{d}x}\mathop{\mathrm{d}y}+\int\limits_{x_0/x}^{1}\int\limits_0^{x_0}\frac{1}{x_0+1} \cdot \frac{1}{x_0+1} \mathop{\mathrm{d}x}\mathop{\mathrm{d}y}\right],\qquad x_0 \leq x \end{cases} $$
where $2$ - comes from two ways of assigning different colors to two balls. W.l.o.g. I assumed blue ball previous intensity is $1$, and red ball had intensity $x_0$, i.e. new intensities should be uniformly picked from $[-1, x_0]$.
Solving, $$ \mathbb{P}(X_1 < x | X_0 = x_0) = \begin{cases} \frac{x}{(x_0+1)^2},\qquad\qquad\qquad\,\,\, x_0 >x \\ \frac{2}{(x_0+1)^2} \left[x_0 - \frac{x_0^2}{2x}\right],\qquad x_0 \leq x \end{cases} $$
Taking a derivative I can calculate a pdf, $$ f_{x_0}(x) = \begin{cases} \frac{1}{(x_0+1)^2},\qquad\,\,\,\, x_0 > x \\ \frac{x_0^2}{x^2(x_0+1)^2},\qquad x_0 \leq x \end{cases} $$
Now, denote the expected number of steps until both balls are colored in the same color as $g(x_0)$ for the initial ratio of intensities $x_0$. Then we get an equation $$ g(x_0) = 1 + \int\limits_0^{\infty} f_{x_0} (x) g(x) \mathop{\mathrm{d}x} $$
My question is, how can I solve this integral equation to get $g(x)$?