Trying to solve for an unknown function, numerically or otherwise I would like so find the function $n(z)$ that solves the following equation:
$$
n(z) = \frac{.2 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}{.5 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}
$$
That is, $n(z)$ is defined in terms of it's own integral.  Even if there's no nice analytic solution to this, Numerical approximations of some sort are good enough.  In fact this is just a parametrized instance of a more general equation I am interested in.  What methods should I be using to find such a solution?  Does Mathematica or Matlab has nice built in routines for this kind of thing?
For your further information I am interested only in $n(z)$ on $[1,\infty]$, and it is required to be between 0 and 1, so those integrals lie between 0 and 1, so $n(z)$ should be monotonically increasing.  It seems like it should be well behaved, but on the face of it Mathematica doesn't like it.

Background: it is the steady state of a dynamic system in a network; think epidemic diffusion.  $n(z)$ is the infection rate among agents with z links, and in the steady state it is
$$
n(z)  = (g + (1-g)t v z )/(g + r + (1-g) t v z ),
$$
where $g$ is the new infection rate, $r$ is the cure rate,and $v$ is the new infections from your peers in a networks, and $f(z)$ is the distribution of that networks; the proportion of peers withg $z$ links.    One such distribution often of interest is power-law, which is 
$$
f(z) = 2 z^{-3}
$$
and t is the infection rate among a random peer, given by
$$
t = \int_1^{\infty} z n(z) f(z)dz,
$$
which, along with a paramtetrization of the above rates, gave the problem above.
Other distributions of interest are geometric, where $f(z) = Log(4) 2^{-z}$.  So we are solving for the steady state level of infection among guys with z peers. 
 A: Well you know how $n(z)$ looks like. Note that $\displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$ is just a number. Call it $a$.
So now we have
$$n(z) = \frac{0.2+0.24 a z}{0.5+0.24 a z} = 1 - \frac{0.3}{0.5+0.24 a z}$$
where $a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$.
Use the above equation to get an equation solely in $a$ which you can solve for numerically.
I have done the solving part below.
$a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz = \int_{1}^{\infty} \frac{1}{z^2} dz - \int_{1}^{\infty} \frac{0.3}{0.5 z^2 + 0.24 a z^3} dz = 1 - \int_{1}^{\infty} \frac{15}{25 z^2 + 12 a z^3} dz$
$$a = \frac{2}{5} + \frac{36}{125}a \log(\frac{25}{12a}+1)$$
You can now try to solve this numerically or plug it in mathematica as I did, which gives me $a \approx 0.673338$.
As you expected, $a \in (0,1)$.
So the function now is $$n(z) \approx 1 - \frac{15}{25+8.080056z} = \frac{10+8.080056z}{25+8.080056z}$$
and the function $n(z) \in (0,1)$, $\forall z \in \mathbb{R}^+$. In fact when $z \in [1,\infty)$, $n(z) \in (\frac{6}{11},1)$.
