Hazard rate of a mixed Poisson process. In section 5.4.3 of Ross' Introduction to Probability Models, he introduces the mixed Poisson processes which is a regular Poisson process conditional on some distribution $L(\lambda)$ of the rate, $\lambda$.

Suppose that $L(\lambda)$ is continuous with density function $g(\lambda)$. 
$P\{N(t+s) - N(s) = n \} = \int_0^{\infty} P\{N(t+s) - N(s) = n | L = \lambda \}\ g(\lambda)\ d\lambda = \int_0^{\infty} e^{-\lambda t}\frac{(\lambda t)^n}{n!}\ g(\lambda)\ d\lambda$

In example 5.29, he shows that when $L$ follows a Gamma distribution, $N(t)$ follows a negative binomial distribution.
Now, a regular Poisson process has a constant hazard rate ($=\frac{f(t)}{S(t)}$ where $f(t)$ is the PDF of the interarrival distributions and $S(t)$ is the survival function), $\lambda$ at any time, making it memory-less. For the mixed Poisson process, it seems that the hazard rate as a function of time should become:
$$\lambda(t) = \int_0^{\infty} \lambda\ g(\lambda)\ d\lambda = E(L)$$
It seems to suggest that the hazard rate is still a constant. But this would imply that the inter-arrival distribution of the point events is still exponential. And that would imply a Poisson number of arrivals in any time period, not negative binomial. What am I missing here?
 A: Adding an answer since I derived the expression for the hazard rate through alternate means. Would still be nice to know what the problem with the approach in the question is though.
Since conditional on $L(\lambda)$, it becomes a Poisson process, the inter-arrival time is exponential distributed, the inter-arrival time of this mixed Poisson process is an exponential mixture. The PDF is (if $g_\Lambda$ is the distribution from which $\lambda$ is sampled):
$$f(t) = \int\limits_0^\infty \lambda e^{-\lambda t} g_\Lambda(\lambda)d \lambda$$
And the survival function is:
$$S(t) = \int\limits_0^\infty e^{-\lambda t}g_\Lambda(\lambda d \lambda$$
The hazard rate function then becomes:
$$\lambda(t) = \frac{f(t)}{S(t)}$$
For example, if $g_\Lambda$ is a gamma distribution, we get:
$$f(t) = \int\limits_0^\infty \lambda e^{-\lambda t} \theta e^{-\theta \lambda}\frac{(\theta \lambda)^{m-1}}{(m-1)!}d \lambda = \frac{\theta ^m}{(m-1)!}\int\limits_0^\infty e^{-(t+\theta)\lambda}\lambda^m d \lambda = \frac{\theta^m m}{(t+\theta)^{m+1}}$$
This is the Lomax distribution, meaning the hazard rate function is:
$$\lambda(t) = \frac{m}{t+\theta}$$
Still curious as to why directly integrating the hazard rate function doesn't work.
