I am trying to show that the posterior of the following model is proper. Assuming, $p(\alpha, \beta) \propto 1$, and $n,x, y$ given.
$$ \begin{align} p(\alpha, \beta) & \propto p(\alpha, \beta | n, x) p(y|\alpha, \beta, n, x) \\ & \propto \prod_i^k \bigg[\frac{e^{\alpha + \beta x}}{1 + e^{\alpha + \beta x}} \bigg]^{y_i} \bigg[\frac{1}{1 + e^{\alpha + \beta x}} \bigg]^{n_i-y_i} \end{align} $$
I reason that by virtue of results in analysis (if $f,g$ riemann integrable then $fg$ is riemann integrable), it is enough to show, for $(\alpha, \beta) \in (-\infty, + \infty)$
$$ \int\int \begin{align} \bigg[\frac{e^{\alpha + \beta x}}{1 + e^{\alpha + \beta x}} \bigg]^{y} \bigg[\frac{1}{1 + e^{\alpha + \beta x}} \bigg]^{n-y} \end{align} d\alpha d\beta < \infty $$
I apply the following change of variables,
$$ \begin{align} \theta_1 & = g_1(\alpha, \beta) = \alpha + \beta x \\ \theta_2 & = g_2(\alpha, \beta) = \beta x\end{align} $$
To get,
$$ \frac{1}{x} \int\int \begin{align} \bigg[\frac{e^{\theta_1}}{1 + e^{\theta_1}} \bigg]^{y} \bigg[\frac{1}{1 + e^{\theta_1}} \bigg]^{n-y} \end{align} d\theta_1 d\theta_2$$
I can show that the inner integral is a beta integral (applying a change of variable again), however, I must have the regions of integration wrong, since I would the conclude that for some $K$, I would be left with,
$$ \frac{1}{x} \int K d\theta_2 = + \infty $$
I would greatly appreciate some help.
Origin of the question (Bayesian Data Analysis, 3rd Edition, CH3 P14).