On moment generating function of generalized gamma distribution

Question

I'm reading An Intermediate Course in Probability by Gut. I am confused about a statement made concerning the generalized gamma distribution and its existence of a moment generating function.

I quote:

Another class of distributions that possesses moments of all orders but not a moment generating function is the class of generalized gamma distributions whose densities are $$f(x)=Cx^{\beta-1}e^{-x^{\alpha}},\quad x>0,\tag1$$where $\beta>-1$, $0<\alpha<1$, and $C$ is a normalization constant (that is chosen such that the total mass equals $1$).

It is clear that all moments exist, but, since $\alpha<1$, we have $$\int_{0}^\infty e^{tx}x^{\beta-1}e^{-x^\alpha}\, dx=\infty\tag2$$ for all $t>0$, so that the moment generating function does not exist.

First, I don't think it is clear that all moments exist. Integrating $(1)$ and making the substitution $y=x^\alpha$, and rewriting the integral in terms of a gamma integral, I get that $C=\Gamma(\beta/\alpha)$. The only condition I get is $\beta/\alpha>0$ (this is so the gamma integral converges), so I don't see that the restriction $\beta>-1$ makes sense. I also don't see why $\alpha<1$ would make sense. According to Wikipedia, $\alpha$ and $\beta$ are simply positive, but in view of my finding that $\beta/\alpha>0$, could they both be negative too?

Second, why does $(2)$ diverge to infinity for all $t>0$? If we now modify so that $\alpha$ is not bounded above by $1$, does the integral still not converge?

Answer 1

I got $C=\alpha/\Gamma(\beta/\alpha)$.

I think $\beta>-1$ is needed for the $r$th moment existence, where $(\beta+r)/\alpha$ will show up in place of $\beta/\alpha$. For example, if $\beta=-1$ and $r=1$, we'll have problems with the $\Gamma$.

For $\beta$ and $\alpha$ both being negative, it could be a valid distribution, with a very common example being the inverse gamma distribution, frequently used in Bayesian statistics.

For the $\alpha<1$, I think the quote meant that it is only a condition for (2) to be $\infty$, not a condition for existence of moments. When $\alpha<1$, for sufficiently large $x$, the $tx^1$ will dominate $x^\alpha$ in the exponent. Thus, the integrand of (2) will keep increasing in its tale, making the integral $\infty$.