In Casella and Berger (2002) I found a proof for the moment-generating function (mfg) of a lognormal distribution not being existent (see exercise 2.36 on page 81 and the answer provided here on page 2-12).
Starting point is the following lognormal pdf (with $\mu = 0$ and $\sigma^2 = 1$):
$f(x) = \int_0^\infty \frac 1{2\pi x}e^{-(ln(x))^2/2}dx$
The mgf of the lognormal distribution is therefore
$M_x(t) = \int_0^\infty \frac {e^{tx}}{2\pi x}e^{-(ln(x))^2/2}dx$
In reference to l'Hopital rule, they then point out that
$\lim \limits_{x \to \infty} e^{tx-(ln(x))^2} = \infty.$
They conclude the proof by stating that for any $k > 0$ there is a constant $c$ such that
$\int_k^\infty \frac {e^{tx}}{x}e^{-(ln(x))^2/2}dx\ge c\int_k^\infty \frac 1xdx = c\ln|_k^\infty = \infty.$
Hence, the $M_x(t)$ does not exist.
I think I get the proof. Key is when $t>0$ and $x \to\infty$ then $e^{tx}$ tends to blow up (also see here).
What I do not get is what constant $c$ is supposed to represent. It must be related to $e^{tx}e^{-(ln(x))^2/2}$. However, why can one integrate out any term based on $e^{tx}e^{-(ln(x))^2/2}$ if it is dependent on $x$? Can we integrate out such a term if if we evaluate it accordingly? For example, something along the lines of $()|_k^{\infty}$?
Best Answer
The $c$ represents a positive real number such that $$\color{blue}{e^{tx}e^{-(\ln x)^2/2} \ge c\quad \forall x\ge k}$$ (the $c$ can depend on $t$, but is independent of $x$). Make sure you can explain/show why such a $c$ exists! (It's related to the fact that for any $t>0$, we have $e^{tx}e^{-(\ln x)^2/2} \to \infty$ as $x\to \infty$.) Because of this inequality, since $k > 0$, we get the integral inequality obtained in your post.