Proving that the lognormal distribution has no moment generating function I need to prove that the lognormal distribution has no mgf, that is,
$\int_0^\infty \frac{e^{tx}}{\sqrt{2\pi}x}e^\frac{{-(ln(x))^2}}{2} = \infty$.
What is the best way to start this off?
 A: For every $t\gt0$, $tx-\frac12(\ln x)^2-\log x\gt0$ for every $x$ large enough, say every $x\geqslant \xi(t)$, hence
$$
\int_0^\infty\frac{\mathrm e^{tx}}{\sqrt{2\pi}x}\mathrm e^{-(\ln x)^2/2}\mathrm dx\gt\int_{\xi(t)}^\infty\frac1{\sqrt{2\pi}}\mathrm dx=+\infty.
$$
A: Hint: when $t > 0$ and $x$ is large, the most important part of the integrand is
the $e^{tx}$, which blows up.  You might, for example, show that, for any fixed $t > 0$, when $x$ is
large enough, $e^{tx/2}/ x > 1$ and $e^{tx/2} e^{-(\ln x)^2/2} > 1$.  
A: There's another cool way to prove this, let's see....
With $c\in\mathbb{R^+},\:X$ ~ LN$(\mu,\sigma^2)\:$ we have $\:Y=\ln(X)$ ~ N$(\mu,\sigma^2)\:$
$M_X(t)=E\large[$$e^{tX}\large]$$=E\large[$$e^{te^Y}\large]=$
$\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R}}}\normalsize e^{\large{^{te^{^{y}}}}}e^{^{-\large\frac{1}{2}(\frac{y-\mu}{\sigma})^2}}$d$y$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R}}}\normalsize e^{\large{^{te^{^{y}}}}}e^{^{-\large\frac{1}{2\sigma^2}(y^2-2\mu y+\mu^2)}}$d$y$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R^+}}}\normalsize e^{\large{{t{(1+y+y^2/2+y^3/6)-\frac{1}{2\sigma^2}(y^2-2\mu y+\mu^2)}}}}$d$y$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sim c\LARGE\int_{_{\Large\mathbb{R^+}}}\normalsize e^{\large{^{y^3}}}$d$y\:\longrightarrow \:\infty$
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