A distribution is sub-Gaussian with parameter $\sigma^2$ if and only if $$\mathbf{E} e^{tX} \leq e^{\frac{\sigma^2t^2}{2}}$$ for all $t$ real.
I need to use this characterization to show that the square of a standard normal distribution (ie a chi-square distribution with one degree of freedom) is not sub-Gaussian. Its pdf is $$f(x) = \frac{1}{\sqrt{2 \pi x}} e^{-x/2}$$
The LHS of the characterization becomes $$\mathbf{E} e^{tX} = \int_0^\infty e^{tx} f(x) dx = \frac{1}{\sqrt{2 \pi}} \int_0^\infty \frac{e^{(t-1/2)x}}{\sqrt{x}}dx$$ And I get stuck because the integral does not seem to converge.
Oh, it converges; the integrand diverges at $0$, but in a harmless way, like $\int_0^\epsilon x^{-1/2}dx=2\epsilon^{1/2}$ for $\epsilon\ge0$. The Gamma function $\Gamma(s):=\int_0^\infty x^{s-1}e^{-x}dx$ satisfies $\Gamma(1/2)=\sqrt{\pi}$. You can easily show $\Bbb Ee^{tX}$ is $(1-2t)^{-1/2}$ for $t<\tfrac12$, otherwise it's infinite. Needless to say, this is not the behaviour of a sub-Gaussian MGF.