As a part of independent research, I saw that the Cauchy distribution is a useful example of a case, where the characteristic function exists, but the mgf does not exist.
I wanted to show that for the Cauchy distribution, the MGF does not exist. To do this, I reasoned that $$ \begin{split} M(t) = \mathbb{E}\left[e^{tX}\right] &= \int_{-\infty}^\infty \frac{e^{tx} dx}{\pi \left(1+x^2\right)} \ge \frac{1}{\pi} \int_2^\infty \frac{e^{tx} dx}{1+x^2} \ge \frac{1}{\pi} \int_2^\infty x^{-3} e^{tx} dx \end{split} $$ but I am having trouble showing the last integral diverges. Is this the right track, and if so, how do I complete the argument, or is there an easier way to do this?
I know that one can prove that all the moments don't exist (like in this question), but the problem is, that only shows derivatives at zero don't exist, not that the function does not. This is exactly what happens with the characteristic function -- it exists, and even is differentiable everywhere except at zero...
Thanks for your help.
More simply, you can argue that the integrand is unbounded, because $\displaystyle \lim_{x\rightarrow+\infty} \frac{e^{tx}}{1+x^2} = +\infty$ when $t>0$, while $\displaystyle \lim_{x\rightarrow-\infty} \frac{e^{tx}}{1+x^2} = +\infty$ when $t<0$.
Nevertheless, the moment generating function exists at $t=0$, with $M_X(0) = \mathbb{E}[1] = 1$, in order to ensure that the Cauchy distribution is normalized.