I want to show that, for every $t\neq 0$, the moment generating function of the standard Cauchy distribution is equal to $+\infty$, i.e.
$$M_X(t) = \int_{-\infty}^{+\infty} \frac{e^{tx}}{1+x^2}dx = +\infty$$
I'm really rusty in term of probabilities and integrals, but here is my first try, could anyone validate (or not) this? \begin{align*} \int_{-\infty}^{+\infty} \frac{e^{tx}}{1+x^2}dx &= \int_{-\infty}^{0} \frac{e^{tx}}{1+x^2}dx + \int_{0}^{+\infty} \frac{e^{tx}}{1+x^2}dx \\ & \geq \int_{0}^{+\infty} \frac{(e^{x})^t}{1+x^2}dx\\ & \geq \int_{0}^{+\infty} \frac{1}{1+x^2}dx\\ &= \lim_{x\rightarrow +\infty} \arctan(x)\\ &= +\infty \end{align*}