Evaluate integral with exponential and polynomial How can I show that
\begin{align}
\int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\infty
\end{align}
for $t\neq0$.
I started as follows:
\begin{align}
\int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\int_{-\infty}^0 e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx+\int_{0}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx.
\end{align}
Here I can 'see' that the right integrand goes to infinity, and the left one would just be a real value I'd say, because the integrand goes to 0.
Is there a way for me to evaluate these integrals in a somewhat rigorous way?

Suppose first that $t>0$. You have $$\lim_{x\to\infty}\frac{e^{tx}}{\pi(1+x^2)}=\infty.$$
As the integrand is positive everywhere, 
\begin{align}
\int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx\geq
\int_0^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\infty.
\end{align}
For $t<0$, you use that $$\lim_{x\to-\infty}\frac{e^{tx}}{\pi(1+x^2)}=\infty,$$
and now
\begin{align}
\int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx\geq
\int_{-\infty}^0 e^{tx}\frac{1}{\pi(1+x^2)}\,\mathrm dx=\infty.
\end{align}
