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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?

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    It should suffice to notice that the first integral on your right-hand side has a non-negative integrand on its domain of definition, so is non-negative, while $e^{tx}/(\pi(1+x^2)) \ge x$ for large enough $x > 0$ (exponentials dominate all polynomials), so the second integral diverges to infinity. – 2017-01-31
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    @Mr.Chip Ohhh, of course, thank you. – 2017-01-31

2 Answers 2

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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}

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If the result holds for all $t > 0$, then it holds for all $t < 0$ using the $u$-substitution $u = -x$. So we can assume $t > 0$. In that case, the mean value theorem gives $e^{tx} > tx$ for all $x \ge 0$. Thus

$$\int_{-\infty}^\infty e^{tx}\frac{1}{\pi(1 + x^2)}\, dx \ge \int_0^\infty e^{tx}\frac{1}{\pi(1 + x^2)}\, dx > \frac{t}{\pi}\int_0^\infty \frac{x}{1 + x^2}\, dx = \frac{t}{\pi}\lim_{b\to \infty} \frac{1}{2}\log(1 + b^2) = \infty.$$