Test the improper integral ∫ e-x2 for convergence

Test the following improper integral for convergence:

$\int_{-\infty}^{\infty} e^{-x^2} \, dx.$

The integral converges.

Proof. We have

$\begin{align*} \int_{-\infty}^{\infty} e^{-x^2} \, dx &= \int_{-\infty}^0 e^{-x^2} \, dx + \int_0^{\infty} e^{-x^2} \, dx \\[9pt] &= 2 \int_0^{\infty} e^{-x^2} \, dx &(e^{-x^2} \text{ is an even function})\\[9pt] &= 2 \left( \int_0^1 e^{-x^2} \, dx + \int_1^{\infty} e^{-x^2} \, dx \right). \end{align*}$

The first integral converges since it is a proper integral. For the second integral, since $x \geq 1$ we have $e^{-x^2} \leq e^{-x}$ . But, we know $\int_1^{\infty} e^{-x} \, dx$ converges (by Example #4 on page 418 of Apostol with $s = 0$ ). Hence, we have established the convergence of