Answer
The integral $\int_{-\infty}^\infty e^{-x^2} dx$ converges by the comparison test.
Work Step by Step
The given integral can be re-written as sum of two integrals as:
$\int_{-\infty}^{\infty} e^{-x^2} dx=\int_{-\infty}^{1} e^{-x^2} dx+\int_{-1}^{1} e^{-x^2} dx+\int_{1}^{\infty} e^{-x^2} dx$
We can see that the first and third integral above defines an improper integral and so, we compare it with $e^{-|x|}$
For $x\geq 1$, or, $x \leq -1$, we have: $e^{-x^2} \leq e^{-|x|}$
and $\int_{1}^{\infty} e^{-x^2} dx \leq \int_1^\infty e^{-|x|} dx $
Now, $\int_{-\infty }^{-1} e^{-|x|} dx =\int_{\infty}^1 e^{-|x|} (-dx) \\=\int_1^{\infty} e^{-|x|} dx\\= \lim\limits_{R \to \infty}[-e^{-x}]_1^R\\=e-\lim\limits_{R \to \infty} e^{-R} \\=e$.
So, the integral $\int_{1}^{\infty} e^{-|x|} dx$ converges to $e$.
Hence, the integral $\int_{-\infty}^\infty e^{-x^2} dx$ converges as well by the comparison test.
