Answer
The integral $\int_{1}^{\infty} e^{-x^2} dx$ converges to $e$.
Hence, the integral $\int_0^\infty e^{-x^2} dx$ converges as well by the comparison test.
Work Step by Step
The given integral can be re-written as sum of two integral as:
$\int_{0}^{\infty} e^{-x^2} dx=\int_{0}^{1} e^{-x^2} dx+\int_{1}^{\infty} e^{-x^2} dx$
We can see that the second integral above defines an improper integral and so, we compare it with $e^{x}$
For $x\geq 1$, we have: $e^{-x} \geq e^{-x^2}$
and $\int_{1}^{\infty} e^{-x^2} dx \leq \int_1^\infty e^{-x} dx \\=\lim\limits_{R \to \infty}[-e^{-x}]_1^R\\=e-\lim\limits_{R \to \infty} e^{-R} \\=e$.
So, the integral $\int_{1}^{\infty} e^{-x^2} dx$ converges to $e$.
Hence, the integral $\int_0^\infty e^{-x^2} dx$ converges as well by the comparison test.
