SEE EDIT BELOW.

Show that $\displaystyle \int_0^\infty e^{-x^2}$ converges.

$\displaystyle e^{-x^2} < e^{-x}$ for $\displaystyle x>1$.

$\displaystyle \int_0^\infty e^{-x}$ converges to 1.

So, by the Comparison Theorem, $\displaystyle \int_0^\infty e^{-x^2}$ converges.

Does that work? I'm only wary of it because I used the fact that $\displaystyle e^{-x^2} < e^{-x}$ for $\displaystyle x>1$ which doesn't consider x between 0 and 1... Any help?

EDIT: Sorry, I misread the problem! I'm supposed to do $\displaystyle \frac{1}{2}\int_0^\infty e^{-x^2}$ which fixes my issue.