Proving this integral converges

**paupsers** · February 9th 2010, 08:22 AM

SEE EDIT BELOW.

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

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

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

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

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

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

**Krizalid** · February 9th 2010, 08:52 AM

well, we simply split this thing on two sets: $[0,1]$ and $[1,\infty),$ and we can proceed from there.

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