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Thread: Proving this integral converges

  1. #1
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    Proving this integral converges

    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.
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  2. #2
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    Krizalid's Avatar
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    well, we simply split this thing on two sets: $\displaystyle [0,1]$ and $\displaystyle [1,\infty),$ and we can proceed from there.
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