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I want to show the convergence of the following improper integral $\int_0^\infty e^{-x^2}dx$. I try to use comparison test for integrals $x≥0$, $-x ≥0$, $-x^2≥0$ then $e^{-x^2}≤1$. So am ending with the fact that $\int_0^\infty e^{-x^2}dx$ converges if $\int_0^\infty dx$ converges but I don’t appreciate this. Thanks

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    Please use LaTeX. Is this integral of $\int_0^{\infty}e^{-x^2}\,dx$ (`\int_0^{\infty}e^{-x^2}dx`), $\int_0^{\infty}(e-x^2)\,dx$ (`\int_0^{\infty}(e-x^2)dx$`, or what?2012-01-06
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    I assume you mean $\int_0^\infty e^{-x^2}\ dx$. Hint: compare to $\int_0^\infty e^{-x}\ dx$.2012-01-06
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    @neemy: Yes the integral $\int_{0}^{\infty} e^{-x^2} \ dx$ converges and it's value is $\sqrt{\pi}$2012-01-06
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    There is a wonderful math markup language called LaTeX which this site supports and allows you to communicate math much more clearly and legibly. Learn it; it will help you here and elsewhere. I have edited your question with proper formatting. If the new version is not the question you were trying to ask, just let me know.2012-01-06
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    @Chandrasekhar You probably meant to write the value as $\sqrt{\pi}/2$, not $\sqrt{\pi}$.2012-01-06
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    Related: http://math.stackexchange.com/questions/9286/proving-int-0-infty-e-x2-dx-frac-sqrt-pi22012-01-09

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