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
[Math] Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge
integrationreal-analysis
Best Answer
Write
$$\int_0^\infty e^{-x^2} \, dx = \int_0^1 e^{-x^2} \, dx + \int_1^\infty e^{-x^2} \, dx$$
The first integral on the right hand side is of a bounded function over a bounded interval, hence finite. For the second notice that we have $x^2 > x$ when $x > 1$. Therefore, $e^{-x^2} < e^{-x}$ for $x > 1$ and it follows that
$$\int_1^\infty e^{-x^2} \, dx < \int_1^\infty e^{-x} \, dx$$ $$= \lim_{x \to \infty} -e^{-x} + e^{-1} = 1/e < \infty$$