Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge?

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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