The improper integral of $e^{-x^2}$ over $\mathbb{R}$ is finite and positive

Given: The real exponential function and the mixed-improper convention on the real line.

For every real $u$, $1+u\le \exp (u)$ ($1+x\le \exp (x)$ for every real $x$, hence $(1-p{)}^m\le \exp (-mp)$).

The improper integral ${\int }_1^{\mathrm{\infty }}{x}^{-p}dx$ converges exactly when the rational $p>1$ (The improper $p$-test for rational exponents).

For every real $u$, $\exp (u)>0$ and $\exp (-u)=1\mathrm{/}\exp (u)$ (The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$).

Improper integrals at $-\mathrm{\infty }$ and $+\mathrm{\infty }$ must converge separately before they are added (Improper integrals with several singular ends).

If $0\le u\le v$ toward a singular end and the improper integral of $v$ converges there, then the improper integral of $u$ converges there (Comparison tests for improper integrals).

The exponential function is strictly increasing on $\mathbb{R}$ (The exponential function is strictly increasing).

If an integrable function $h$ satisfies $m\le h(x)\le M$ on $[a,b]$, then $m(b-a)\le {\int }_a^{b}h\le M(b-a)$ (If $m\le f\le M$ on $[a,b]$ then $m(b-a)\le L(f,P)\le \underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P)\le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with ${\int }_a^{b}c=c(b-a)$).

Every continuous function on a compact interval is Riemann integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

A valid $C^1$ substitution carries a convergent improper integral to the corresponding transformed improper integral (Change of variable in an improper integral).