Convergence as a result of symmetry

Question

I have the following improper integral: $$\int _{-\infty }^{+\infty }\:e^{-x^2}\,{\rm d}x$$ Now, I just proved it is convergent for $[0;+\infty]$. The thing is I admit I'm a bit lazy and I don't want to repeat the whole same process again for the other interval $[-\infty;0]$. I was thinking that as $e^{-x^2}$ is an even function, which means it is symmetric for the y-axis, and as the previous integral converges for $[0;+\infty]$ then it would be pretty logical to say that it converges for $[-\infty;0]$. I was wondering if this reasoning is valid, and if not, which things would you change for it to be. Thanks in advance!

Answer 1

Yes; if you have shown that the integral $$\int_0^{\infty}e^{-x^2}\,{\rm d}x,$$ converges, then by the substitution $y:=-x$ it follows that the integral $$\int_{-\infty}^0e^{-y^2}\,{\rm d}y,$$ also converges, and to the same value.