Proving $\int_{0}^{\infty} e^{-x^2}dx=\sqrt{n}\int_{0}^{\infty} e^{-nx^2}dx$

Question

Prove: $\int_{0}^{\infty} e^{-x^2}dx=\sqrt{n}\int_{0}^{\infty} e^{-nx^2}dx$ for every natural number

I tried a lot of things, I think induction is the way to go here but I couldn't really get anywhere after the initial stage.

Help would be appreciated

Answer 1

By the change of variable $x=\sqrt{n}\:u$, giving $\,dx=\sqrt{n}\:du$, then you get

$$ \int_{0}^{\infty} e^{\large-x^2}dx=\sqrt{n}\int_{0}^{\infty} e^{\large-nu^2}du, $$

as announced.