The evaluation of the Gaussian integral is standard in Undergraduate Calculus ( Calculus II ?). I am not really asking how to evaluate this integral, which can be done by appealing to Fubini's theorem. Rather I do not understand this statement.
Note that $f(x) = \exp(-x^2)$ is integrable on $[0,+\infty)$. Indeed, $f(x) = O\left(\frac{1}{x^2}\right)$, thus ...
This statement preceeds, the line before Fubini is applied. I am not sure of the significance of "Big-Oh" here. I do not know how to use the above method to show a function is integrable. Could someone help me ?
Thank you for your time.
I think there might be a mistake, and the author meant $x^2$ instead of $x$. Indeed note that $$ \lim_{x \to \infty} \frac{x^2}{e^{x^2}} = 0$$ so $f(x) = O(\frac{1}{x^2})$. Now for any $h > 0$ we have $$ \int_h^\infty \frac{dx}{x^2} = \frac{1}{h}$$ now by the above there exists $m > 0$ such that for $x > m$ we have $$ e^{-x^2} \leq \frac{M}{x^2}$$ and now it follows that $$ \int_0^\infty e^{-x^2} \ dx = \int_0^m e^{-x^2} \ dx + \int_m^\infty e^{-x^2} \ dx$$ and again by the above and since $e^{-x^2}\le1$ for every $x$, $$ \int_0^\infty e^{-x^2} \ dx \leq \int_0^m e^{-x^2} \ dx + \frac{M}{m}\leq m + \frac{M}{m}$$ so the indefinite integral converges.