Clarification of the standard proof showing that the normal distribution p.d.f integrates to $1$.

Question

Considering the standard normal distribution p.d.f $f(x) = \frac {1}{\sqrt{2\pi}} e^{\frac {-x^2}{2}}$, it seems that most introductory probability/statistics textbooks show that $I = \int_{-\infty}^{\infty}f(x) = 1$ by showing that $I^2 = 1$, but this reasoning implies that $I$ as an improper integral exists in the first place. How can I conclude that $I$ exists?

Answer 1

You only need to check $\int_1^\infty f(x) \mathop{dx} < \infty$ and $\int_{-\infty}^{-1} f(x) \mathop{dx} < \infty$ are true (or something similar). Comparison with an integrable function like $g(x) = \frac{1}{x^2}$ should suffice.

Answer 2

Check endpoints with asymptotic functions(or you can take limit to see it if its finite). The limit when $x\to+\infty$ is $0$, so you only check for $x\to-\infty$.