TITLE: Show that $\frac{1}{2}-\frac{1}{2e}<\int_0^{+\infty}e^{-x^2}dx<1+\frac{1}{2e}$
QUESTION [7 upvotes]: Show that 
$$\frac{1}{2}-\frac{1}{2e}<\int_0^{+\infty}e^{-x^2}dx<1+\frac{1}{2e}$$
I know that one way to do this is to evaluate the integral in the middle, and then compare these three numbers. I wonder how can we do this without explicitly compute the integral?

REPLY [7 votes]: Proof for the upper bound:
$$\int_0^{\infty}e^{-x^2}dx =\int_{0}^{1}e^{-x^2}dx+\int_{1}^{\infty} e^{-x^2} dx <\int_{0}^{1}1 dx+\int_{1}^{\infty} x e^{-x^2} dx 
=1+\frac{1}{2e}$$
where the last integral is easily calculated by substituting $u=-x^2$.

Proof for the lower bound:
$$\int_0^{\infty}e^{-x^2}dx > \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1} xe^{-x^2} dx =\frac{1}{2}-\frac{1}{2e}$$
where the last integral is easily calculated by substituting $u=-x^2$.