Why does the limit of $ f(x) = \int_{0}^{x} e^{-t^2/2} dt$ exist? For a function
$$ f(x) = \int_{0}^{x} e^{-t^2/2} dt$$
for $x \ge 0$ one has to argue whether $ \lim_{x \to \infty} f(x) $ exists or not.
I thought about the following:
$$ f'(x) = e ^ {-x^2/2} $$
$$ f''(x) = -x\ e ^ {-x^2/2} $$
Given $x \ge 0$, it's obvious that $f(x)$ is strictly increasing and concave. Also, it is clear that $ e^{-x^2/2} \to 0$ as $x \to \infty$, so $f$ has to converge.
Is this enough of a 'proof', or is there more to it?
 A: If you want to compute the limit as $x$ goes to infinity, the standard argument is as follows. 
We have 
$$
f(x)^2 = \left( \int_0^x e^{-t^2/2} \, dt \right) \left( \int_0^x e^{-s^2/2} \, ds \right) = \int_0^x \int_0^x e^{-t^2/2} e^{-s^2/2} dt ds = \iint_{[0,x] \times [0,x]} e^{-(t^2+s^2)/2} \, dt\, ds. 
$$
We have $[0,x] \times [0,x] \subseteq \{ (u,v) \, | \, u,v \ge 0, \, \sqrt{u^2 + v^2} \le \sqrt 2 \, x \} \subseteq [0,\sqrt 2 \, x] \times [0, \sqrt 2 \, x]$. These inclusions imply that if we call the left-most region $A_x$, the middle one $B_x$ and the right-most one $C_x$, 
$$
\iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \le \iint_{B_x} e^{-(t^2+s^2)/2} \, dt\, ds \le \iint_{C_x} e^{-(t^2+s^2)/2} \, dt\, ds
$$
and since all double integrals are increasing functions of $x$, the limit is defined (i.e. it is either positive real or infinite), and together with $C_x = A_{\sqrt 2 \, x}$, we have the inequalities
$$
\lim_{x \to \infty} f(x)^2  = \lim_{x \to \infty} \iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
                            \le \lim_{x \to \infty} \iint_{B_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
                            \le \lim_{x \to \infty} \iint_{C_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
                            = \lim_{x \to \infty} \iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
                            = \lim_{x \to \infty} f(x)^2. 
$$
Thus, it suffices to compute
$$
 \lim_{x \to \infty} \iint_{B_x} e^{-(t^2 + s^2)/2} \, dt \, ds.
$$
Using a polar change of coordinates : $t = r \cos \theta$, $s = r \sin \theta$, $dt \, ds = r dr d\theta$, we get
$$
\iint_{B_x} e^{-(t^2 + s^2)/2} \, dt \, ds = \int_0^{\pi/2} \int_0^x r e^{-r^2/2} \, dr d\theta = \frac{\pi}2 \left( 1 - e^{-x^2/2} \right).
$$
by the change of variables $u = r^2/2$. Letting $x \to \infty$ leads to 
$$
\lim_{x \to \infty} f(x)^2 = \frac{\pi}2,
$$
hence by continuity of the square root function, $\lim_{x \to \infty} f(x) = \sqrt{\frac{\pi}2}$.
Hope that helps,
P.S. : This has probably been already computed on this website.
