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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?

bonifaz
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    Just to be clear, the limit is $\lim_{x \to \infty} f(x)$? Also, you make several wrong claims. For example, $f$ is strictly increasing ($e^{-x^2/2} > 0$). – Najib Idrissi Jun 20 '13 at 09:37
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    Try comparison of the integrand to $e^{-t}$, or some simple variant thereof. – Gerry Myerson Jun 20 '13 at 09:39
  • you can also compare it with $\frac{1}{t^2}$. – Ma Ming Jun 20 '13 at 09:52
  • for $1 \leq x$ we have $0 \leq e^{-x^{2}/2} \leq \frac{1}{x^2}$ and the limits exists for $\int_1^x \frac{1}{x^2}=1-\frac{1}{x}$ – Averroes Jun 20 '13 at 09:52
  • As a physicist, I can assure you that, essentially, $\int_{0}^{x} e^{-t^2/2} dt=\text{erf}(x)=(2/\pi)\tan^{-1}(c\cdot x)$, for some $c\approx 1$. – Nikolaj-K Jun 20 '13 at 10:01

1 Answers1

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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.

Patrick Da Silva
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