# How to Show that the Error Function has an Upper Bound?

Here is the error function:

$$\mathrm{erf}(x)=\frac{2}{\sqrt\pi}\int^x_0e^{-t^2} dt$$

Here is the question:

Show that the odd function erf is bounded, by using the fact that:$$e^{-t^2} \le te^{-t^2} , t \ge1$$ Remark: The normalisation of erf is chosen so that: $$\lim_{x→∞}\mathrm{erf}(x) = 1.$$

I think that in order to show that the error function is bounded I need to do a substitution into the error function, however I am not sure how to go from there.

• you have to bound $e^{-t^2}$ by an integrable function. and your inequatliy for $t \ge 1$ is just... trivial and useless – reuns May 28 '16 at 4:18
• @user1952009 Calm down, the inequality is trivial (which is good) and useful to show the claim (even though other arguments exist). – Did May 28 '16 at 6:11
• @Did : didn't want to be offending, just that the hint given by the teacher is one of the worst possible – reuns May 28 '16 at 6:22
• @user1952009 "one of the worst possible" Pfff... If this is "one of the worst possible hints" that you saw in your life, you live in a very secluded mathematical environment. Once again, the hint is easy to prove, easy to use, and it yields the result. – Did May 28 '16 at 6:30
• @Did : why are you upset ? I said I didn't want to be offending, only to the teacher. are you the teacher ? ok so calm down :) and for the mathematical part, the student have to learn that on $[a,\infty[$, $\mathcal{O}(t^{-2}) \implies$ integrable, in particular $e^{-t}$ is very very integrable. – reuns May 28 '16 at 6:37

$$erf(x)=\frac{2}{\sqrt\pi}\int^x_0e^{-t^2} dt \le \frac{2}{\sqrt\pi}\left[\int_0^1 1 dt +\int^x_1 te^{-t^2} dt\right]=\frac{2}{\sqrt\pi}\left[1-\frac{1}{2}\int_1^x \frac{d}{dt}(e^{-t^2})dt\right]$$
$$\frac{2}{\sqrt\pi}\left[1 -\frac{1}{2}\left(e^{-x^2} - e^{-(1^2)} \right) \right] \le \frac{2}{\sqrt\pi}\left[1 -\frac{1}{2}\left(0 - 1 \right) \right] \le \frac{3}{\sqrt{\pi}}=constant$$
• I prefer much more $e^{-t^2} < e^{-t}$ for $t > 1$ – reuns May 28 '16 at 4:35