How do I prove the error function $$ \mbox{erf}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-t^{2}} dt. $$ is entire?

Could you give me some scratch proof?

  • $\begingroup$ What does 'entire' mean? Do you mean normalized? $\endgroup$ – Phil H Sep 28 '12 at 10:39
  • $\begingroup$ You can ude Morera's theorm. $\endgroup$ – Mhenni Benghorbal Sep 28 '12 at 10:46
  • $\begingroup$ I'm sure I saw a question about the error function being entire just recently, but I can't find it. Was that yours? Did you delete it? $\endgroup$ – joriki Sep 28 '12 at 10:48
  • $\begingroup$ @PhilH What I mean by 'entire' is 'analytic everywhere (the whole complex plane)'. $\endgroup$ – jachilles Sep 28 '12 at 10:53
  • $\begingroup$ @MhenniBenghorbal Thanks! I will try using it. $\endgroup$ – jachilles Sep 28 '12 at 10:54

A related problem. First make the change of variables $ t=zy $, then advance with the proof as in this answer. Changing the variables results in the following integral

$$ \text{erf}(z) = \frac{2}{\pi}\int_{[0,z]} e^{-t^2}\,dt=\frac{2}{\sqrt{\pi}}\,{\int _{0}^{1}\!{z\,{\rm e}^{-{z}^{2}{y}^{2}}}{dy}} \,.$$


Let $[0,z]$ denote the straight-line path from $0$ to $z$. Then we define $$\text{erf}(z) = \frac{2}{\pi}\int_{[0,z]} e^{-t^2}\,dt.$$

Now note, using Cauchy's theorem (and the analyticity of $e^{-t^2}$), that $$\frac{\text{erf}(z+h) - \text{erf}(z)}{h} = \frac{2}{\pi h}\int_{[0,z+h] - [0,z]} e^{-t^2}\,dt = \frac{2}{\pi h}\int_{[z,z+h]} e^{-t^2}\,dt.$$ Finally, this last expression tends to $(2/\pi)e^{-z^2}$ as $h\to 0$, so that $\text{erf}$ is differentiable with derivative $\text{erf}^\prime(z) = 2e^{-z^2}/\pi$. Indeed, let $\epsilon>0$. Choose $h$ small enough that $e^{-t^2}$ differs from $e^{-z^2}$ by less than $\epsilon$ as long as $|t-z|\leq|h|$. Then $$\left|\frac{1}{h}\int_{[z,z+h]} e^{-t^2}\,dt - e^{-z^2}\right| = \left|\frac{1}{h}\int_{[z,z+h]} (e^{-t^2} - e^{-z^2})\,dt\right| \leq \frac{1}{h} \int_{[z,z+h]} |e^{-t^2} - e^{-z^2}|\,dt \leq \epsilon.$$

  • $\begingroup$ Thanks for your answer. I've almost got your answer. But I didn't exactly get your Mean Value Theorem; may I ask you eactly which Mean Value Theorem you used? $\endgroup$ – jachilles Sep 28 '12 at 11:17
  • $\begingroup$ Because it is a line integral, I'm quite confused about using Mean Value Theorem... $\endgroup$ – jachilles Sep 28 '12 at 11:23
  • $\begingroup$ You can make a change of variables to make the domain of integration $[0,1]$, if you like. Alternatively, see the direct argument that I just added to the answer. $\endgroup$ – Sean Eberhard Sep 28 '12 at 11:55
  • $\begingroup$ Okay I clearly got your last part. Thanks! But just to make sure my knowledge is right: there is no MVT that can be used line integral right? And that's why we need to make the domain of integration to [0,1] right, if we want to use MVT, rifht? $\endgroup$ – jachilles Sep 28 '12 at 12:05
  • $\begingroup$ I'll leave you to formulate and to prove a version of MVT for line integrals, possibly by relating a line integral to an integral over $[0,1]$. $\endgroup$ – Sean Eberhard Sep 28 '12 at 12:31

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