Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What does 'entire' mean? Do you mean normalized? –  Phil H Sep 28 '12 at 10:39
    
You can ude Morera's theorm. –  Mhenni Benghorbal Sep 28 '12 at 10:46
    
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? –  joriki Sep 28 '12 at 10:48
    
@PhilH What I mean by 'entire' is 'analytic everywhere (the whole complex plane)'. –  julypraise Sep 28 '12 at 10:53
    
@MhenniBenghorbal Thanks! I will try using it. –  julypraise Sep 28 '12 at 10:54

2 Answers 2

up vote 2 down vote accepted

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

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

share|improve this answer

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 $$\left|\frac{1}{h}\int_{[z,z+h]} e^{-t^2}\,dt - e^{-z^2}\right| = \left|\frac{1}{h}\int_{[z,z+h]} (e^{-t^2} - e^{-z^2})\,dt\right| \leq \frac{1}{h} \int_{[z,z+h]} |e^{-t^2} - e^{-z^2}|\,dt \leq \epsilon.$$

share|improve this answer
    
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? –  julypraise Sep 28 '12 at 11:17
    
Because it is a line integral, I'm quite confused about using Mean Value Theorem... –  julypraise Sep 28 '12 at 11:23
    
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. –  Sean Eberhard Sep 28 '12 at 11:55
    
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? –  julypraise Sep 28 '12 at 12:05
    
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 $[0,1]$. –  Sean Eberhard Sep 28 '12 at 12:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.