# Homework Help: Error Function

1. Jan 22, 2010

### Bachelier

How do we prove that the error function erf(x) and the Fresnal integral are odd functions?

2. Jan 22, 2010

### Dick

By using the definition of each one. They are all integrals of some even function from 0 to x. Isn't that always odd?

3. Jan 22, 2010

### Bachelier

OK, let me ask the question is a different way:

$${erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$$

How do I prove that?
$${erf}(-x) = - {erf}(x)$$

Last edited: Jan 22, 2010
4. Jan 22, 2010

### Dick

Ok, let me pose the solution in a different way. f(t) is even, like e^(-t^2), i.e. f(-t)=f(t). Let F(x)=integral f(t)*dt from 0 to x. Do a change of variable from t to u=(-t). What happens? Don't you get F(x)=(-F(-x))? Isn't that odd?

5. Jan 22, 2010

### Bachelier

Thanks Dick.

After deep thought, I think I got it now. I was missing one piece of information. I didn't know that the integral of an even function on 0 to infinity is an odd function.

I am going to explain my understanding below and please correct me if I am wrong, thanks in advance. :)

$$\int_0^x f(t) dt = F(x) - F(0)$$

Based on the fundamental theorem of calculus.

F(0) = 0

so we have now:

$$\int_0^x f(t) dt = F(x)$$

$$\int_0^x f(t) dt = -(-F(x))$$

$$\int_0^x f(t) dt =-(F(-x))$$

$$F(-x) = - \int_0^x f(t) dt$$

What do you think chief?

6. Jan 23, 2010

### Dick

No offence, but I think that's crap. How did F(x) become -F(-x)?? You just assumed what you want to prove. You have to prove something, not just slide stuff around. Like I said, use the substitution u=(-t). F(x)=integral f(t) from 0 to x. Do the substitution. I promise you, if you actually work through this you will understand it, if you don't you won't.

Last edited: Jan 23, 2010
7. Jan 23, 2010

### apoptosa

..
$${Erf}(x)\equiv \frac{1}{\sqrt{2\pi }}\int_0^x e^{-t^2} \, dt$$

$$\text{Let} t\to -u s.t d (-u)=-\text{du}=\text{dt}$$

$$u_2=-t_2=-(-x)=x;$$
$$u_1=-t_1=0;$$

$$\text{Erf}(-x)=\frac{1}{\sqrt{2\pi }}\int_0^x e^{-(-u)^2} \, d(-u)$$
$$\text{Erf}(-x)=-\frac{1}{\sqrt{2\pi }}\int_0^x e^{-u^2} \, du$$

$$i.e \text{Erf}(-x)=-\text{Erf}(x) \text{by} \text{defn}.$$

Last edited: Jan 23, 2010
8. Jan 23, 2010

### Bachelier

Thank you. You both cleared up things for me real good.

Well appreciated.