Error Function and the Fresnal integral

[Bachelier]

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

 

[Dick]

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

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

 

[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?
$${erf}(-x) = - {erf}(x)$$

 

[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?

 

[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?

 

[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.

 

[apoptosa]

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

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

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

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

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

 

[Bachelier]

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

Well appreciated.