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Error Function and the Fresnal integral

  1. Jan 22, 2010 #1
    How do we prove that the error function erf(x) and the Fresnal integral are odd functions?
  2. jcsd
  3. Jan 22, 2010 #2


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    By using the definition of each one. They are all integrals of some even function from 0 to x. Isn't that always odd?
  4. Jan 22, 2010 #3
    OK, let me ask the question is a different way:

    [tex]{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt[/tex]

    How do I prove that?
    [tex]{erf}(-x) = - {erf}(x)[/tex]
    Last edited: Jan 22, 2010
  5. Jan 22, 2010 #4


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    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?
  6. Jan 22, 2010 #5
    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. :)

    [tex]\int_0^x f(t) dt = F(x) - F(0)[/tex]

    Based on the fundamental theorem of calculus.

    F(0) = 0

    so we have now:

    [tex]\int_0^x f(t) dt = F(x)[/tex]

    [tex]\int_0^x f(t) dt = -(-F(x))[/tex]

    [tex]\int_0^x f(t) dt =-(F(-x))[/tex]

    [tex]F(-x) = - \int_0^x f(t) dt [/tex]

    What do you think chief?
  7. Jan 23, 2010 #6


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    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
  8. Jan 23, 2010 #7
    [tex]{Erf}(x)\equiv \frac{1}{\sqrt{2\pi }}\int_0^x e^{-t^2} \, dt[/tex]

    [tex]\text{Let} t\to -u s.t d (-u)=-\text{du}=\text{dt}[/tex]


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

    [tex]i.e \text{Erf}(-x)=-\text{Erf}(x) \text{by} \text{defn}.[/tex]
    Last edited: Jan 23, 2010
  9. Jan 23, 2010 #8
    Thank you. You both cleared up things for me real good.

    Well appreciated. :smile:
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