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D/dx[int(0,x) e^(-t^2) dt] : two methods, two answers

  1. Dec 24, 2011 #1
    There are two methods to take d/dx [ ∫t=0x exp(-t^2) dt].

    First method: using the relationship of integral and antiderivative, one gets
    (exp(-t^2) , from t = 0 to x, so exp(-x^2) - 1.

    Second method: the integral is (1/2)sqrt(pi)*erf(t) from 0 to x, which is (1/2)sqrt(pi)*erf(x), and the derivative of this is exp(-x^2).

    So, which answer is correct, and what is wrong with the other method?

    (Wolfram alpha favors the second answer, but I have another source that favors the first answer.)
     
  2. jcsd
  3. Dec 24, 2011 #2

    jgens

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    The second answer is right. Can you show how you got this so we can help point out your mistake?
     
  4. Dec 24, 2011 #3
    Thank you, gladly. I am using the idea that d(∫abf(t)dt)/dx = f(x)|ab; here a=0 , b = x, and f(t)= exp(-t^2).
     
  5. Dec 24, 2011 #4

    jgens

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  6. Dec 24, 2011 #5
    Ah, I see my mistake. Thank you very much, gjens.
     
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