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Proving F(x) = ∫[0,x] exp(-t^2) is odd.

  1. Jan 24, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?


    2. Relevant equations



    3. The attempt at a solution
    Will it be by showing that ∫[-x,x] exp(-t^2) = 0?
     
  2. jcsd
  3. Jan 24, 2013 #2

    Mark44

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    Wouldn't the definition of "odd function" be relevant?
    No, I don't think so.
     
  4. Jan 24, 2013 #3

    Dick

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    That's not true. You do it by showing F(-x)=(-F(x)).
     
  5. Jan 24, 2013 #4
    I know that a function is odd if F(x)=-F(-x). But how may I show that that is indeed the case for the above function?
     
  6. Jan 24, 2013 #5

    jbunniii

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    Try substituting -x for x and see what you get.
     
  7. Jan 24, 2013 #6

    HallsofIvy

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    It's fairly easy to show that a function is odd if and only if its derivative is even. And, by the "fundamental theorem of Calculus" [itex]F'= e^{-x^2}[/itex].
     
  8. Jan 24, 2013 #7
    Actually, I did initially manage to show that the derivative was even. Problem is, I had no idea that would mean the function itself was odd!
    For you, HallsofIvy, it might be fairly easy. I would have to mull over it. May I demonstrate that by simple integration on both sides, or is it more intricate than that?
     
  9. Jan 24, 2013 #8

    Dick

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    The way you show it is basically the same as jbunniii's suggestion. Write the integral for F(-x) and do the substitution t->(-t) in the integral.
     
  10. Jan 24, 2013 #9
    I did that too, Dick! Leaves me with a proof that int[0,x]exp(-t^2)=-int[0,-x]exp(-t^2), which I am not sure how to approach.
    In my previous reply I was wondering how to demonstrate even derivative implies odd function.
     
  11. Jan 24, 2013 #10

    Dick

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    Take int[0,-x]exp(-t^2)dt. Do the u-substitution u=(-t), du=(-dt). And don't forget to change the limits. If t goes from 0 to -x, what are the limits for u? Since F'(x)=exp(-t^2) you can substitute any even function for exp(-t^2) and the same proof works.
     
  12. Jan 24, 2013 #11
    Wouldn't that leave me to prove that int[0,x] exp(-u^2)du + int[0,-x]exp(-u^2)du=0?
     
  13. Jan 24, 2013 #12

    Dick

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    That would be true. Why are you expressing it as a sum?
     
  14. Jan 24, 2013 #13
    I am sorry, but I fail to see how that got me closer to proving the original statement. I mean, how is this integral different than the one with dt? What's eluding me?
     
  15. Jan 24, 2013 #14

    Dick

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    I don't know. Can you show me what you get if you follow the suggestions in post 10? You should just get one transformed integral with a u variable instead of t. What is it?
     
  16. Jan 24, 2013 #15
    Could you please explain to me why I could substitute any even function? I understand that the proof would still work, but would that be rigorous enough? What would the formal justification for such a maneuver be?
     
  17. Jan 24, 2013 #16

    Dick

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    Mmm. I just meant that once you prove it then if they give you f(t) and tell you f(t) is even instead of exp(-t^2) the same proof works. I wasn't trying to say anything fancy.
     
  18. Jan 24, 2013 #17

    D H

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    I disagree with the "if" part. That f'(x) is an even function is a necessary but not sufficient condition for f(x) being an odd function. Example: f(x)=1 has an even derivative but obviously isn't an odd function.

    Having f(0)=0 is a sufficient but not necessary condition for f(x) being an odd function. Example: f(x)=1/x is odd, but f(0) does not exist.
     
  19. Jan 24, 2013 #18
    Okay, then I am still left with the seeming difficulty in seeing how the du substitution facilitated things. I believe you have expressed your approval of the expression I obtained having performed that substitution (post 12).
     
  20. Jan 24, 2013 #19

    Dick

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    It wasn't approval. It was a confession that the equality you wrote is true even though I have no idea how you got it. What I wanted to know is what you got for the du integral following the steps in post 10.
     
  21. Jan 24, 2013 #20

    Curious3141

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    ##F(x) = \int_0^x e^{-t^2}dt##

    Hence ##F(-x) = \int_0^{-x} e^{-t^2}dt##

    What you need to do now is to evaluate ##\int_0^{-x} e^{-t^2}dt##

    To do that, substitute u = -t. You haven't actually done that yet. You will end up with an integral with respect to u. You also need to take into account what happens to the bounds of definite integration when you make the substitution (remember, everything needs to be expressed in terms of u).

    Do this and write what you get.
     
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