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Help with a definite integral property

  1. Jan 7, 2010 #1
    Hello,

    Saw this expression:

    [tex]^{inf}_{-inf} \int e^{x^{2}} = 2 \times ^{inf}_{+0} \int e^{x^{2}}[/tex]

    I am unable to understand this change of base. You can change from -inf to 0 and 0 to +inf but do not see how that equals to 2 * the integral from 0 to infinity.

    I hope someone will help me see the light :)

    Thanks,

    Luc
     
  2. jcsd
  3. Jan 7, 2010 #2
    it's because [tex]e^{x^2}[/tex] is an even function, that is, it satisfies the equation f(-x) = f(x). if you look at the graph of the function it's symmetric about the y-axis
     
  4. Jan 7, 2010 #3
    That makes sense. Thanks for that. I still have one confusion though:

    [tex]\int^{\infty}_{-\infty} e^{x^{2}} = \int^{0}_{-\infty} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}[/tex]

    Now using the function symmetry we have:

    [tex]\int^{\infty}_{-\infty} e^{x^{2}} = \int^{-\infty}_{0} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}[/tex]

    However, the limits are still messed up in the first term, right? The upper limit still has [tex]-\infty[/tex]

    Or have I done this completely wrong?

    Thanks,

    Luca
     
  5. Jan 7, 2010 #4
    [itex]\int_{-\infty}^{\infty} e^{x^2} dx = \int_{-\infty}^{0} e^{x^2} dx + \int_{0}^{\infty} e^{x^2} dx[/itex] but since [itex]e^{x^2}[/itex] is even, [itex]\int_{-\infty}^{0} e^{x^2} dx = \int_{0}^{\infty} e^{x^2} dx[/itex] so [itex]\int_{-\infty}^{\infty} e^{x^2} dx = 2\int_{0}^{\infty} e^{x^2} dx[/itex]
     
  6. Jan 7, 2010 #5
    Ahhhh thanks. Did not know you could change the integral limits for an even function like that.

    Cheers,
    Luca
     
  7. Jan 7, 2010 #6

    LCKurtz

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    And, of course, all the integrals mentioned above diverge unless the integrand has a negative sign:

    [tex]e^{-x^2}[/tex]

    :rolleyes:
     
  8. Jan 7, 2010 #7
    ARGHHHH i don't know how i missed that! i guess it's too late to delete everything i've added to this thread :mad: luckily for me math isn't an exact science
     
    Last edited: Jan 7, 2010
  9. Jan 8, 2010 #8
    Of course. Checking for convergence or divergence is another business all-together.

    However, the change of limits for the integral still holds, right?

    Thanks,
    Luca
     
  10. Jan 8, 2010 #9
    yes, because [itex]\int_{0}^{\infty} f(x) dx = \lim_{M \rightarrow \infty}\int_{0}^{M} f(x) dx[/itex]
     
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