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How would I know if my integrand is odd?

  1. Jul 21, 2013 #1
    So, I'm studying normalization (Quantum Mechanics). I came across examples of integrands which they say are odd. So, they say the integral of an odd function is zero. How would I know my integrand is odd? Thanks :)
     
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  3. Jul 21, 2013 #2
    Replace [itex]x\rightarrow -x[/itex]. If the integrand transforms [itex]f(x)\rightarrow f(-x)=-f(x)[/itex] then it is odd.
     
  4. Jul 21, 2013 #3
    Why is the integral of an odd function ZERO?
     
  5. Jul 21, 2013 #4

    SteamKing

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    Take the definite integral of sin(x) w.r.t. x between 0 and 2*pi. (Or between -pi and pi). What is its value?
    Plot sin(x) between 0 and 2*pi (or between -pi and pi). What do you notice about the plot?
     
  6. Jul 21, 2013 #5
    How would you take the derivative of this?


    [itex]\int^{∞}_{0} e ^{\frac{-2amx^2}{\hbar}}[/itex]


    What's the trick in doing integrals having infinity as one of its limit? Or what if the limit is from ∞ to -∞?
     
    Last edited: Jul 21, 2013
  7. Jul 21, 2013 #6
    The derivative of an even function is odd and the derivative of an odd function is even (a simple proof, just use the definition). So, if you have an odd integrand [itex]f(-x)=-f(x)[/itex], then for [itex]F(x)[/itex] such that [itex]F'(x)=f(x)[/itex] [itex]F(-x)=F(x)[/itex] and so

    [tex]\int_{-a}^{a} f(x) \mathrm{d}x = F(a) - F(-a) = F(a) - F(a) = 0[/tex]

    which is what is usually used when an integral over a symmetric interval is said to be zero because the integrand is odd.
     
  8. Jul 21, 2013 #7

    CompuChip

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    The key in Deldeals post that has been ignored until now: it works only if the integration interval is symmetric. In QM it usually is, by convenient choice of x = 0, but in general it is something you should check.
     
  9. Jul 21, 2013 #8
    Adding to this, it's exactly why in jhosamelly's latest post the symmetry properties won't be of much use ** – basically the value of [itex]\int_0^\infty e^{-x^2}\mathrm{d}x[/itex] needs to be known from elsewhere – but were it integrated from -∞ to ∞, the integral would evaluate to twice the value of [itex]\int_0^\infty e^{-x^2}\mathrm{d}x[/itex], as the integrand is even.

    ** And there are no specific "tricks" related to symmetry properties when it comes to having infinity as a limit for integrals like these, although the limit obviously must exist – you can't just say that [itex]\int_{-\infty}^{\infty} x \mathrm{d}x =0[/itex] even though [itex]f(x)=x[/itex] is odd. That would be the so-called Cauchy principal value of the integral, although one that I don't think is very useful.
     
    Last edited: Jul 21, 2013
  10. Jul 21, 2013 #9
    Thanks for the responses guys. I'm learning.

    But, let me show you the example I'm having difficulty understanding.

    So, in normalization.

    [tex]\int_{-∞}^{∞} |ψ(x)|^{2} \mathrm{d}x = 1[/tex]

    (I get this)

    Say our

    [tex] ψ(x) = A e^{-amx^2 / \hbar} [/tex]

    So we have

    [tex]\int_{-∞}^{∞} |A e^{-amx^2 / \hbar}|^{2} \mathrm{d}x = 1[/tex]
    [tex]A^2\int_{-∞}^{∞} e^{-2amx^2 / \hbar} \mathrm{d}x = 1[/tex]

    The example I'm looking at suddenly have 2 in front.
    [tex]2A^2\int_{-∞}^{∞} e^{-2amx^2 / \hbar} \mathrm{d}x = 1[/tex]

    Then,

    [tex]2 A^2 \left(\frac{1}{2}\sqrt{\frac{\pi}{\frac{2am}{\hbar}}} \right)= 1[/tex]

    I assume the ones on the parenthesis is the integral. Any help understanding this more?
     
  11. Jul 21, 2013 #10

    SteamKing

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    It's not clear what the variable of integration is in this expression.

    And, take the derivative w.r.t. which variable?
     
  12. Jul 21, 2013 #11
    The lower limit was probably changed to zero, which is what I mentioned in #8: The integrand is even, so

    [tex]\int_{-a}^a f(x) \mathrm{d}x = F(a)-F(-a) = F(a) -(-F(a)) = 2F(a)[/tex]

    holds. The book/material probably lists or assumes (maybe even proves!) the result [itex]\int_0^\infty e^{-rx^2} \mathrm{d}x=\frac{\sqrt{\pi}}{2\sqrt{r}}[/itex] or something similar. Using that,

    [tex]A^2 (\int_{-\infty}^{\infty} e^{-rx^2} \mathrm{d}x) = A^2(2\int_0^\infty e^{-rx^2} \mathrm{d}x)[/tex]

    (because of the symmetric interval & evenness)

    [tex] = A^2 (2 \frac{\sqrt{\pi}}{2\sqrt{r}})= 2A^2 (\frac{1}{2} \frac{\sqrt{\pi}}{\sqrt{r}})[/tex]

    Did that help at all?
     
  13. Jul 21, 2013 #12

    YES!!!


    Certainly! Thanks!!!!!!
     
  14. Sep 14, 2016 #13
    These are known as Gaussian Integrals, which are integration over Gaussian function. I can perceive that you did Griffiths Quantum Mechanics. Make sure you even learn about Gamma functions and few other special functions which you come across.
     
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