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Double check the derivation integral representation of Bessel Function

  1. Jul 9, 2013 #1
    I am reading the article Mirela Vinerean:

    http://www.math.kau.se/mirevine/mf2bess.pdf

    On page 6, I have a question about
    [tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}][/tex]

    I think there is a mistake at the last term. If you look at n=0, the equation will be:
    [tex]J_0(x)[e^0+(-1)^0e^{-0}]\;=\;j_0(x)[1+1]\;=\;2J_0(x)[/tex]
    Which is not correct. The problem is n=0 is being repeated in both the n=+ve and n=-ve.

    The equation should be:
    [tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^ne^{-jn\theta}][/tex]

    With this, the first term covers the original n=+ve from 0 to ∞. The second term covers the original from -∞ to -1. Now you only have one term contain n=0. Am I correct?

    Thanks
     
  2. jcsd
  3. Jul 9, 2013 #2
    Also in the same page right below the equations of the first post:
    [tex]\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;=\;\frac{\pi}{2}\delta_{mn}[/tex]

    1) What is ##\delta_{mn}##?
    2) If m≠n, the result should be zero. But [itex]\int_0^{\pi} \sin n\theta \sin m\theta d\theta\;≠0[/itex] because the integration is from 0 to ##\pi##, not from -##\pi## to +##\pi##.

    Thanks
     
  4. Jul 9, 2013 #3
    I can't help you with your first post. [itex] \delta_{mn} [/itex] is the Kronecker delta. It is 1 is m=n and 0 if m≠n. https://en.wikipedia.org/wiki/Kronecker_delta

    That integral is zero if m≠n. One to show that is using complex exponentials [tex] \int_0^\pi sin(n\theta)sin(m\theta)d\theta = \int_0^\pi \frac{e^{in\theta}-e^{-in\theta}}{2i}\frac{e^{im\theta}-e^{-im\theta}}{2i} d\theta= -\frac{1}{4} \int_0^\pi e^{i(m+n)\theta}-e^{i(m-n)\theta}-e^{i(n-m)\theta}+e^{-i(m+n)\theta}d\theta [/tex]. Since we know that this integral will be real, we can just consider the real parts of each of the expressions. [itex] \mathrm{Re}(e^{ix})=cos(x) [/itex] so we can write the integral as [tex] -\frac{1}{4} \int_0^\pi cos((m+n)\theta)-cos((n-m)\theta)-cos((m-n)\theta)+cos(-(m+n)\theta) d\theta [/tex]. This is zero since cosine is odd about [itex] \pi/2 [/itex] in sense that [itex] cos(\pi/2+x)=-cos(\pi/2-x) [/itex]. Note that you need m≠n. Else, the [itex] cos((n-m)\theta) [/itex] is constantly one and does not integrate to zero.
     
    Last edited: Jul 9, 2013
  5. Jul 9, 2013 #4
    Thanks for the Kronecker delta, I forgot about this one.

    I know the cosine part is fine. But if you look at page 6 of the provided link in the first post, the author claimed orthogonality properties with the sine function. That's the part I am challenging.

    Thanks
     
  6. Jul 9, 2013 #5
    I am confused. Didn't I just show that [itex] \int_0^\pi sin(mx)sin(nx) dx =0 [/itex] where [itex] m \neq n [/itex], which is the orthogonality condition?
     
    Last edited: Jul 9, 2013
  7. Jul 9, 2013 #6
    An integral does not have to be real, I don't think you can assume it's real and get rid of the imaginary part.

    A very simple test is not use exponential and just go with integration with n=0 and m=1

    [tex]\int_0^{\pi} \sin \theta d\theta =-\cos\theta|_0^{\pi}=-[-1-1]=2[/tex]
     
    Last edited: Jul 9, 2013
  8. Jul 9, 2013 #7
    It is real. If you wanted to verify this, you could expand all of the [itex] e^{ik\theta} [/itex] (where k is one of the linear combinations of m and n) terms as [itex] cos(k\theta)+isin(k\theta) [/itex] and find that all of the imaginary terms drop out. I skipped this step because the original integral with just sines must be real, and the integral with complex exponentials is only different by a factor of -1/4, so it too must be real.

    Edit: I just saw your edit. If n=0, then the integrand is 0 and the integral is trivially zero.
     
    Last edited: Jul 9, 2013
  9. Jul 9, 2013 #8
    The forum is so slow that we are cross posting. I just edited the last post, I don't even use exponential, just use n=0 and m=1. that will be just integrate a sine function and show it's not zero.
     
  10. Jul 9, 2013 #9
    Yes, it is a bit annoying. See my last edit. Look more carefully at what happens to the integrand if one of m or n is zero.
     
  11. Jul 9, 2013 #10

    micromass

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    You're taking the integral of a real function. That will always be real.
     
  12. Jul 9, 2013 #11
    Yes, I was wrong on the integration. I kept thinking if n=0, then it's an integration of a sine function and it's not zero. But if n=0, the whole thing is zero to start with!!!

    Thanks


    Can you help with my first post?
     
  13. Jul 9, 2013 #12
    Sorry, I can't as I don't know anything about Bessel functions. I hope that you get the answer you need though.
     
  14. Jul 9, 2013 #13
    Actually the question has not much to do with Bessel Function. All you need to know is if n is an integer, ##J_{-n}(x)=(-1)^n J_n(x)##, the rest is a series problem. It should be

    [tex]\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}][/tex]

    Not as the article that:
    [tex]\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}]\;=\;\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=0}^{\infty}[(-1)^nJ_n(x)e^{-jn\theta}][/tex]

    According to the article, n=0 are being repeated in both and result in twice the value for n=0.
     
  15. Jul 12, 2013 #14
    Anyone please?
     
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