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Integral of Bessel Function of the First Kind

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to show that the definite integral (from 0 to infinity) of the Bessel function of the first kind (i.e.Jo(x)) goes to 1.


    2. Relevant equations
    All of the equations which I was given to do this problem are shown in the picture I have attached. However, I believe the result shown in part (b) is most relevant.


    3. The attempt at a solution
    I have fully worked through parts (a) and (b) of this question; however, I am unsure about how to attack this last problem. I was wondering if Parseval's theorem may be useful here but that has been my most recent thoughts on it.

    Any help would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2012 #2

    lanedance

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    am i missing something, or what is that theta in the 2nd last formula?

    I think you're probably on the right track, parseval's coupled with the eveness of the function should do it
     
  4. Jan 24, 2012 #3
    that theta is just my prof's way of writing a Heaviside step function
     
  5. Jan 24, 2012 #4

    lanedance

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    i thought so, but bit confusing with the notation, probably confused other responder stoo... however that should simplify the integral

    starting with parseval's and symmetry (even), you have:
    [tex] 2\int_0^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx
    = \int_{-\infty}^{\infty}\tilde{J}_0(k)dk = \int_{-\infty}^{\infty}dk\theta(1-|k|)\frac{2}{\sqrt{1-k^2}}[/tex]

    now for a given k, where is the heaviside function 0?
     
  6. Jan 24, 2012 #5
    Doesn't Parseval's relation relate to the Modulus Squared of the function and its fourier transform?
     
  7. Jan 24, 2012 #6

    lanedance

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    good point, and sorry for the silly mistake, was running out the door to meet the missus, now as an idea that needs work.... starting with what you want to evaluate
    [tex] \int_0^{\infty}dx J_0(x) = \int_0^{\infty}dx \int_0^{\infty}dk e^{-ikx} \tilde{J}_0(k) [/tex]

    now assuming we can change the order of integration...
    [tex] = \int_0^{\infty}dk \int_0^{\infty}dx e^{-ikx} \tilde{J}_0(k)[/tex]

    which gives
    [tex] = \int_0^{\infty}dk ( e^{-ikx} \tilde{J}_0(k) )|_0^{\infty}[/tex]

    the
    [tex] = \int_0^{\infty}dk \tilde{J}_0(k)[/tex]

    so it's worth seeing what that integrates to and whether the order change is valid...
     
    Last edited: Jan 24, 2012
  8. Jan 24, 2012 #7
    I seem to see a problem with the first integration, over x. There should be i and k terms coming down from that integration, sadly. Unless i'm making a silly mistake.
     
    Last edited: Jan 24, 2012
  9. Jan 24, 2012 #8

    lanedance

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    glad you're on the ball tonight... more ideas
    [tex] = \int_0^{\infty}dk \int_0^{\infty}dx \\int_0^{\infty} \frac{1}{\pi} ( e^{-ikx} \tilde{J}_0(k) )[/tex]

    [tex] = \int_0^{\infty}dk \int_0^{\infty}dx\frac{1}{\pi} (cos(kx) - isin(kx)) \tilde{J}_0(k) )[/tex]

    and as we know the function is real, we only need keep the real part for the integral...
    [tex] =\frac{1}{\pi} \int_0^{\infty}dk \int_0^{\infty} cos(kx) \tilde{J}_0(k) [/tex]
     
    Last edited: Jan 25, 2012
  10. Jan 24, 2012 #9
    I think that this might work, but I think the way that we should approach it is to use the fact that we know that the function and the fourier transform are even, and then we are allowed to do the following:

    [tex] \int_0^{\infty}dx J_0(x) = \frac{1}{2} \int_{-\infty}^{\infty}dx J_0(x) [/tex]

    Then we can say:

    [tex] \frac{1}{2} \int_{-\infty}^{\infty}dx J_0(x) = \frac{1}{2} \int_{-\infty}^{\infty}dx \frac{1}{2\pi} \int_{-\infty}^{\infty}dk e^{ikx} \tilde{J_0} [/tex]

    Now, switch the order of integration, and the J-tilde doesn't depend on x, so we can write

    [tex] = \frac{1}{4\pi} \int_{-\infty}^{\infty}dk \tilde{J_0} \int_{-\infty}^{\infty}dx e^{ikx} [/tex]

    The second integral is the fourier transform of 1 which is

    [tex] 2\pi \delta(k) [/tex]

    So integrating over k pulls out the value of J-tilde at k = 0, which is 2 (see above posts).
    Then we have the desired result

    [tex] \int_0^{\infty}dx J_0(x) = \frac{1}{4\pi} 2*2*\pi = 1 [/tex]
     
  11. Jan 25, 2012 #10
    Thanks for all the help! That was an interesting problem
     
  12. Mar 29, 2012 #11
    Hi All
    I need your support for solution for this equation in attached
    please I need fast reply for my ask
     

    Attached Files:

  13. Mar 30, 2012 #12
    any body answer me because I relley need this solution or factorial for this eq. in attahced
     
  14. Mar 31, 2012 #13

    lanedance

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    Hi mkk123 - welcome to PF!

    This is an old closed post, no point re-opening. You should post new questions as new questions in the forum - way more people will look at it that way
     
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