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Proof that Bessel functions tend to zero when x approaches infinity

  1. Sep 6, 2011 #1
    I am aware that Bessel functions of any order [itex]p[/itex] are zero in the limit where x approaches infinity. From the formula of Bessel functions, I can't see why this is. The formula is:

    [tex]J_p\left(x\right)=\sum_{n=0}^{\infty} \frac{\left(-1\right)^n}{\Gamma\left(n+1\right)\Gamma\left(n+1+p\right)}\left(\frac{x}{2}\right)^{2n+p}[/tex]

    Does anyone know a proof of why this is? That is, why is it that

  2. jcsd
  3. Sep 6, 2011 #2


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    I don't a have proof right now, but you may find the following integral identity more useful in understanding the limiting behavior than the infinity sum identity you've been considering.

    [tex]J_n(x) = \frac{1}{\pi} \int_0^\pi \cos (n \lambda - x \sin \lambda) d\lambda[/tex]
  4. Sep 6, 2011 #3
    Have you considered the differential equation that the bessel functions solve?
  5. Sep 8, 2011 #4
    Thanks for your help so far, but I'm still struggling with this one. From the representation of the Bessel function involving the integral, I still can't prove it.

    Looking at Bessel's differential equation:
    [tex]x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - p^2)y = 0[/tex]
    we can take the limit of this as x tneds to infinity, substitute [itex]J_p\left(x\right)[/itex] and divide by [itex]x^2[/itex], giving

    [tex]\lim_{x\to\infty}J_{p}''\left(x\right)+\frac{J_{p}'\left(x\right)}{x} +\left(1-\frac{p^2}{x^2}\right)J_p\left(x\right)=0[/tex]

    But now, to finish the proof, I'd need to show that [itex]J_{p}''\left(x\right)[/itex] goes to zero as x approaches infinity and that [itex]J_{p}'\left(x\right)[/itex] is finite. I'm not sure this approach is helpful.
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