Proof that Bessel functions tend to zero when x approaches infinity

[Monsterman222]

I am aware that Bessel functions of any order $p$ 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:

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

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

$$\lim_{x\to\infty}J_p\left(x\right)=0$$

 

[uart]

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.

$$J_n(x) = \frac{1}{\pi} \int_0^\pi \cos (n \lambda - x \sin \lambda) d\lambda$$

 

[deluks917]

Have you considered the differential equation that the bessel functions solve?

 

[Monsterman222]

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:
$$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - p^2)y = 0$$
we can take the limit of this as x tneds to infinity, substitute $J_p\left(x\right)$ and divide by $x^2$, giving

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

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