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.

 

[CellCoree]

I can provide a mathematical proof for why Bessel functions tend to zero when x approaches infinity.

Firstly, let's recall the definition of Bessel functions. Bessel functions of order p are solutions to the differential equation:

x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - p^2)y = 0

Using this definition, we can rewrite the formula for Bessel functions as:

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

Now, let's consider what happens to this sum as x approaches infinity. We can rewrite the sum as:

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

As x approaches infinity, the term \left(\frac{x}{2}\right)^{2n+p} will become larger and larger for each term in the sum. However, the other terms in the sum, \frac{(-1)^n}{n!\Gamma(n+1)}, will become smaller and smaller. This is because the factorials in the denominator will grow faster than the exponential in the numerator.

In fact, we can show that for any n, the term \frac{(-1)^n}{n!\Gamma(n+1)} will approach 0 as x approaches infinity. This is because the factorial in the denominator grows faster than the exponential in the numerator, causing the fraction to approach 0.

Therefore, as x approaches infinity, each term in the sum will approach 0, and the entire sum will approach 0. This shows that \lim_{x\to\infty}J_p(x) = 0, proving that Bessel functions tend to zero when x approaches infinity.

In conclusion, the reason why Bessel functions tend to zero when x approaches infinity is due to the behavior of the terms in the sum that defines them. As x gets larger, the exponential term in the numerator becomes negligible compared to the factorial in the denominator, causing the entire sum to approach 0.