Bessel function integral

Let $J_0$ denote the Bessel function of the first kind. Prove that

$\int_0^\infty J_0(x) \, \mathrm{d}x=1$

Solution

We recall that

$J_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{4^n (n!)^2}$

Hence,

$\begin{align*} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}\\ &=\frac{1}{p}\sum_{n=0}^{\infty} \binom{-1/2}{n}\frac{1}{p^{2n}} \\ &= \frac{1}{\sqrt{1+p^2}} \end{align*}$

Then,

$\lim_{p \rightarrow 0^+} \int_{0}^{\infty}J_0(x)e^{-px}\,\mathrm{d}x = \lim_{p \rightarrow 0^+} \frac{1}{\sqrt{1+p^2}} =1$

Using the fact that the $J_0(x)$ looks like an ‘almost periodic’ function with decreasing amplitude. If we denote by $\{\alpha_k\}_{k \geq 0}$ the zeros of $J_0$ then $\alpha_k \nearrow \infty$ as $k \to \infty$ and furthermore