Bounded bessel functions in an s-set projection proof

Question

The following is an extract from Falconer's Geometry of Fractal Sets about the proof of:

"...Using the definition of a Bessel function $J_0=\frac{1}{2\pi}\int^{2\pi}_0 \cos(u \cos \theta) d\theta$...

I really cannot understand how the integral of $J_0$ is bounded? This corresponds to the inequality in the section above. I have tried to look at properties of Bessel functions without joy. Furthermore I think this is complicated as it is a $J$ type bessel function. (I may be wrong).

I have tried to find justification in Watson's A treatise on the theory of bessel functions but to no avail.

Answer 1

The Laplace transform of $J_0(x)$ is well-known to be $1/\sqrt{1+s^2}$, i.e. $$ \int_0^{+\infty} J_0(x)e^{-sx}\,dx = \frac{1}{\sqrt{1+s^2}}. $$ If you let $s\to 0^+$ (justify that this is valid), you find that $$ \int_0^{+\infty} J_0(x)\,dx = 1. $$ Hence, $$ \int_{-\infty}^{+\infty} J_0(x)\,dx = 2, $$ since $J_0$ is even.

Answer 2

Isn't $J_0(u)$ given by a cosine integral? For real $u$, it seems clear that $|J_0(u)|\leq 1$. Do you mean the integral of $J_0(u)$?