On the absolute integrability of Bessel functions

Question

Reading "How do you integrate a Bessel function", it didn't seem like it was an easy task. Thinking more about Bessel functions, speficially $J_0(x)$, it occurred that it looked a lot like the sinc function. Since I've had to work with sinc functions in the past, my first curiosity was: is there a $p$, for which the following holds? If so, what is the lowest $p$?

$$\int_{-\infty}^\infty\vert J_0(x)\vert^p dx<\infty$$

I couldn't think of a way to do it analytically. Firing up mathematica/wolfram alpha to do it also didn't result in anything. My guess is that looking at the curves of $|J_0(x)|$ (blue) and $|\text{sinc}(x)|$ (red) below, $J_0$ is generally above the sinc for the most part. So, in general, $|J_0(x)|^p>|\text{sinc}(x)|^p$ and so if there is such a $p$, it has to be $\geq 2$. However, this is totally a guess, and not backed by anything.

Is this known somewhere? I'm cool with a pointer to the reference. If not, how to attack this problem?

Answer 1

The integral converges iff $p>2$. This follows from this asymptotic expansion of $J_0$:

$$J_0(x)=\sqrt{\frac{2\pi}x}\cos\left(x-\frac\pi4\right)\left(1+O(1/x)\right)\;,$$

which yields

$$|J_0(x)|^p=\left|\frac{2\pi}x\right|^{p/2}\left|\cos\left(x-\frac\pi4\right)\right|^p\left(1+O(1/x)\right)\;.$$

The integral over the $O(1/x)$ term converges for any $p > 0$, since it is dominated by $|x|^{-p/2-1}$ for sufficiently large $x$. The integral over the leading term converges for $p>2$, since in that case it is dominated by $|x|^{-p/2}$ with $p/2>1$ for sufficiently large $x$. It diverges for $p\le2$, since in that case we can divide it into chunks of length $2\pi$, in each of which the cosine factor has the same form, leading to a constant factor that we can lump together with $(2\pi)^{p/2}$, and the factor $x^{-p/2}$ dominates $|x|^{-1}$ (since $p/2\le1$), so we get a series that dominates a multiple of the harmonic series and thus diverges.

Answer 2

I have been presented with a similar problem, that is why I am answering in this section. Joriki, do you use the p-test to test for convergence? If you have two functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)\geq0$, then if the integral of $g(x)$ diverges, so should the integral over $f(x)$. And if the integral over $f(x)$ converges, then this should also be the case for the integral over $g(x)$. In this case, we have $\frac{|cos(x)|^p}{|x|^{p/2}} \leq \frac{1}{|x|^{p/2}}$. As a result, if and only if $\frac{p}{2}>1$ then the integral over our function converges, because the integral over $\frac{1}{|x|^{p/2}}$ converges. However, if $p\leq2$ then the integral over $\frac{1}{|x|^{p/2}}$ diverges, but we cannot say anything about the integral over $\frac{|cos(x)|^p}{|x|^{p/2}}$. From your answer, (at least that is what I have understood) it seems that you use this argument which only gives information in the case of $p\geq2$. Could you please clarify your line of thought? (To clarify things a bit, I am referring to integrals from 1 to infinity, but I wanted to stress out the thinking part).