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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.

enter image description here

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

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Why do you say it would have to be $\ge2$? Since $|\text{sinc}(x)|=|\sin x|/|x|\le1/|x|$, $\text{sinc}$ is in $L^p$ for any $p>1$. –  joriki Aug 5 '11 at 1:23

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up vote 5 down vote accepted

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.

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sorry for not accepting earlier. i wasn't around. your answer is helpful! –  user7815 Sep 6 '11 at 21:51

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