TITLE: Calculation of integral with Bessel function
QUESTION [5 upvotes]: I have a trouble with to calculating (or bounding from above) the following integral:
$$
\int_{-\infty}^{\infty}\left(\frac{J_2(x)}{x^2}\right)^p\, dx, \quad p\geq 1,
$$
where $J_2(x)$ is a Bessel function {http://en.wikipedia.org/wiki/Bessel_function}.
Any ideas would be very helpful.
Thank you.

REPLY [2 votes]: A simple idea is just to use the Taylor series of $\frac{J_2(x)}{x^2}$:
$$f(x)=\frac{J_2(x)}{x^2}=\sum_{m=0}^{+\infty}\frac{(-1)^m}{4^{m+1}\,m!(m+2)!}x^{2m}$$
then approximate the LHS with the function $\frac{1}{8}\exp\left(-\frac{x^2}{12}\right)$:
$$g(x)=\frac{1}{8}\exp\left(-\frac{x^2}{12}\right)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{8\cdot 12^m\,m!}x^{2m}.$$
The approximation is quite good:
$\hspace1in$
$\hspace2in\quad\;\; \frac{J_2(x)}{x^2}\text{ against }\frac{1}{8}e^{-x^2/12}$
and it leads to:
$$\int_{-\infty}^{+\infty}f(x)^p\,dx \approx \int_{-\infty}^{+\infty}g(x)^p\,dx = \frac{1}{2^{3p-1}}\sqrt{\frac{3\pi}{p}}.\tag{1}$$
It is possible to check that $\frac{f(x)}{g(x)}\in(0,1]$ for any $x\in I=[-5,5]$: since the two integrals outside $I$ are very small, it is expected that $(1)$ always holds as an upper bound:

$$\int_{-\infty}^{+\infty}\left(\frac{J_2(x)}{x^2}\right)^p\,dx \leq\frac{1}{2^{3p-1}}\sqrt{\frac{3\pi}{p}}.$$