Asymptotic limit of the following integral?

Question

I am interested in the asymptotic limit of the following integral for $a\rightarrow\infty$, $$\int_0^1\mathop{\mathrm{d}x}J_2(ax)x^n,$$ where $n>-1$ and $J_2(x)$ is the Bessel function of first kind. Does any one know how to proceed? Thanks

Answer 1

As stated here: https://en.wikipedia.org/wiki/Bessel_function

For large real $z$, $J_r(z) = \sqrt{\frac{2}{\pi z}}\left(\cos(z-\frac{r\pi}{2}-\frac{\pi}{4})+O(z^{-1})\right) $ so

$\begin{array}\\ J_2(z) &= \sqrt{\frac{2}{\pi z}}\left(\cos(z-\pi-\frac{\pi}{4})+O(z^{-1})\right)\\ &= \sqrt{\frac{2}{\pi z}}\left(\cos(z-\frac{5\pi}{4})+O(z^{-1})\right)\\ &= -\sqrt{\frac{1}{\pi z}} (\cos(z)+\sin(z))+O(z^{-1})\\ &= \sqrt{\frac{2}{\pi z}} \left(-\frac{\sqrt{2}}{2}(\cos(z)+\sin(z))+O(z^{-1})\right)\\ &= -\sqrt{\frac{1}{\pi z}} \left((\cos(z)+\sin(z))+O(z^{-1})\right)\\ \end{array} $

so, for large $a$,

$\begin{array}\\ J_2(az) &= -\sqrt{\frac{1}{\pi az}} \left((\cos(az)+\sin(az))+O((az)^{-1})\right)\\ &= -\sqrt{\frac{1}{\pi az}} (\cos(az)+\sin(az))+O((az)^{-3/2})\\ \end{array} $

Therefore we need to estimate, where $f$ is $\sin$ or $\cos$,

$\begin{array}\\ \int_0^1x^ndx\left(\frac{f(ax)}{\sqrt{ax}}+O(a^{-3/2}x^{-3/2})\right) &=\int_0^1x^ndx\frac{f(ax)}{\sqrt{ax}}+\int_0^1x^ndxO(a^{-3/2}x^{-3/2})\\ &=\frac1{\sqrt{a}}\int_0^1x^{n-1/2}f(ax)dx+O(\frac1{a^{3/2}}\int_0^1x^{n-3/2}dx)\\ &=\frac1{\sqrt{a}}\int_0^1x^{n-1/2}f(ax)dx+O(\frac1{(n-1/2)a^{3/2}})\\ \end{array} $

By the Riemann-Lebesgue lemma, $\int_0^1x^{n-1/2}f(ax)dx \to 0 $ like $\frac1{a}$ for both $\sin$ and $\cos$, so the overall function goes to zero like $\frac1{a^{3/2}}$.