When the order, or argument, of a Bessel function is large, the asymptotics are well known, however, I am having issue finding the leading and subleading orders in terms of $r$ as $r\rightarrow\infty$ for the following integral.
Let $1/2<d<1$ and ℝ $\ni a>0$,
$$I=\int^{r^{-d}}_{-r^{-d}}J^2_{|x|r}(xr+ar)\,dx.$$
The argument of the Bessel function is always large and of order $r$, however at maximum, the order of the order is $r^{1-d}$ but can be zero too. The large argument Hankel expansion (https://dlmf.nist.gov/10.17) requires the order be fixed, which is not the case, though the order of the Bessel function's order is much smaller than the argument.
Any suggestions how to treat this would be greatly appreciated.
Edit: Missed $dx$ in the integral.
Best Answer
A heuristic argument is as follows. Using the approximations proposed in my comment without the error terms, and the formula $2\cos^2 w =1+\cos(2w)$, we find $$ I \sim \frac{1}{{\pi r}}\int_{ - r^{ - d} }^{r^{ - d} } {\frac{{\mathrm{d}x}}{{x + a}}} + \frac{2}{{\pi r}}\operatorname{Im} \int_{ - r^{ - d} }^{r^{ - d} } {\frac{{\exp (\mathrm{i}r(2x - \pi \left| x \right| + 2a))}}{{x + a}}\mathrm{d}x} . $$ The first term is $$ \frac{1}{{\pi r}}\log \left( {\frac{{a + r^{ - d} }}{{a - r^{ - d} }}} \right) \sim \frac{2}{{\pi ar^{1 + d} }}. $$ The integral in the second term is $\mathcal{O}(r^{-1})$ by a stationary phase argument. Thus the final approximation is $$ I \sim \frac{2}{{\pi ar^{1 + d} }} $$ as $r\to +\infty$, provided $a>0$ is bounded away from $0$ and $d$ is bounded away from $1$. If, for example, $d=3/4$, $a=2$ and $r=10^5$, then $I=5.46\ldots \times 10^{-10}$, whereas the asymptotic formula gives $5.66\ldots \times 10^{-10}$. If $d=2/3$, $a=4$ and $r=10^4$, then $I=3.42869\ldots \times 10^{-8}$, whereas the asymptotic formula gives $3.42888\ldots \times 10^{-8}$.