Asymptotic Expansion of the Bessel-Integral Function

Solution 1:

Two options. Either, as suggested in the comments, replace $J_\nu(t)$ with an integral representation like (Gradshteyn-Ryzhik 8.411.11) $$ J_\nu(t) = \frac{2}{\pi} \int_0^\infty \sin\Bigl(t \cosh \xi - \frac{\nu \pi}{2} \Bigr) \cosh(\nu \xi) \, d \xi $$ and play stationary phase/saddle point method/Laplace method games.

Or, if you're lazy, take the word of someone (like the DLMF) who has already done this to $J_\nu(t)$ itself, getting (for $t$ large compared to $\nu$) $$ J_\nu(t) \approx \sqrt{\frac{2}{\pi t}} \cos\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4}\Bigr ). $$ Per the DLMF asymptotic, the error in this approximation saves a power of $t$ compared to this leading term (i.e., the error is $O(t^{-3/2})$) whence integrating it against $1/t$ contributes $O(x^{-3/2})$.

Insert this into $$ \mathrm{Ji}_\nu(x) \approx \int_{\infty}^x \frac{J_\nu(t)}{t} \, d t = \int_{\infty}^x \sqrt{\frac{2}{\pi t}} \frac{1}{t} \cos\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4}\Bigr ) \, d t $$ and integrate by parts to get $$ \mathrm{Ji}_\nu(x) \approx \sqrt{\frac{2}{\pi x}} \frac{\sin(x - \frac{\nu \pi}{2} - \frac{\pi}{4} )}{x} + \int_\infty^x \sqrt{\frac{2}{\pi t^5}} \sin\Bigl( t - \frac{\nu \pi}{2} - \frac{\pi}{4} \Bigr) \, d t. $$ Note that the integral at the end is $O(x^{-3/2})$, and so is the error from the asymptotic expansion of the Bessel function by the aforementioned power savings.