What is the error of asymptotic relation of large order behavior of Bessel function The NIST Handbook of Mathematical functions eq. 10.19.1 [ref] gives for the asymptotic behavior of the Bessel function $J_\nu(z)$ for large real order $\nu\to\infty$ for fixed argument $z$ as
$$
J_\nu(z) \sim \frac{1}{\sqrt{2\pi \nu}}\big(\frac{ez}{2\nu}\big)^\nu
$$
without providing any error.  Is it known what the corrections are to this asymptotic relation?
 A: We can take the integral representation
$$J_\nu(z) =
\frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}}
 \int_{-1}^1 (1 - t^2)^{\nu - 1/2} \cos z t \, dt,$$
which is valid for $\nu > -1/2$. The change of variables $-u^2 = \ln( 1-  t^2)$ gives
$$f(u) = \frac
 {\left| u \right| e^{-u^2/2} \cos z \sqrt {1 - e^{-u^2}}}
 {\sqrt {1 - e^{-u^2}}}, \\
J_\nu(z) =
\frac {z^\nu} {2^\nu \sqrt \pi \, \Gamma {\left( \nu + \frac 1 2 \right)}}
 \int_\mathbb R f(u) e^{-\nu u^2} du.$$
Expanding the gamma function around $\nu = \infty$ and expanding $f(u)$ around $u = 0$, we obtain as many terms in the asymptotic series as we want:
$$\frac {z^\nu} {2^\nu \hspace {1px} \Gamma {\left( \nu + \frac 1 2 \right)}} =
\frac 1 {\sqrt {2 \pi}} \left( \frac {e z} {2 \nu} \right)^\nu \left(
 1 + \frac 1 {24} \nu^{-1} + O(\nu^{-2}) \right), \\
f(u) = 1 - \frac {2 z^2 + 1} 4 u^2 + O(u^4), \\
\frac 1 {\sqrt \pi} \int_{\mathbb R} f(u) e^{-\nu u^2} du =
\nu^{-1/2} - \frac {2 z^2 + 1} 8 \nu^{-3/2} + O(\nu^{-5/2}),\\
J_\nu(z) = \frac 1 {\sqrt {2 \pi}} \left( \frac {e z} {2 \nu} \right)^\nu \left(
 \nu^{-1/2} - \frac {3 z^2 + 1} {12} \nu^{-3/2} + O(\nu^{-5/2}) \right).$$
