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For a formal inverse-power series ${\sum }_{n\ge 0}{a}_n{z}^{-n-1}$, divide its coefficients by $n!$ to form a Borel transform $\sum a_n{t}^n/n!$ and integrate it against $e^{-zt}$. Singularities on the integration ray require a specified contour prescription. Different lateral contours can produce exponentially small differences and the Stokes phenomenon.

A divergent asymptotic expansion should usually be stopped near its smallest term rather than summed indefinitely. For factorial terms $n!z^{-n}$ the cutoff has $n=|z|+O(1)$. This exposes exponentially small contributions and the Stokes phenomenon. The smallest term alone does not always bound the remainder, especially when a saddle meets a pole.

Take the logarithmic derivative of the proposed asymptotic expansion. Through order $z^{-1}$, $u^{\prime }/u=\lambda +\mu /z+O(z^{-2})$ and $u^{\prime \prime }/u={\lambda }^2+2\lambda \mu /z+O(z^{-2})$. Matching the constant and inverse-power terms givesThe second formula assumes a simple root, $2\lambda +f_0\ne 0$. At a repeated root the ansatz can fail: the next equation first requires $f_1\lambda +g_1=0$, and if that holds higher orders determine the remaining behaviour.

For the Bessel differential equation, $\lambda =\pm i$ and $\mu =-1/2$. Set $u=e^{\lambda z}{z}^{-1/2}v(z)$. Direct substitution leaves $v^{\prime \prime }+2\lambda v^{\prime }+(1/4-{\nu }^2)z^{-2}v=0$. With $v=1+a/z+O(z^{-2})$, matching $z^{-2}$ gives $a=(1/4-{\nu }^2)/(2\lambda )$. ThusTwo constant multiples of the Hankel functions normalize their leading coefficients to one and realize these solutions in their appropriate sectors.

For generic $\nu$, these are sectorial expansions, not unchanged expansions for a single branch over arbitrarily continued $\arg z$. The factors $z^{-1/2}$ require a branch, and crossing sector boundaries can introduce the other exponential by Stokes phenomenon and analytic continuation. More precisely, the standard normalized first Hankel solution has its expansion for $-\pi +\delta \le \arg z\le 2\pi -\delta$, and the second for $-2\pi +\delta \le \arg z\le \pi -\delta$, with any fixed positive $\delta$ inside these sectors. In particular both are valid in the common slit sector $-\pi <\arg z<\pi$, away from its boundaries. These sector statements are recorded in NIST's Hankel asymptotics. Exceptional half-integer orders have terminating elementary expressions, so a blanket assertion of nonzero Stokes mixing for every order would be too strong.

Let $b=e^{i\theta }$ and use the method of steepest descents with phase $\Phi (z)=-z^2$. For $z=x+iy$, $\mathrm{Re}\Phi =y^2-x^2$ and $\mathrm{Im}\Phi =-2xy$. The simple saddle point is $z=0$: the real axis is its descending path and the imaginary axis its ascending path. The quadratic complex endpoint descent path from $b$ satisfies $\mathrm{Im}(z^2)=\sin 2\theta$ and can be parametrized bywith the square-root branch chosen continuously from $b$. Increasing $s$ reduces the real part of the exponent by $\lambda s$.

For $0\le \theta <\pi /2$, that endpoint path goes to positive real infinity. Deform the original contour into the real path from $-d$ through the saddle point towards $+\infty$, followed backwards along the endpoint descent path to $b$. The integrand is an entire function, so finite contour deformations are legitimate; connecting arcs in the far decaying sector vanish. If $T_b$ denotes the oriented integral from $b$ to that infinity, this givesThe last expression follows by setting $z^2=b^2+s$: $dz/ds=1/(2z)$, so the endpoint integral is $e^{-\lambda b^2}{\int }_0^{\infty }{e}^{-\lambda s}/[2z(s)]ds$. This derives its phase and orientation directly.

For $\pi /2<\theta \le \pi$, the descent path from $b$ goes to negative real infinity. Deform from $-d$ towards that infinity and then backwards along this endpoint path to $b$. There is no saddle point segment, anduses the same formula with the negative square-root branch. At $\theta =\pi /2$ the contour can run from $-d$ to zero and then up the imaginary axis. It gives a half-Gaussian contribution and the dominant imaginary endpoint term $i{e}^{\lambda }/(2\lambda )$. The change in the subdominant saddle point contribution is a Stokes phenomenon; it does not change the leading endpoint term on either side of that angle.

The Gaussian saddle and endpoint dominance by angle follows by comparing exponential sizes and gives all fixed-angle leading behaviours:At $\theta =\pi /4$, the endpoint term has magnitude $O({\lambda }^{-1})$, so the saddle point term of order ${\lambda }^{-1/2}$ still dominates. In the middle interval the endpoint term grows exponentially for $\pi /4<\theta <3\pi /4$, is oscillatory of order ${\lambda }^{-1}$ at $3\pi /4$, and decays exponentially for $3\pi /4<\theta <5\pi /6$. On the latter interval it nevertheless decays more slowly than the fixed-endpoint term. Equality of their exponential rates means $\cos 2\theta =1/2$ on the left half-plane, hence $\theta =5\pi /6$. At that angle both terms are required; their coefficient magnitudes are unequal, so their sum has no complete destructive cancellation.

These fixed-angle asymptotic expansions are not uniform if $\theta$ is allowed to vary with $\lambda$ near a dominance transition. The endpoint/saddle balance near $\pi /4$ involves $\lambda (\theta -\pi /4)=O(\log \lambda )$; the two-endpoint competition near $5\pi /6$ has an angular scale $O({\lambda }^{-1})$.

The following sketches show the contour orientations, the saddle point and both endpoints. The hyperbolic endpoint paths are followed backwards when returning from infinity to $b$; the dotted far-end connectors indicate the vanishing connecting arc.

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.