Using the well known integral formula for the Hermite polynomials ( see Szego, Orthogonal Polynomials 1967 ):
$H_n(x)=\frac{n!}{2\pi i} \int_{\mid w \mid =1} e^{2xw-w^2-(n+1)logw}dw $
I am trying to compute the main order asymptotics of $H_n(x)$, as $n\rightarrow \infty$ in the case where $x=2^{\frac{1}{2}}(n+1)^{\frac{1}{2}}+2^{-\frac{1}{2}}(n+1)^{-\frac{1}{6}}t$, where t belongs to a bounded region of the complex plane. We may first assume, for simplicity that $t$ is real and bounded.
I am struggling to derive the result for $t$ non zero: For $t=0$, we may let $w=(\frac{n+1}{2})^{\frac{1}{2}}\omega $ to find that $2xw-w^2-(n+1)logw =(n+1)(2\omega-\frac{\omega^2}{2}-log\omega) -\frac{n+1}{2}log\frac{n+1}{2} $
Hence we have that $H_n(x)=\frac{n!}{2\pi i} (\frac{n+1}{2})^{-\frac{n}{2}} \int_{\mid \omega \mid =1} e^{(n+1)(2\omega-\frac{\omega^2}{2}-log\omega)}d\omega $. This is in standard form! Letting $F(\omega)=2\omega-\frac{\omega^2}{2}-log\omega$, we have that the saddle point ( where $F^{'}(\omega)=0$ ) is at $\omega =1$. Its important to note that $F^{''}(1)=0$ too so we cannot use Laplaces method.
Letting $\omega =1+ re^{i\theta}$ we expand $F(\omega)$ at 1 to find, for fixed $\theta$ and small $r$: $F(r)= F(1)+\frac{F^{'''}(1)(re^{i\theta})^3}{6}+O(r^4)$
The directions of steepest descent are where $\Im(F(\omega))=0$: we have that $\Im(F(\omega))=-\frac{1}{3}r^3sin(3\theta) +O(r^4)$. Thus the directions of steepest descent are at $\theta =0,+-\pi/3$
Deforming the contour $\mid \omega \mid = 1$ onto the lines $\theta = +-\pi/3$ and showing that the rest of the integrals are $o(1)$ gives the asymptotics for $t=0$.
Namely, $H_n(x) =e^{\frac{x^2}{2}}2^{\frac{n+1}{2}}e^{\frac{n}{2}logn-\frac{n}{2}}\pi^{\frac{1}{2}}n^{\frac{1}{6}}(Ai(0)+o(1))$
Where $Ai(t) = \int_0^{\infty} cos(\frac{1}{3}r^3+tr)dr$ is the Airy function of the first kind.
I would appreciate any input for the case for bounded and real $t$. I hope the above method is extendable!
I am thinking maybe we need to play with the limits of integration around the saddle point in order to obtain the $O(tr)$ term needed in the exponential in the integrand, to give us Airy! Possibly giving an explicit bound $0\leq r \leq n^{\epsilon}$ for some explicit $\epsilon$.
Any help will be much appreciated.