I'm interested in the limit of large $|z|$ for parabolic cylinder function $D_{\nu}(z)$ with $\nu=-i\lambda,\; z = e^{i\frac{\pi}{4}}x, \; \lambda >0, \; x \in \mathbb{R}$.
On the one hand, there exists well known approximation (see e.g here or Whittaker and Watson "A course of modern analysis"(1990, p. 347-349)):
\begin{equation}
D_{\nu}(z) = z^{\nu}e^{-\frac{z^2}{4}}\left[ \sum\limits_{n=0}^N \frac{\left(-\frac{\nu}{2}\right)_n \left(\frac{1}{2} - \frac{\nu}{2}\right)_n}{n!\left(-\frac{z^2}{2}\right)^n} + \mathcal{O}\left(\frac{1}{|z^2|^{N+1}}\right)\right], \; |\text{arg}(z)|<\frac{3\pi}{4} \;\;(1)
\end{equation}
\begin{equation}
D_{\nu}(z) = z^{\nu}e^{-\frac{z^2}{4}}\left[ \sum\limits_{n=0}^N \frac{\left(-\frac{\nu}{2}\right)_n \left(\frac{1}{2} - \frac{\nu}{2}\right)_n}{n!\left(-\frac{z^2}{2}\right)^n} +\mathcal{O}\left(\frac{1}{|z^2|^{N+1}}\right)\right] + \frac{\sqrt{2\pi}e^{-\nu\pi i + \frac{z^2}{4}}}{\Gamma(-\nu)}z^{-\nu-1}\left[\sum\limits_{n=0}^N \frac{\left(\frac{\nu}{2}\right)_n \left(\frac{1}{2} + \frac{\nu}{2}\right)_n}{n!\left(\frac{z^2}{2}\right)^n} + \mathcal{O}\left(\frac{1}{|z^2|^{N+1}}\right)\right], \\ -\frac{5\pi}{4} < \text{arg}(z) <-\frac{\pi}{4} \; \; (2)
\end{equation}
where $(\alpha)_0 = 1, \; (\alpha)_{n\neq 0} = \alpha(\alpha+1)\ldots(\alpha+n-1)$. In the common range of validity $-\frac{3\pi}{4}<\text{arg}(z)<-\frac{\pi}{4}$ the second term in (2) is exponentially small, so it's not a problem.
On the other hand, there is a way to estimate the function using the method of steepest descent. Let me just show the way it is done in this work https://iopscience.iop.org/article/10.1088/0305-4470/5/12/007/pdf (sorry me, I can't find free available source of it). We start from the representation: \begin{equation} D_{\nu}(z) = \frac{\Gamma(1+\nu)}{2\pi i}e^{-\frac{z^2}{4}}z\int \exp\left(z^2(v-\frac{1}{2}v^2) - (1+\nu)\log(zv) \right)dv, \; \arg(z) \in (-\pi,\pi)\;\;\; (3) \end{equation} where the contour goes around the cut $[0,e^{i(\pi-\arg(z))}\infty)$ in the comlex $v$-plane. We assume that both $|z|$ and $|\nu|$ are large enough. Saddle points are: \begin{equation} f(v)=z^2(v-\frac{1}{2}v^2) - (1+\nu)\log(zv) \end{equation} \begin{equation} f'(v)=0 \Rightarrow 2zv_{\pm} = z \pm \sqrt{z^2-4(1+\nu)}, \;\arg(zv_{\pm}) \in (-\pi,\pi). \end{equation} Then the authors provide the general result, which contains contributions from these saddle points. I've derived for my particular $z$ and $\nu$ the following: \begin{equation} D_{\nu}(z) \simeq \frac{1}{\sqrt{2}} \sqrt{1+\frac{1}{2} \frac{x}{\sqrt{\lambda+\frac{x^2}{4}}}} e^{ \pi \frac{\lambda}{4} + i\frac{\lambda}{2} - i\lambda \log \left[\sqrt{\lambda+\frac{x^2}{4}} + \frac{x}{2}\right] -ix\sqrt{\lambda+\frac{x^2}{4}} + i\varphi_1 } + \frac{1}{\sqrt{2}} \sqrt{1-\frac{1}{2} \frac{x}{\sqrt{\lambda+\frac{x^2}{4}}}} e^{-3 \pi \frac{\lambda}{4} + i\frac{\lambda}{2} - i\lambda \log \left[\sqrt{\lambda+\frac{x^2}{4}} - \frac{x}{2}\right] + ix\sqrt{\lambda+\frac{x^2}{4}} + i\varphi_2 }, \; \; (4) \end{equation} where $\varphi_1,\;\varphi_2$ are some negligible phases. I investigate the case $|x| \gg \lambda \gg 1$ and have the following problem: for $x<0, \; \arg(z) = -\frac{3\pi}{4}$ expressions (2) and (4) match very well (if we expand gamma function in (2) and square roots in (4)). However, for $x>0$ asymptotic (1) gives only the first term in (4). Yes, the second term in (4) in this limit is suppressed by $e^{-\pi \lambda}$ and decays with $x$, but we pose $\lambda$ to be fixed and some terms in (1) decay faster!
It seems (1) is true, because in this case the exact relation holds: \begin{equation} D_{\nu}(z) = e^{-\nu \pi i}D_{\nu}(-z) + \frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{-\frac{1}{2}(\nu+1)\pi i} D_{-\nu-1}(iz) \end{equation} (actually, (2) is obtained using this relation in books).
I may be wrong, but I don't see the reason why the second saddle point does not contribute when $x>0$ and contributes when $x<0$. I provide the following trial: here are lines of constant phase $Im(f(v))=Im(f(v_{\pm}))$, which pass through the saddle points ($v_{-}$ is intersection point of orange lines when $x<0$, $v_{+}$ is near $0$, and they swap when $x>0$, green line is the cut). As I understand, we should deform the contour around the cut to some contour along these lines (am I right?). 
I see good contour for $x<0$: orange line parallel to the cut and then the blue line. Also it can be seen, that the orange line closest to the cut gives desired result for $x>0$. But still it's not clear: first, does smth strange happening when the lines go around the cut near $0$ make extra complexity? And why the orange line on the second picture is the correct contour (if all this consideration is correct)? Could anyone help me with this problem? Thank you.