We aim to show $$\int_{-\infty}^\infty e^{i\lambda x^2} x^l e^{-x^2}\, dx \sim \lambda^{-(l+1)/2} \sum_{j=0}^\infty c_j^{(l)} \lambda^{-j} \tag{9}$$ for even non-negative integers $l$, as done in Chapter 8 of Stein's Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. I shall ask two questions, while quoting the proof in the text.
The left side of $(9)$ is $$\int_{-\infty}^\infty e^{-(1-i\lambda) x^2} x^l \, dx;$$ setting $z = (1-i\lambda)^{1/2}\, x$ and noting that the rapid decay of $e^{-z^2}$ allows us to replace the contour $(1-i\lambda)^{1/2}\cdot \Bbb R$ by $\Bbb R$, we see that the above integral equals $$(1-i\lambda)^{-1/2 - l/2} \int_{-\infty}^\infty e^{-x^2} x^l \, dx$$
- What role does the rapid decay of $e^{-z^2}$ play in replacing the contour $(1-i\lambda)^{1/2}\cdot \Bbb R$ by $\Bbb R$? I'm not sure how to show that $$\int_{(1-i\lambda)^{1/2}\cdot \Bbb R} e^{-x^2} x^l \, dx = \int_{\Bbb R} e^{-x^2} x^l \, dx$$
Here, we have fixed the principal branch of $z^{-(l+1)/2}$ in the complex plane slit along the negative half-axis. With this determination, we have $$(1-i\lambda)^{-(l+1)/2} = \lambda^{-(l+1)/2} \cdot (\lambda^{-1} - i)^{-(l+1)/2} $$ if $\lambda > 0$. Thus, the power series expansion of $(w-i)^{-(l+1)/2}$ (on the open unit disc) gives us the asymptotic expansion $(9)$ with $w = \lambda^{-1} \to 0$.
- What exactly does the author mean by the symbol $\sim$ in this context? Following the above argument, we would get $$\int_{-\infty}^\infty e^{i\lambda x^2} x^l e^{-x^2}\, dx = \lambda^{-(l+1)/2} \sum_{j=0}^\infty c_j^{(l)} \lambda^{-j}\left(\int_{-\infty}^\infty e^{-x^2} x^l \, dx\right) \tag{$\ast$}$$ and I wonder why the term $\int_{-\infty}^\infty e^{-x^2} x^l \, dx$ is dropped from the expression, even though it is dependent on $l$.