I want to prove Theorem 2.5 on page 9 from these lecture notes by following the guidelines provided in Exercise 13 on page 18. Using complex integration one shows that \begin{equation}\int_{\mathbb{R}} e^{i\lambda x^2}e^{-x^2}x^m\text{d}x=(1-i\lambda)^{-\frac{m+1}{2}}\int_{\mathbb{R}} e^{-x^2}x^m \text{d}x=\lambda^{-\frac{m+1}{2}}\left(\frac{1}{\lambda}-i\right)^{-\frac{m+1}{2}}\int_{\mathbb{R}} e^{-x^2}x^j \text{d}x\;\;\;\;\;(1)\end{equation} Then the hint is to write $$\left(z-i\right)^{-\frac{m+1}{2}}=\sum_{k=0}^{\infty}c_{k}^{(m)}z^{k}$$ as a power series for $|z|<1$ and set $z=\frac{1}{\lambda}$. Until here I think I understand all steps but what I don't see is how to conclude $$\int_{\mathbb{R}} e^{i\lambda x^2}e^{-x^2}x^m\text{d}x-\lambda^{-1/2}\sum_{j=0}^{N-1}a_j\lambda^{-j/2}=\mathcal{O}(\lambda^{-N/2})$$ for suitable coefficients $a_j$ and $N\in\mathbb{N}$.
Edit: Solved thanks to advice in the comments. Using $\left(\frac{1}{\lambda}-i\right)^{-\frac{m+1}{2}}=\sum_{k=1}^{N-1}c_k^{(m)}\lambda^{-k}+\mathcal{O}(\lambda^{-N})$ we get that the right side of (1) euqals (for convenience I set $\hat{c}_j=\int e^{-x^2}x^j \text{d}x$) $$\hat{c}_j\lambda^{-\frac{m+1}{2}}\sum_{k=1}^{N-1}c_k^{(m)}\lambda^{-k}+\mathcal{O}(\lambda^{-N-\frac{m+1}{2}})=\hat{c}_j\lambda^{-\frac{1}{2}}\sum_{k=1}^{N-1}c_k^{(m)}\lambda^{-\frac{2k+m}{2}}+\mathcal{O}(\lambda^{-N-\frac{m+1}{2}})$$ Rename the coefficients and we are done.