\exp\bigg\{-\frac{(k-k_0)^2}{4\sigma^2} - i k V_0\bigg\}$$ with $k_0$, $\sigma$ and $V_0$ the central frequency, bandwidth and central position, respectively. In the narrow bandwidth limit ($k_0 \gg \sigma$), the integration over $k$ can be approximately calculated to a very good accuracy.

When $k_0 \gg \sigma$, the Gaussian wave packet $f(k; k_0, \sigma, V_0)$ is significantly nonzero only for positive $k$, so the range of integration of $k$ can be extended to $(-\infty, \infty)$ without introducing large errors. Secondly, since $f(k; k_0, \sigma, V_0)$ is well localized around $k_0$, those values of $A_{k\omega}$ and $B_{k\omega}$ only near $k_0$ are relevant. Writing [@Downes13] $$\frac{1}{\sqrt{k}}\bigg( \frac{k}{a}\bigg)^{i \omega/a} \approx \frac{1}{\sqrt{k_0}}e^{i\frac{\omega}{k_0}\frac{k}{a}}e^{i\frac{\omega}{a}[\ln(\frac{k_0}{a})-1]}$$ and then expanding $A_{k\omega}$ and $B_{k\omega}$ around $k_0$ yields

$$\begin{aligned}
\label{A}
A_{\omega} \equiv \int_0^{\infty} d k f(k) A_{k \omega} 
&\approx& i \sqrt{\frac{\sigma}{\pi \omega k_0}} \bigg(\frac{1}{2\pi}  \bigg)^{1/4} \sqrt{2\sinh(\pi \omega/a)} \Gamma(1-i\omega/a)
e^{i\frac{\omega}{a}\ln(\frac{k_0}{a})} e^{-i k_0 V_0}\exp\bigg\{-\frac{\sigma^2(\omega/a-k_0 V_0)^2}{k_0^2}\bigg\},  \nonumber \\\end{aligned}$$

$$\begin{aligned}
\label{B}
B_{\omega} \equiv \int_0^{\infty} d k f(k) B_{k \omega}
&\approx& -i \sqrt{\frac{\sigma}{\pi \omega k_0}} \bigg(\frac{1}{2\pi}  \bigg)^{1/4} \sqrt{2\sinh(\pi \omega/a)} \Gamma(1+i\omega/a)
e^{-i\frac{\omega}{a}\ln(\frac{k_0}{a})} e^{-i k_0 V_0}\exp\bigg\{-\frac{\sigma^2(\omega/a+k_0 V_0)^2}{k_0^2}\bigg\} \nonumber \\\end{aligned}$$

up to first order in $k-k_0$.

Using Eq. (\[ParticleNumber:1\]) and $$|\Gamma(1-i\omega/a)|^2 = |\Gamma(1+i\omega/a)|^2  = \frac{\pi \omega/a}{\sinh(\pi \omega/a)}$$ the expectation value $N(f) = \langle 0_M | \hat{a}^{\dagger}(f)\hat{a}(f)|0_M \rangle$ of the Gaussian mode particle number 