Expanding out \(\alpha\) we get
\begin{align} \label{eq:alpha} \alpha & =a^{-1/3}\left(\frac{b^{2}}{3a}-\frac{2b^{2}}{3a}+c\right),\notag \\ & =a^{-1/3}\left(c-\frac{b^{2}}{3a}\right),\notag \\ & \label{eq:alpha}=\left(\frac{\ddot{f_{k}}}{3}\right)^{-1/3}\pi^{2/3}\left(2(f_{k}-f)-\frac{\dot{f_{k}}^{2}}{\ddot{f_{k}}}\right).\\ \end{align}Expanding out \(\beta\) (and substituting in \(d=0\) from the start) we get
\begin{align} \label{eq:beta} \beta & =-\frac{b}{3a}\left(\frac{b^{2}}{9a}-\frac{b^{2}}{3a}+c\right),\notag \\ & =\frac{b}{3a}\left(\frac{2b^{2}}{9a}-c\right),\notag \\ & \label{eq:beta}=\frac{2\pi\dot{f_{k}}}{\ddot{f_{k}}}\left(\frac{\dot{f_{k}}^{2}}{3\ddot{f_{k}}}-(f_{k}-f)\right).\\ \end{align}Putting this back into equation \ref{eq:fourier2} gives
\begin{equation} \label{eq:fourier3} \label{eq:fourier3}H_{k}(f)=y_{k}\left(\frac{\pi}{3}\ddot{f_{k}}\right)^{-1/3}e^{i\left(\phi_{k}-\pi f\Delta t+\beta\right)}\int_{z_{1}}^{z_{2}}\exp{\left[i\left(z^{3}+\alpha z\right)\right]}{\rm d}z,\\ \end{equation}where
\begin{align} z_{1} & =\left[\left(\frac{\pi}{3}\right)^{4/3}\dot{f_{k}}\ddot{f_{k}}^{1/3}-\frac{\Delta t}{2}\right]\left(\frac{\pi}{3}\ddot{f_{k}}\right)^{1/3},\notag \\ z_{2} & =\left[\left(\frac{\pi}{3}\right)^{4/3}\dot{f_{k}}\ddot{f_{k}}^{1/3}+\frac{\Delta t}{2}\right]\left(\frac{\pi}{3}\ddot{f_{k}}\right)^{1/3}.\\ \end{align}