function is holomorphic, $$\begin{aligned}
\left\vert \int_{\gamma^u_{R}} h(z) \mbox{d}z \right\vert \leq 2 R B\quad 
\mbox{and} \quad 
\left\vert \int_{\gamma^d_{R}} h(z) \mbox{d}z \right\vert \leq 2 R B \;\; ,\end{aligned}$$ where the path $\gamma^d_{R}$ is the half-circle of radius $R$ in the lower complex plane and $\gamma^u_{R}$ the half-circle of radius $R$ in the upper complex plane. From this we find $$\begin{aligned}
\lim_{R \to \infty} \vert \int_{\gamma^u_{R}} \frac{h(z)}{z^2} \mbox{d}z \vert 
&=& 0 \\
\lim_{R \to \infty} \vert \int_{\gamma^d_{R}} \frac{h(z)}{z^2} \mbox{d}z \vert 
&=& 0 \ .\end{aligned}$$ Our integral of interest can now be expressed as $$\begin{aligned}
\int_{-\infty}^{\infty}
e^{- \mathrm{i}\lambda \varepsilon z} h(z) \; \mbox{d} z = 
- \frac{1}{\lambda^2} \partial^2_{\varepsilon} \int_{-\infty}^{\infty}
e^{- \mathrm{i}\lambda \varepsilon z} 
\frac{h(z)}{z^2} \; \mbox{d} z  \; \; ,\end{aligned}$$ which allows us to make use of the residue theorem. Now for $\varepsilon > 0$ we may close the contour in the lower plane. To compute the integral we shift the pole into the lower plane. This yields $$\begin{aligned}
\int_{-\infty}^{\infty}
e^{-\mathrm{i}\lambda \varepsilon z} \frac{h(z)}{z^2} 
\; \mbox{d} z = - 2\pi \left( \lambda \varepsilon h + \mathrm{i} \partial_z h
\right) \Theta(\varepsilon) \ .\end{aligned}$$ With this we now get $$\begin{aligned}
\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega'&=&
\lim_{\varepsilon \to 0} \frac{2\pi}{\lambda} h(0) \delta(\varepsilon) \\ 
&=& \lim_{\varepsilon \to 