$x$ in the complex plane and obtain $$\int_0^{\infty} \vert\beta^1_{\omega'\omega}\vert^2 \; {\mathrm{d}}\omega' = 
\lim_{\epsilon \to 0}
\frac{d_0^2}{4\pi^2 c_1^2} \int_{0-\mathrm{i}\epsilon}^{\infty-\mathrm{i}\epsilon} 
\frac{\omega'}{\omega} 
\Bigg|\int_{x_0}^{x_0-\mathrm{i}\infty} e^{-\mathrm{i}\omega_+ x} 
x^{-\mathrm{i}\omega\lambda} \; {\mathrm{d}}x \; \Bigg|^2
\;{\mathrm{d}}\omega' \; .$$ Introducing $\tilde{\omega}_+ = \tilde{\omega} + 
\omega' \frac{d_0}{c_1}$ and $\tilde{\omega} := \omega + \varepsilon$, we regularize the product of the inner integrations by a frequency split, $$\begin{aligned}
&&\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega' = 
\lim_{\epsilon \to 0 \atop \varepsilon \to 0}
\frac{1}{4\pi^2 c_1^2} \times \nonumber \\
&&\int_{0-\mathrm{i}\epsilon}^{\infty-\mathrm{i}\epsilon} 
\frac{\omega'}{\omega} 
\int_{x_0}^{x_0-\mathrm{i}\infty} e^{-{\mathrm{i}} \omega_+ x} 
x^{-\mathrm{i}\omega \lambda} \; {\mathrm{d}}x \; 
\int_{x_0}^{x_0+\mathrm{i}\infty} 
e^{\mathrm{i}\tilde{\omega}_+^* \tilde{x}} 
\tilde{x}^{\mathrm{i}\tilde{\omega} \lambda} \; \mathrm{d}\tilde{x} 
\;{\mathrm{d}}\omega' \; , \label{57}\end{aligned}$$ where we took the limit partially. In the first of the inner integrals we further substitute $\mathrm{i}\omega_+ x =:y$ and in the second $-\mathrm{i}\tilde{\omega}^*_+\tilde{x} =:\tilde{y}$, $$\begin{aligned}
 \label{55}
&&\int_0^{\infty} \vert \beta^1_{\omega'\omega} \vert^2 \; \mathrm{d}\omega' =
\lim_{\epsilon \to 0 \atop \varepsilon \to 0}
\frac{1}{4\pi^2 c_1^2 \omega} e^{- \pi \omega \lambda}\times \\
&&\int_{0-\mathrm{i}\epsilon}^{\infty-\mathrm{i}\epsilon} \omega' 
\frac{\omega_+^{\mathrm{i}\lambda\omega}}{\omega_+}
\frac{(\tilde{\omega}_+^*)^{-\mathrm{i}\lambda\tilde{\omega}}}
     {\tilde{\omega}_+^*}
\int_{\mathrm{i}\omega_+ x_0}^{\mathrm{i}\omega_+ x_0+\infty} e^{-y} 
y^{-\mathrm{i}\omega \lambda} \; {\mathrm{d}}y \; 
\int_{-\mathrm{i}\tilde{\omega}_+^* x_0}^{-\mathrm{i}\tilde{\omega}_+^* x_0+
\infty} e^{-\tilde{y}} 
\tilde{y}^{\mathrm{i}\tilde{\omega}\lambda} \; {\mathrm{d}}\tilde{y} \;
{\mathrm{d}}\omega' \; . \nonumber\end{aligned}$$ Since $$\begin{aligned}
(\omega_+^* + \varepsilon)^{-\mathrm{i}\tilde{\omega}\lambda}=
(\omega_+^*)^{-\mathrm{i}\tilde{\omega}\lambda }+O(\varepsilon) \;\;,\end{aligned}$$ and ignoring terms ${\cal O}(\varepsilon)$, we can write $$\begin{aligned}
&&\int_0^{\infty} \vert\beta^1_{\omega'\omega}\vert^2 \; {\mathrm{d}}\omega' = 
\lim_{\epsilon \to 0 \atop \varepsilon 