& \quad +2i\int_{0}^{\varkappa /2}d\beta \int d\alpha \;e^{i\lambda
(x+y)}F(\alpha ,\beta )  \label{eq5.3.1} \\
& \equiv H_{1}(x+y)+H_{2}(x+y),  \notag\end{aligned}$$where $$F(\alpha ,\beta )\equiv \frac{1}{2\pi }\zeta _{z,t}(\lambda )(2i\lambda
)^{n-1}\frac{S(\lambda )}{B(\lambda )}\partial _{\overline{\lambda }}G(\alpha ,\beta ).$$

Due to the rapid decay of $e^{8i\lambda^3t}$ as $\lambda\to\infty$ along $\mathbb{R}+ih$, the function $F(\alpha,\beta)$ is subject to the conditions of Proposition \[pr4.1\] and hence the integral operator with kernel $H_1$ is trace class. Our analysis of the integral operator with kernel $H_2$ is based upon the following lemma.

\[lem5.1\] Let $F(\alpha ,\beta )$ be such that for some $h>0$ $$\int_{0}^{h}\left( \int_{-\infty }^{\infty }\left\vert F\left( \alpha ,\beta
\right) \right\vert d\alpha \right) \frac{d\beta }{\beta }<\infty .
\label{star}$$Then the integral Hankel operator $\mathbb{H}$ with the kernel ($\lambda
=\alpha +i\beta $) $$H(x)=\int_{0}^{h}d\beta \int_{-\infty }^{\infty }\frac{d\alpha }{2\pi }e^{i\lambda x}F(\alpha ,\beta )$$is trace class and $$\left\Vert \mathbb{H}\right\Vert _{\mathfrak{S}_{1}}\leq \frac{1}{2}\int_{0}^{h}\frac{d\beta }{\beta }\int_{-\infty }^{\infty }d\alpha
\left\vert F(\alpha ,\beta )\right\vert .$$

We have $$\begin{aligned}
H(x+y)& \eqsim \int_{0}^{h}d\beta e^{-\beta x}\int_{-\infty }^{\infty
}d\alpha e^{i\alpha x}F(\alpha ,\beta )  \notag \\
& =\int_{0}^{h}e^{-\beta (x+y)}\widehat{F_{\beta }}(x+y)dx,
\label{eq5.2.1too} \\
\widehat{F_{\beta }}(x+y)& \equiv\int_{-\infty }^{\infty }e^{i\alpha
(x+y)}F(\alpha +\beta )d\alpha  \notag \\
& \eqsim \widehat{F_{\beta }^{1/2}}\ast \widehat{F_{\beta }^{1/2}}(x+y) 
\notag \\
& =\int_{-\infty }^{\infty }F_{\beta }^{1/2}(x+s)F_{\beta }^{1/2}(y-s)ds.
\label{eq5.2.2}\end{aligned}$$ Here we have used the convolution theorem. Inserting into implies that 