f_2
\\
f_3
\end{pmatrix}:=\widetilde \Psi(x)
\begin{pmatrix}
1
\\
0
\\
0
\end{pmatrix}, \qquad
{\mathbf}{h}(y)=\begin{pmatrix}
h_1
\\
h_2
\\
h_3
\end{pmatrix}
:=
\frac{1}{2 \pi i}
\widetilde \Psi^{-t}(y) \begin{pmatrix}
0
\\
1
\\
1
\end{pmatrix}.$$

With the function $F$ defined in , we have $$\label{eq:derivatives}
\frac{\partial}{\partial s}F(s;\rho)=\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}s} \ln \det(I-K_{s,\rho}^{\mathrm{Pe}})
=-\textrm{tr}\left((I-K_{s,\rho}^{\mathrm{Pe}})^{-1}
\frac{{\,\mathrm{d}}}{{\,\mathrm{d}}s}K_{s,\rho}^{\mathrm{Pe}}\right)=-R(s,s)-R(-s,-s),$$ where $R(u,v)$ stands for the kernel of the resolvent operator, that is, $$R=\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}-I=K_{s,\rho}^{\mathrm{Pe}}\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}=\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}K_{s,\rho}^{\mathrm{Pe}}.$$ Since the kernel of the operator $K_{s,\rho}^{\mathrm{Pe}}$ is integrable in the sense of [@IIKS90], its resolvent kernel is integrable as well; cf. [@DIZ97; @IIKS90]. Indeed, by setting $$\label{def:FH}
{\mathbf}{F}(u)=
\begin{pmatrix}
F_1 \\
F_2 \\
F_3
\end{pmatrix}:=\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}{\mathbf}{f}, \qquad {\mathbf}{H}(v)=\begin{pmatrix}
H_1 \\
H_2 \\
H_3
\end{pmatrix}
:=\left(I-K_{s,\rho}^{\mathrm{Pe}}\right)^{-1}{\mathbf}{h},$$ we have $$\label{eq:resolventexpli}
R(u,v)=\frac{{\mathbf}{F}^t(u){\mathbf}{H}(v)}{u-v}.$$

We could also represent $\frac{\partial}{\partial \rho}F(s;\rho)$ in terms of ${\mathbf}{F}$ and ${\mathbf}{H}$ defined in . To proceed, we note from and that $$\label{def:fderivative}
\frac{\partial {\mathbf}{f}}{\partial \rho}(x)
=
\frac12
\begin{pmatrix}
0 & 0 & 1
\\
x & \rho & 0
\\
1 & x & \rho
\end{pmatrix}
{\mathbf}{f}(x).$$ Moreover, by taking derivative with respect to $\rho$ on both sides of $\Psi \cdot \Psi^{-1}=I$, it is readily seen from that $$\frac{\partial \Psi^{-1}}{\partial \rho} = -\frac{\Psi^{-1}}{2}
\begin{pmatrix}
0 & 0 & 1
\\
z & \rho & 0
\\
1 & 