)
\end{equation}
where $\chi_I$ is the characteristic function of a collection of finite intervals $I:= [a_1,a_2] \cup [a_3, a_4] \cup \ldots \cup [a_{N-1}, a_{N}]$, which assumes values $0$ or $1$ depending on the variable being respectively outside or within $I$.

\begin{oss}
The Bessel operator is not trace-class on an infinite interval. Thus, it is meaningless to consider the operator restricted to an infinite interval. 
\end{oss}

\begin{oss}
Defining $K_a := K_B(x,y) \chi_{[0,a]}(y)$, then we have
\begin{equation}
K_B (x,y) \cdot \chi_{I}(y) := \sum_{j=1}^N (-1)^j K_{a_j}(x,y)
\end{equation}
\end{oss}

Our goal is to setup a Riemann-Hilbert problem associated to the Fredholm determinant of $K_B\chi_I$. 

\begin{thm}\label{SB1}
The following identity between Fredholm determinants holds
\begin{equation}
\det \left(\operatorname{Id} - K_B\chi_I \right) = \det \left( \operatorname{Id} - \mathbb{B} \right)
\end{equation}
where $\mathbb{B}$ is a trace-class integrable operator 
acting on $L^2(\gamma \cup \hat \gamma)$ with kernel
\begin{subequations}
\begin{align}
&\mathbb{B}(s,t):= \frac{\vec{f}(s)^T \cdot \vec{g}(t)}{s-t} \label{Bintegrableker} \\
&\vec{f}(s) = \frac{1}{2\pi i} \left[ \begin{array}{c} e^{\frac{a_1s}{2}-\frac{1}{4s}}s^{-\nu} \\ 0 \\ 0 \\ \vdots \\ 0   \end{array} \right]\, \chi_{\gamma} (s) + \frac{1}{2\pi i} \left[ \begin{array}{c} 0 \\  e^{-a_1 s + \frac{1}{4s}}s^{\nu} \\ - e^{-a_2 s + \frac{1}{4s}}s^{\nu} \\ \vdots \\ (-1)^{N} e^{-a_{N-1}s + \frac{1}{4s}}s^{\nu} \\ (-1)^{N+1} e^{-a_Ns + \frac{1}{4s}}s^{\nu} \end{array} \right] \, \chi_{\hat \gamma}(s) \label{Bintegrableker3} \\
&\vec{g}(t) = \left[ \begin{array}{c} 0 \\ e^{\frac{a_1t}{2}} \\ e^{t\left(a_2-\frac{a_1}{2}\right)} \\ \vdots \\ e^{t\left( a_N-\frac{a_1}{2}\right)}  \end{array} \right] \, \chi_{\gamma}(t) + \left[ \begin{array}{c} 1 \\0 \\ 0 \\ \vdots \\0  \end{array} \right] \, \chi_{\hat \gamma}(t) \label{Bintegrableker2}
\end{align} 
\end{subequations}
\end{thm}


\begin{proof}
We work on a single kernel $K_{a_j}$ and we will later sum them up, thanks to the linearity of the operations that we are going to perform. 


First of all, we can notice that, if $x<0$ or $y<0$, $K_B(x,y)\equiv 0$; then, using Cauchy's theorem, we can write
\begin{gather}
 K_{a_j}(x,y)= K_B(x,y)  \chi_{[0,a_j]}(y) = \nonumber \\
 =   \int_{i\mathbb{R}+\epsilon} \frac{d\xi}{2\pi i} \, \frac{e^{\xi (a_j-y)}}{\xi - s} \iint_{\hat \gamma \times \gamma} \frac{e^{ xt - \frac{1}{4t} -a_js + \frac{1}{4s}}}{t-s} \left(\frac{s}{t}\right)^\nu   \frac{dt\, ds}{(2\pi i)^2} = \nonumber \\
 =  \int_{i\mathbb{R}+\epsilon} \frac{d\xi}{2\pi i} \, e^{- \xi y} \int_{i\mathbb{R} + \epsilon} \frac{dt}{2\pi i}\, e^{xt} 
 \int_{\hat \gamma} \frac{ds}{2\pi i} \, \frac{e^{ \xi a_j - \frac{1}{4t} -a_j s + \frac{1}{4s}}}{(\xi - s)(t-s)} \left(\frac{s}{t}\right)^\nu 
\end{gather}
where $i\mathbb{R} + \epsilon$ ($\epsilon>0$) is a translated imaginary axis and we continuously deformed the curve $\gamma$ into such translated imaginary axis, in order to make the Fourier operator defined below more explicit. This is possible thanks to the analyticity of the kernel. We also discarded the conjugation term  $\left(\frac{y}{x}\right)^{\nu/2}$, due to the invariance of the Fredholm determinant under conjugation by a positive function.


Defining the following Fourier transform: 
\begin{equation}
\renewcommand\arraystretch{2}
\begin{array}{c|c}
\mathcal{F}: L^2(\mathbb{R}) \rightarrow L^2(i\mathbb{R}+\epsilon) & \mathcal{F}^{-1}: L^2(i\mathbb{R}+\epsilon) \rightarrow L^2(\mathbb{R}) \\
f(x) \mapsto \frac{1}{\sqrt{2\pi i}} \int_\mathbb{R}  f(x) e^{\xi x} dx & h(\xi) \mapsto \frac{1}{\sqrt{2\pi i}} \int_{i\mathbb{R}+\epsilon} h(\xi) e^{-\xi x}d\xi
\end{array}\label{Fourier}
\end{equation} 
it is straightforward to deduce that 
\begin{equation}
K_B \chi_I = \mathcal{F}^{-1} \circ \mathcal{K}_B \circ \mathcal{F}
\end{equation}
with $\mathcal{K}_B = \sum_j (-1)^j \mathcal{K}_{a_j}$ and $\forall \, j=1,\ldots, N$ $\mathcal{K}_{a_j}$ is an operator on $L^2(i\mathbb{R} +\epsilon)$ with kernel
\begin{equation*}
\mathcal{K}_{a_j} (\xi, t) =  \int_{\hat \gamma} \frac{ds}{2\pi i} \,  \frac{e^{ \xi a_j - \frac{1}{4t} -a_j s + \frac{1}{4s}}}{(\xi - s)(t-s)} \left(\frac{s}{t}\right)^\nu
\end{equation*}

In order to ensure convergence of the Fourier-transformed Bessel kernel, we conjugate $\mathcal{K}_B$  by a suitable function
\begin{align}
e^{\frac{a_1t}{2} - \frac{a_1\xi}{2}} \mathcal{K}_B (\xi, t) &=  \sum_{j=1}^N (-1)^j  \int_{\hat \gamma} \frac{ds}{2\pi i} \,  \frac{e^{ \xi \left(a_j-\frac{a_1}{2}\right) + \frac{a_1 t}{2}  - \frac{1}{4t} -a_j s + \frac{1}{4s}}}{(\xi - s)(t-s)} \left(\frac{s}{t}\right)^\nu \nonumber \\
&=: \sum_{j=1}^N (-1)^j \mathcal{K}_{a_j} 
\end{align}
and we continuously deform the translated imaginary axis $i\mathbb{R} + \epsilon$ into its original shape $\gamma$; note that $a_j-\frac{a_1}{2} >0$, $\forall \, j=1, \dots, N$.
With abuse of notation, we call the new kernels $\mathcal{K}_B$ and $\mathcal{K}_{a_j}$ as well.

\begin{lemma}\label{lemma21}
For each $j=1,\ldots, N$, $\mathcal{