There is a relationship between these two concepts. Let the complex function $f(z)$ be analytic on and inside a simple closed curve $C$ in the complex plane. Then Cauchy's integral formula states that for any point $z$ inside $C$ we have
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=2\pi j f(z)\tag{1}$$
For $z$ outside $C$ we have
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=0\tag{2}$$
Furthermore, for $z$ on the curve $C$ the following holds:
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=\pi j f(z)\tag{3}$$
A proof of Eq. $(3)$ can be found in any text on complex analysis.
Let's now choose the curve $C$ as the real line from $-R$ to $R$ and the semi-circle $Re^{j\phi}$, $\phi\in[0,\pi]$, such that $C$ is closed. For $R\to\infty$, the curve $C$ "encloses" the upper half-plane. If $f(z)$ is analytic in the upper half-plane, and if it decays to zero sufficiently fast such that the contribution of the integral along the semi-circle with radius $R$ vanishes as $R\to\infty$, Eq. $(3)$ becomes
$$\int_{-\infty}^{\infty}\frac{f(\zeta)}{\zeta -x}d\zeta=\pi j f(x)\tag{4}$$
Note that the integral in $(4)$ is along the real line. With $z=x+jy$ and $f(z)=f_R(x,y)+jf_I(x,y)$, Eq. $(4)$ can be written as
$$\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)+jf_I(\zeta,0)}{\zeta -x}d\zeta=\pi j \big[f_R(\zeta,0)+jf_I(\zeta,0)\big]\tag{5}$$
Splitting Eq. $(5)$ into real and imaginary parts yields
\begin{align} f_I(\zeta,0) & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)}{\zeta-x}d\zeta = \mathscr{H}\big\{f_R(\zeta,0)\big\} \tag{6}\\ f_R(\zeta,0) & = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_I(\zeta,0)}{\zeta-x}d\zeta = -\mathscr{H}\big\{f_I(\zeta,0)\big\}\tag{7} \end{align}
where $\mathscr{H}\{\cdot\}$ denotes the Hilbert transform.
Equations $(6)$ and $(7)$ are identical to Eqs $(1)$ and $(2)$ in the question with $x_R(t)=f_R(t,0)$ and $x_I(t)=f_I(t,0)$.
[To be continued .......]