Double line integral of a feynman propagator

Question

I'm having trouble in finding the result of the following double integral of a Feynman propagator in Euclidean signature: \begin{equation} I = -\frac{i}{(2\pi)^2}\oint_\gamma dx \oint_\gamma dy\,\frac{1}{(x-y)^2}. \tag{1} \end{equation} The integrals are on the same contour (a loop only on spatial directions), and $(x-y)^2=(x^0-y^0)^2+(x^1-y^1)^2+(x^2-y^2)^2+(x^3-y^3)^2$. I tried to do these integrals by making the replacement: \begin{equation} \oint dy \,\frac{1}{(x-y)^2}\leadsto \int d^4w \oint dy\,\frac{\delta^{(4)}(y-w)}{(x-w)^2}. \end{equation} But I ended up with \begin{equation} I = -\frac{1}{4\pi} \oint dx \oint dy\,\frac{\delta(|\vec{x}-\vec{y}|)}{|\vec{x}-\vec{y}|}, \end{equation} wich seems not to be well behaved. When the loop extends into the temporal direction by a length $T$ such that $T \gg R$ (where $R$ is the spatial extent of the loop), the following result is found: \begin{equation} I \simeq -\frac{1}{4\pi R}T. \end{equation} I couldn't find any resources on the internet about how to deal with (1) when the loop is purely spatial, so I'm really stuck on this problem.