Demonstration of an identity between a simple and a double integral of a Gaussian-Lorentzian product

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In the context of a physics problem, I found that a double integral of Gaussian and Lorentzian products has values that are extremely close to that of a single integral.

By defining $D(\sigma) = \frac{1}{\sqrt{2 \pi} \sigma} \int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} e^{-\frac{(x-y)^2}{2 \sigma^2}} \dfrac{1}{1 + x^2} \dfrac{1}{1 + y^2} dx dy$

and $S(\sigma) = \frac{1}{\sqrt{2 \pi} \sigma} \int_{- \infty}^{+ \infty} e^{-\frac{x^2}{2 \sigma^2}} \dfrac{1}{1 + x^2} dx $

I find numerically that $ D(2\sigma) \approx S(\sigma)\times \pi/2 $, evaluated for $\sigma \in (10^{-2}, 10^2)$ and the difference $< 10^{-4}$.

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Moreover, I was able to check that in the limits $\sigma \rightarrow 0$ and a simple equivalent when $\sigma \rightarrow \infty$, the equality is true, which suggests that this could in general be a strict equality.

It really looks like a nice problem, but I did not find any way to prove the equality true (or true within some smart approximation). Do any of you have any idea how to prove this? (Interestingly there is an analytical expression for $S(\sigma)$).