After learning that both the improper integrals $\int_{-\infty}^{\infty} e^{-x^2} \, dx$ and $\int_{-\infty}^{\infty} \frac{1}{x^2 + 1} \, dx$ converge to $\sqrt{\pi}$ and $\pi$ respectively, I hypothesized that the product of their integrands would also converge. Since both integrands are always between $0$ and $1$, the product should be less than both of them at any point meaning, $$\int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1} \, dx$$ should converge to a value less than $\sqrt{\pi}$.
However, I have had little success solving this integral. I tried to solve it like solving the Gaussian integral, letting $$I=\int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1} \, dx$$ $$I^2=\int_{-\infty}^{\infty} \frac{e^{-x^2}}{x^2 + 1} \, dx\int_{-\infty}^{\infty} \frac{e^{-y^2}}{y^2 + 1} \, dy$$ $$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{e^{-y^2}}{y^2 + 1} \, \frac{e^{-x^2}}{x^2 + 1} \, dxdy$$ However, I have had trouble simplifying the denominator down from here. If a closed form exists, I hope someone can help me find it.
