Mainly for purposes of comparison, let me flesh out what I called "a cheap version of Laplace".

Choose $\rho\in (0,1)$. Let $g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then $$g''(y) = \frac{-\sigma}{(x^2+y^2)^{\frac{\sigma}{2} + 1}} + \frac{\sigma \left(\frac{\sigma}{2}+1\right) \cdot 2 y^2}{(x^2+y^2)^{\frac{\sigma}{2} + 2}}$$ and so $|g''(y)|\leq \sigma (\sigma+1)/(x^2+y^2)^{\sigma/2+1}$. Let $I$ be the interval $\lbrack (1-\rho) y_0, (1+\rho) y_0\rbrack$. Then, for $y\in I$, $|g''(y)|\leq \sigma (\sigma+1)/((1-\rho) l_0)^{\sigma+2}$, where $l_0 = \sqrt{x_0^2+y_0^2}$, and so $$g(y) = g(y_0) + g'(y_0) (y- y_0) + O^*\left(c_0 (y-y_0)^2\right),$$ where $c_0 = \sigma (\sigma+1)/2 ((1-\rho) l_0)^{\sigma+2}$. Thus, by cancellation, $$\int_I g(y) e^{-(y-y_0)^2/2} dy = \int_I g(y_0) e^{-(y-y_0)^2/2} dy + O^*\left(\int_I c_0 (y-y_0)^2 e^{-(y-y_0)^2/2} dy\right).$$ Since $g'(y)<0$ for $y\geq 0$, we also know that $g(y)<g(y_0) + c_0 (y-y_0)^2$ for $y>(1+\rho) y_0$. We conclude that $$\begin{aligned}\int_{(1-\rho) y_0}^{\infty} g(y) e^{-(y-y_0)^2/2} dy &\leq g(y_0) \int_{-\infty}^\infty e^{-y^2/2} dy + c_0 \int_{-\infty}^\infty y^2 e^{-y^2/2} dy\\ &= g(y_0) \sqrt{2\pi} + c_0 \sqrt{2\pi} = \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} .\end{aligned}$$

It remains to consider $y\leq (1-\rho) y_0$. Since $g(y)\leq g(0) = 1/x_0^\sigma$, $$\begin{aligned}\int_{-\infty}^{(1-\rho) y_0} g(y) e^{-(y-y_0)^2/2} dy &= \frac{1}{x_0^\sigma} \int_{-\infty}^{-\rho y_0} e^{-y^2/2} dy \\ &\leq \frac{1}{x_0^\sigma (\rho y_0)} \int_{-\infty}^{-\rho y_0} y e^{-y^2/2} dy = \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}. \end{aligned}$$

Thus we obtain $$\int_{-\infty}^\infty \frac{e^{-(y-y_0)^2/2}}{(x_0^2+y^2)^{\sigma/2}} dy \leq \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} + \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}$$ for any $0<\rho<1$.

Hardly very powerful or elegant, but I wonder: (a) is the above qualitatively optimal? That is, are the lesser-order terms of the right order? (b) can one give an even quicker proof of the same or a stronger bound?

Mainly for purposes of comparison, let me flesh out what I called "a cheap version of Laplace". Write $\sigma = 2 r$.

Choose $\rho\in (0,1)$. Let $g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then $$g''(y) = \frac{-\sigma}{(x^2+y^2)^{\frac{\sigma}{2} + 1}} + \frac{\sigma \left(\frac{\sigma}{2}+1\right) \cdot 2 y^2}{(x^2+y^2)^{\frac{\sigma}{2} + 2}}$$ and so $|g''(y)|\leq \sigma (\sigma+1)/(x^2+y^2)^{\sigma/2+1}$. Let $I$ be the interval $\lbrack (1-\rho) y_0, (1+\rho) y_0\rbrack$. Then, for $y\in I$, $|g''(y)|\leq \sigma (\sigma+1)/((1-\rho) l_0)^{\sigma+2}$, where $l_0 = \sqrt{x_0^2+y_0^2}$, and so $$g(y) = g(y_0) + g'(y_0) (y- y_0) + O^*\left(c_0 (y-y_0)^2\right),$$ where $c_0 = \sigma (\sigma+1)/2 ((1-\rho) l_0)^{\sigma+2}$. Thus, by cancellation, $$\int_I g(y) e^{-(y-y_0)^2/2} dy = \int_I g(y_0) e^{-(y-y_0)^2/2} dy + O^*\left(\int_I c_0 (y-y_0)^2 e^{-(y-y_0)^2/2} dy\right).$$ Since $g'(y)<0$ for $y\geq 0$, we also know that $g(y)<g(y_0) + c_0 (y-y_0)^2$ for $y>(1+\rho) y_0$. We conclude that $$\begin{aligned}\int_{(1-\rho) y_0}^{\infty} g(y) e^{-(y-y_0)^2/2} dy &\leq g(y_0) \int_{-\infty}^\infty e^{-y^2/2} dy + c_0 \int_{-\infty}^\infty y^2 e^{-y^2/2} dy\\ &= g(y_0) \sqrt{2\pi} + c_0 \sqrt{2\pi} = \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} .\end{aligned}$$

It remains to consider $y\leq (1-\rho) y_0$. Since $g(y)\leq g(0) = 1/x_0^\sigma$, $$\begin{aligned}\int_{-\infty}^{(1-\rho) y_0} g(y) e^{-(y-y_0)^2/2} dy &= \frac{1}{x_0^\sigma} \int_{-\infty}^{-\rho y_0} e^{-y^2/2} dy \\ &\leq \frac{1}{x_0^\sigma (\rho y_0)} \int_{-\infty}^{-\rho y_0} y e^{-y^2/2} dy = \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}. \end{aligned}$$

Thus we obtain $$\int_{-\infty}^\infty \frac{e^{-(y-y_0)^2/2}}{(x_0^2+y^2)^{\sigma/2}} dy \leq \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} + \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}$$ for any $0<\rho<1$.

Hardly very powerful or elegant, but I wonder: (a) is the above qualitatively optimal? That is, are the lesser-order terms of the right order? (b) can one give an even quicker proof of the same or a stronger bound?

Mainly for purposes of comparison, let me flesh out what I called "a cheap version of Laplace".

Choose $\rho\in (0,1)$. Let $g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then $$g''(y) = \frac{-\sigma}{(x^2+y^2)^{\frac{\sigma}{2} + 1}} + \frac{\sigma \left(\frac{\sigma}{2}+1\right) \cdot 2 y^2}{(x^2+y^2)^{\frac{\sigma}{2} + 2}}$$ and so $|g''(y)|\leq \sigma (\sigma+1)/(x^2+y^2)^{\sigma/2+1}$. Let $I$ be the interval $\lbrack (1-\rho) y_0, (1+\rho) y_0\rbrack$. Then, for $y\in I$, $|g''(y)|\leq \sigma (\sigma+1)/(\rho l_0)^{\sigma+2}$, where $l_0 = \sqrt{x_0^2+y_0^2}$, and so $$g(y) = g(y_0) + g'(y_0) (y- y_0) + O^*\left(c_0 (y-y_0)^2\right),$$ where $c_0 = \sigma (\sigma+1)/2 (\rho l_0)^{\sigma+2}$. Thus, by cancellation, $$\int_I g(y) e^{-(y-y_0)^2/2} dy = \int_I g(y_0) e^{-(y-y_0)^2/2} dy + O^*\left(\int_I c_0 (y-y_0)^2 e^{-(y-y_0)^2/2} dy\right).$$ Since $g'(y)<0$ for $y\geq 0$, we also know that $g(y)<g(y_0) + c_0 (y-y_0)^2$ for $y>(1+\rho) y_0$. We conclude that $$\begin{aligned}\int_{(1-\rho) y_0}^{\infty} g(y) e^{-(y-y_0)^2/2} dy &\leq g(y_0) \int_{-\infty}^\infty e^{-y^2/2} dy + c_0 \int_{-\infty}^\infty y^2 e^{-y^2/2} dy\\ &= g(y_0) \sqrt{2\pi} + c_0 \sqrt{2\pi} = \left(1 + \frac{\sigma (\sigma+1)}{2 \rho^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} .\end{aligned}$$ It remains to consider $y\leq (1-\rho) y_0$. Since $g(y)\leq g(0) = 1/x_0^\sigma$, $$\begin{aligned}\int_{-\infty}^{(1-\rho) y_0} g(y) e^{-(y-y_0)^2/2} dy &= \frac{1}{x_0^\sigma} \int_{-\infty}^{-\rho y_0} e^{-y^2/2} dy \\ &\leq \frac{1}{x_0^\sigma (\rho y_0)} \int_{-\infty}^{-\rho y_0} y e^{-y^2/2} dy = \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}. \end{aligned}$$

Thus we obtain $$\int_{-\infty}^\infty \frac{e^{-(y-y_0)^2/2}}{(x_0^2+y^2)^{\sigma/2}} dy \leq \left(1 + \frac{\sigma (\sigma+1)}{2 \rho^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} + \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}$$ for any $0<\rho<1$.

Hardly very powerful or elegant, but I wonder: (a) is the above qualitatively optimal? That is, are the lesser-order terms of the right order? (b) can one give an even quicker proof of the same or a stronger bound?

Mainly for purposes of comparison, let me flesh out what I called "a cheap version of Laplace".

Choose $\rho\in (0,1)$. Let $g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then $$g''(y) = \frac{-\sigma}{(x^2+y^2)^{\frac{\sigma}{2} + 1}} + \frac{\sigma \left(\frac{\sigma}{2}+1\right) \cdot 2 y^2}{(x^2+y^2)^{\frac{\sigma}{2} + 2}}$$ and so $|g''(y)|\leq \sigma (\sigma+1)/(x^2+y^2)^{\sigma/2+1}$. Let $I$ be the interval $\lbrack (1-\rho) y_0, (1+\rho) y_0\rbrack$. Then, for $y\in I$, $|g''(y)|\leq \sigma (\sigma+1)/((1-\rho) l_0)^{\sigma+2}$, where $l_0 = \sqrt{x_0^2+y_0^2}$, and so $$g(y) = g(y_0) + g'(y_0) (y- y_0) + O^*\left(c_0 (y-y_0)^2\right),$$ where $c_0 = \sigma (\sigma+1)/2 ((1-\rho) l_0)^{\sigma+2}$. Thus, by cancellation, $$\int_I g(y) e^{-(y-y_0)^2/2} dy = \int_I g(y_0) e^{-(y-y_0)^2/2} dy + O^*\left(\int_I c_0 (y-y_0)^2 e^{-(y-y_0)^2/2} dy\right).$$ Since $g'(y)<0$ for $y\geq 0$, we also know that $g(y)<g(y_0) + c_0 (y-y_0)^2$ for $y>(1+\rho) y_0$. We conclude that $$\begin{aligned}\int_{(1-\rho) y_0}^{\infty} g(y) e^{-(y-y_0)^2/2} dy &\leq g(y_0) \int_{-\infty}^\infty e^{-y^2/2} dy + c_0 \int_{-\infty}^\infty y^2 e^{-y^2/2} dy\\ &= g(y_0) \sqrt{2\pi} + c_0 \sqrt{2\pi} = \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} .\end{aligned}$$

It remains to consider $y\leq (1-\rho) y_0$. Since $g(y)\leq g(0) = 1/x_0^\sigma$, $$\begin{aligned}\int_{-\infty}^{(1-\rho) y_0} g(y) e^{-(y-y_0)^2/2} dy &= \frac{1}{x_0^\sigma} \int_{-\infty}^{-\rho y_0} e^{-y^2/2} dy \\ &\leq \frac{1}{x_0^\sigma (\rho y_0)} \int_{-\infty}^{-\rho y_0} y e^{-y^2/2} dy = \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}. \end{aligned}$$

Thus we obtain $$\int_{-\infty}^\infty \frac{e^{-(y-y_0)^2/2}}{(x_0^2+y^2)^{\sigma/2}} dy \leq \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} + \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}$$ for any $0<\rho<1$.

Hardly very powerful or elegant, but I wonder: (a) is the above qualitatively optimal? That is, are the lesser-order terms of the right order? (b) can one give an even quicker proof of the same or a stronger bound?