While studying pseudo-differential operators of type $\left(\sqrt{\alpha^{2} - \partial_{x}^{2}}-\alpha\right)\operatorname{f}\left(x\right)$, I came across the following integral representation of the operator: $${\rm P.V.}\int_{-\infty}^\infty\mathrm{d}\xi\,\frac{\alpha K_1(\alpha \lvert x-\xi\rvert)}{\left\vert x-\xi\right\vert}\left[f(x)-f(\xi)\right],$$ where $K_1(x)$ is the modified Bessel function of the first kind and $\mathrm{p.v.}$ is the Cauchy principal. While I understand how to derive it, I am having trouble computing the limiting behavior of it. Since $$\Bigl(\sqrt{\alpha^2-\partial_x^2}-\alpha\Bigr)f(x)\xrightarrow{\alpha \to\infty}-\frac{1}{2\alpha}\frac{\partial^2f}{\partial x^2}$$ is proportional to the second derivative, it should be possible to derive $$\mathrm{p.v.}\int_{-\infty}^\infty\mathrm{d}\xi\,\frac{\alpha K_1(\alpha \lvert x-\xi\rvert)}{\lvert x-\xi\rvert}(f(x)-f(\xi))\xrightarrow{\alpha \to\infty}-\frac{1}{2\alpha}\frac{\partial^2f}{\partial x^2},$$ but I don't see how.
I tried to look at the asymptotic expansion of the integral kernel, i.e. $$\sqrt{\frac{\pi\alpha}{2}}\mathrm{p.v.}\int_{-\infty}^\infty\mathrm{d}\xi\,\frac{\mathrm{e}^{-\alpha\lvert x-\xi\rvert}}{\lvert x-\xi\rvert^{\frac{3}{2}}}(f(x)-f(\xi)),$$ and work from there. But I don't know how to continue.
