Suppose I have a convolution integral $$\tilde{F}(x; g)=\int dy F(x-y) K(y;g),$$ with a kernel $$K(x;g)\equiv\frac{1}{\pi}\frac{g}{x^2+g^2}.$$ The result of the integral depends on the parameter $g$ in the kernel, and so we should be able to do an asymptotic expansion in powers of $g$ about $g=0$. Since in the limit $g\rightarrow 0$ the kernel $K$ becomes a Dirac delta function, the lowest order term should just be $F$ itself, $$\tilde{F}(x; g)=F(x)+\mathcal{O}(g),$$ but how do I get the rest of the asymptotic expansion?
If I naively expand the kernel inside the integral $$K(x; g) = \frac{g}{\pi x^2}\sum_{k=0}^\infty \left(-\frac{g^2}{x^2}\right)^k$$ then for one thing I miss the lowest order term, and also I run into convergence problems due to the presence of powers of $x^2$ in the denominator. I'm wondering if there is a useful way to do this asymptotic expansion without specifying the form of the function $F$.
The expansion in $g$ is more well-behaved after a Fourier transform. $$f(u)\equiv \int dx F(x)e^{iux}$$ $$k(u;g) \equiv \int dx K(x;g)e^{iux}=e^{-g|u|}$$ $$\tilde{F}(x;g)=\int \frac{du}{2\pi}e^{-iux}f(u)e^{-g|x|}=\sum_{k=0}^\infty\frac{du}{2\pi}e^{-iux}f(u)\frac{(-g|u|)^k}{k!}$$ The terms involving even powers of $g^k$ are derivatives of delta functions that are missing in the naive expansion in my question, and each term of the expansion converges as long as the Fourier transform $f(u)$ decays sufficiently fast.