Question:
I have integrals like
$$ I(x) = \int_0^\infty K(y) F(x,y) dy $$
where $K(y)$ is some kernel function that is sharply peaked at some special value $y = y_0$ and otherwise trends toward zero away from $y_0$.
Considering $K(y) \approx \delta(y-y_0)$ it's easy to derive
$$I(x) = F(x,y_0),$$
but I would like to derive a correction term to this equation since $K(y)$ is not exactly a delta function. This would probably involve $K''(y_0).$ I am sure this is similar to a Laplace expansion but I am unclear exactly how. Any guidance for an expansion of this integral around the peak of $K(y)$ would be greatly appreciated!
Attempt:
writing
$$I(x) = \int_0^\infty e^{ \ln K(y) } F(x,y) dy $$ and expanding $K$ around $y_0$ and taking the lower limit $ y \rightarrow -\infty$ without introducing much error provides
$$I(x) \approx K(y_0) \int_{-\infty}^\infty \exp\Big[\frac{1}{2}\frac{K''(y_0)}{K(y_0)}(y-y_0)^2\Big] F(x,y) dy, $$
which is now at least a convolution integral with a sharply-peaked Gaussian kernel…
Is there a further approximation I can make to evaluate this integral as a series expansion?
Best Answer
A method to deal with highly peaked kernels is described in this article. Only the domain near $y=y_0$ contributes significantly to the integral. Then, one may use a Talor expansion to represent $F(x,y)$ in this region: \begin{equation} F(x,y)=F(x,y_0)+(y-y_0)\frac{\partial F}{\partial y}(x,y_0)+\frac{(y-y_0)^2}{2!}\frac{\partial^2 F}{\partial y^2}(x,y_0)+\cdots \end{equation} The integral can be written as \begin{align} I(x) &= \int_0^\infty K(y) F(x,y) \,dy\\ &=k_0F(x,y_0)+k_1\frac{\partial F}{\partial y}(x,y_0) +k_2\frac{\partial^2 F}{\partial y^2}(x,y_0)+\cdots \end{align} where the coefficients are related to the moments of the kernel: \begin{equation} k_n=\frac{1}{n!}\int_0^\infty (y-y_0)^n K(y)\,dy \end{equation} for $n=0,1,2\cdots$. These coefficients quickly vanish if the function is highly peaked. The first term is equivalent to the delta contribution. Moreover, for a symmetric peak, the odd indices vanish.