TITLE: Integral kernel of the resolvent operator
QUESTION [6 upvotes]: Suppose we have an explicit formula for the integral kernel $k(x,y)$ of an operator $D$ acting on smooth $\mathbb{C}^n$-valued functions defined on an interval $[0,\beta]$, that is
$$
Df(x) = \int_0^\beta k(x,y)f(y)\,dy \,,\quad f \in C^\infty([0,\beta],\mathbb{C}^n)\,. 
$$
I need to use the resolvent $(D - \lambda)^{-1}$, is there a way to immediately deduce a formula for its kernel, given that I have complete knowledge about $k(x,y)$? 
In case the answer would be too extensive I would also be grateful for a reference. Thank you!

REPLY [2 votes]: Let me suggest the following:
$$
(D-\lambda)^{-1}=\sum_{n=0}^\infty (-1)^{n+1}\lambda^{n+1}D^n,
$$
and $(D^nf)(x)=\int_0^\beta k_n(x,y)f(y)\,dy$, and $k_n$ can be obtained recursively as
$$
(D^2f)(x)=\int_0^\beta k(x,y)\left(\int_0^\beta k(y,z)f(z)\,dz\right)dy=\int_0^\beta\left(\int_0^\beta k(x,y)k(y,z)\,dy\right)f(z)\,dz,
$$
which implies that
$$
k_2(x,y)=\int_0^\beta k(x,z)\,k(z,y)\,dz.
$$
In general
$$
k_n(x,y)=\int_0^\beta\cdots\int_0^\beta f(x,z_1)\,f(z_1,z_2)\cdots f(z_{n_1},y)\,dz_1\cdots dz_{n-1}.
$$