I have obtained (formally) a perturbative solution $$ H(y) = \sum_{n=0}^\infty \delta^n H_n(y) $$ to the following integrodifferential equation ($\delta$ is a small constant, $\nu$ is a L\'evy measure and $F$ is a function of $y$) $$ 0 = \delta F ( d_y H  H + 2 ) + \int \nu(dz) \left( \frac{e^{z d_y }  1}{d_y}  ( e^z  1 ) \right) H + \int \nu(dz) z^2 $$ Namely $$ H_1 = A_0^{1} F , \qquad H_n = A_0^{1} F A_1 H_{n1} = (  A_0^{1} F A_1 )^{n1} H_1 . $$ where the operators $A_0$ and $A_1$ are given by $$ A_0 = \int \nu(dz) \left( \frac{e^{z d_y }  1}{d_y}  ( e^z  1 ) \right), \qquad A_1 = d_y1 . $$ The inverse operator (resolvent) $A_0^{1}$ can be defined through $$ A_0^{1} F = \int d \lambda \frac{1}{\phi_\lambda} (\psi_\lambda,F ) \psi_\lambda \qquad (F,G) = \int \overline{F}(y) G(y) dy $$ where $$ A_0 \psi_\lambda = \phi_\lambda \psi_\lambda, \qquad \phi_\lambda =\int \nu(dz) \left( \frac{e^{ i \lambda z}  1}{i\lambda}  ( e^z  1 ) \right) , \qquad \psi_\lambda =\frac{1}{\sqrt{2\pi}}e^{i\lambda y} $$ Explicitly $$ H_n = (1)^{n1} \underbrace{ \int \cdots \int }_{n} d \lambda_n \frac{\psi_{\lambda_n}}{\phi_{\lambda_n}} (\psi_{\lambda_1},F) \prod_{k=1}^{n1} d \lambda_k \frac{\chi_{\lambda_k}}{\phi_{\lambda_k}} ( \psi_{\lambda_{k+1}},F \psi_{\lambda_k} ) \qquad $$ where $$ A_1 \psi_\lambda = \chi_\lambda \psi_\lambda, \qquad \chi_\lambda = i \lambda  1 $$ As stated above, my solution is purely formal. In order to show convergence of the infinite sum I need to show that either (i) $$ \lim_{n \to \infty} \frac{H_{n+1}}{H_n} = 0 $$ or (ii) that the $H_n(y)$ has alternating signs, $\delta^n H_n$ is strictly decreasing, and $\lim_{n\to\infty}H_n=0$. Obviously, showing either of these will depend on $F$ and possibly on $\nu$. But at the moment, I really do not even know how to begin showing either thing. Any advice or references would be appreciated.
