I'm trying to solve this integral equation $$G(t,t-t')=A(t-t')+\int dt_1 G(t,t-t_1)H(t_1)A(t_1-t')$$ where $A$ and $H$ are knownand $H$ is $\Omega_0$-periodic. So I thought to Fourier transform
$$G(t,t-t_1)= \int \frac{d\omega}{2 \pi} e^{\imath \omega (t-t_1)}G(t,\omega)$$ $$H(t)=\sum_n H_n e^{\imath \Omega_0 t}$$ $$A(t_1-t')=\int \frac{d \tilde{\omega}}{2\pi} e^{\imath \tilde{\omega}(t_1-t')}A(\tilde{\omega})$$
So the first term in the RHS of the equation is easy to transform and the second term becomes $$\int dt_1 \frac{d \omega}{2\pi} \frac{d \tilde{\omega}}{2\pi} \sum_n G(t,\omega)e^{i \omega (t-t_1)}H_ne^{i n \Omega_0 t} A(\tilde{\omega})e^{i \tilde{\omega}(t_1-t')}$$
Integrating over $t_1$ gives $\delta(\tilde{\omega}-\omega+n\Omega_0)$. So now I have an equation for $G(t,\omega)$ that reads
$$G(t,\omega)= A(\omega)+\sum_n G(t,\omega+n\Omega_0)H_nA(\omega)$$
Now this is an algebraic equation. Soppose now that also $G$ is $\Omega_0$-periodic in $t$ so that admits a serie descomposition like $H$, that is
$$G(t,\omega)=\sum_k G_k(\omega)e^{-i k \Omega t }$$
Now instead of solving this infinite coupled set of equations I'd like to have an answer for a given $k$ that is to propose a cut of $k$ for example $k=2$
$$G(t,\omega)\sim \sum_{k=-2}^{2} G_k(\omega)e^{-i k \Omega t }$$
And here is where I'm having troubles because despite of doing this each $G_k$ depends on the others and I can't see how to get $G_k$ as a function only on the lessers $k$
Thanks