Difficulties in obtaining the Fourier series for an integral operator I'm currently studying a paper that derives a numerical scheme to solve via Fourier Transform a certain type of PDE's, Neural-Field Equations.
The equation in study has the following integral operator:
\begin{equation}
A(\mathbf{x},t)=\int_{\Omega} d^2\mathbf{y}\int_0^{\tau_{max}} L(\mathbf{x-y},\tau)S[V(\mathbf{y},t-\tau)] d\tau,
\end{equation}
with $\mathbf{x},\mathbf{y}\in\Omega=[-\frac{l}{2},\frac{l}{2}]\times[-\frac{l}{2},\frac{l}{2}]$
Note: we have two convolutions here, one in space (2-dim) and other in time.
The authors derive the Fourier series of $A$ by stating:
\begin{equation}
A(x,y,t) = \sum_{m,n=-\infty}^{\infty}\left(e^{i(k_mx+k_ny)}\int_0^{\tau_{max}}\tilde{L}_{mn}(T)\widetilde{SV}_{mn}(t-T)\,\,dT\right),
\end{equation}
with $\tilde{L}_{mn}(T)$ and $\widetilde{SV}_{mn}(t-T)$ being the spatial Fourier transform of $L(\mathbf{x},t)$ and the image of the nonlinear functional $S[V(\mathbf{x},t)]$, respectively.
They do not do this demonstration and I really need to demonstrate this passage.
I know the spatial Fourier series of $A$ can be written as:
\begin{equation}
A(x,y,t) = \frac{1}{l^2} \sum_{m,n=-\infty}^{\infty} A_{mn}(t)e^{i(k_mx+k_ny)},
\end{equation}
with $A_{mn}$ being:
\begin{equation}
A_{mn} = \frac{1}{l^2} \int_{\Omega} A(x,y,t)e^{-i(k_mx+k_ny)}.
\end{equation}
But, if I just play around, naively, with these expressions I can't get what the authors state.
Any help?
Best regards,
Tiago
 A: 
Let's assign some new functions so that we can have a better view of our integral.
Let :

*

*$L^{t}( \mathbf{x}):=L( \mathbf{x},t)$

*$H^t( \mathbf{x}) := S[V(\mathbf{x},t)]$

*$A^t(\mathbf{x}):= A( \mathbf{x},t)$
So the expression of $A$ is now changed to (you made a small typo in the integral domain of $d^y$):
$$A^t(\mathbf{x})=\int_{0}^{\tau_{max}} \int_{\Omega} L^{\tau}( \mathbf{x-y})H^{t-\tau}(\mathbf{y})d^2\mathbf{y}d\tau$$
From which you can realize, as you know, immediately the emergence of convolutives terms, which help us even reduce even further the formula of $A$, that is:
$$A^t(\mathbf{x}) = \int_{0}^{\tau_{max}}  \left( L^{\tau}*H^{t-\tau}\right)(\mathbf{x})d\tau$$
From this point, I guess you are clear of what to do next, that is to use a common identity we have in Fourier analysis : $ \widehat{ g*h}= \hat{g}\hat{h}$ (up to a multiplicative constant, depending on the definition of convolution).
$$ \tilde{ A^t}(\mathbf{y})\underbrace{=}_{\text{def.}} \int_{0}^{\tau_{max}} \widetilde{ \left( L^{\tau}*H^{t-\tau}\right)} (\mathbf{y}) d\tau =l\int_{0}^{\tau_{max}} \widetilde{  L^{\tau}}(\mathbf{y}) \widetilde{H^{t-\tau}} (\mathbf{y}) d\tau$$
The rest is yours.
Remark Upon finishing writing this reply, I realize your Fourier formula is
a bit different than mine ( in particular at the multiplicative constant), so while the perspective stays the same, I think you should adapt my reply to your demonstration with suitable adjustment in the multiplicative constants.
