While solving a PDE problem involving the Laplace equation in 3D, I arrive at the following summation relation when i substitute the only non-homogeneous boundary condition available
$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\sinh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) = p_h\bigg(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{m\pi y}{l})\cos(\frac{n\pi x}{L})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) - \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{A_{n,m}b_h}{b_h^2 + n^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg)\bigg[b_h\cos(\frac{n\pi x}{L}) + n\pi\sin(\frac{n\pi x}{L}) \bigg]\bigg) $$
I need to find the Fourier coefficients here, denoted by $A_{n,m}$.
Known
As a starting point i know that if $\cos(\frac{m\pi x}{L})$ and $\cos(\frac{n\pi y}{l})$ is multiplied on the RHS and LHS of the whole equation i could use the known orthogonality of
$\int_{0}^{L} \cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{L})$ and $\int_{0}^{L} \cos(\frac{n\pi x}{l})\cos(\frac{m\pi y}{l})$
But the problem is, I will get a term like $\int_{0}^{L} \cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})$.
I do not know what to do with this term. Additionally, I have another doubt as to how the $A_{n,m}$ occurring on both sides of the equation are to be handled. Also, I have done 2D problems but how to handle this 3D case ? Is a double integration required ?
Note
The question asked has its origin as follows:-
From a solution of the form
$$ T(x,y,z)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\cosh\left(\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}z\right). $$
with bc(s) as:
$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 $
$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0$
$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) $$
$ \frac{\partial T(x,y,0)}{\partial z} = 0. $
Extra information From the physical problem available at hand the following is too known:-
$$\frac{\partial T(0,y,-w)}{\partial z}=p_h\bigg( T(0,y,-w) - T_i\bigg) $$
Attempt
Multiplying both sides by $\cos(\frac{k\pi x}{L})$ and $\cos(\frac{j\pi y}{l})$ assuming that the series converges, converting the summation to an integration, first for $x$ and then for $y$, followed by using orthogonality.
$$ \sum\sum A_{n,m}\sqrt{()}\sinh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg) = p_h\bigg[\sum\sum A_{n,m}\cosh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg)- \sum\sum\frac{A_{n,m}b_h^2}{b_h^2 + n^2 \pi^2}\cosh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg)\bigg] $$
leads to
$$ A_{k,j}\sqrt{()}\sinh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2} \bigg) = p_h\bigg[A_{k,j}\cosh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2}\bigg)-\frac{A_{k,j}b_h^2}{b_h^2 + n^2 \pi^2}\cosh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2}\bigg)\bigg] $$
Hence, $A_{k,j}$ needs to be evaluated. But still I have the problem of $A_{k,j}$ cancelling out from the resulting equation.
I admit that I had failed to check this at the start: you have $A_{m,n}$ on both sides. One solution to your original equation is just $A_{m,n} = 0$ for all $m,n$. And in fact, that is the only solution.
To see this, multiply by $\sin\left(\frac{k\pi x}L\right)$ and take the $\int_0^{2L}\ \ dx$ integral of both sides (note the $2L$ - you have to integrate over a full period, and that is $2L$, not $L$). Every term with $\cos\left(\frac{n\pi x}L\right)$ in it becomes $0$. And the only $\sin\left(\frac{n\pi x}L\right)$ term that survives is $n = k$. Now, $\int_0^{2L}\sin^2\left(\frac{k\pi x}L\right)\,dx = L$, so the result is
$$0 = -p_h\left[\sum_{m=1}^{\infty}\frac{A_{k,m}b_h}{b_h^2 + k^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\left(w\sqrt{\frac{k^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\right)k\pi L \right]$$
Then we multiply through by $\cos\left(\frac{j\pi y}l\right)$ and take the $\int_0^{2l}\ \ dy$ integral, which similarly leaves us with
$$0 =-p_h\left[\frac{A_{k,j}b_h}{b_h^2 + k^2 \pi^2 }l\cosh\left(w\sqrt{\frac{k^2\pi^2}{L^2}+\frac{j^2\pi^2}{l^2}}\right)k\pi L \right]$$ $\cosh$ is never $0$. If we assume that $p_h, b_h, L$, and $l$ are all non-zero, this solves to $A_{k,j} = 0$.
Either you have an error in deriving the equation posted, or else your solution is $0$.