Solving Laplace's Equation in the left plane For the Laplace problem, 
$$
{\partial^{2}\,\mathrm{u}\left(x,y\right) \over \partial x^{2}} + {\partial^{2}\,\mathrm{u}\left(x,y\right) \over \partial y^{2}} = 0\,;\qquad
x > 0\,,\quad y \in \mathbb{R}\,;\qquad
\left\{\begin{array}{lcl}
\displaystyle{\mathrm{u}\left(0,y\right)} &\displaystyle{=} & \displaystyle{0} 
\\[3mm]
\displaystyle{\left.\partial\mathrm{u}\left(x,y\right) \over \partial x\right\vert_{\ x\ =\ 0}} &\displaystyle{=} & \displaystyle{\sin\left(ny\right) \over n} 
\end{array}\right.
$$

I am trying to solve the above problem using the method of separation of variables, but after the usual procedure, obtaining two second order ODE's in say $\,\mathrm{X}\left(x\right)$ and $\,\mathrm{Y}\left(y\right)$, I am stuck. 
That is, I am unsure on how to proceed with these boundary conditions in a way that will simplify the problem. 
Any help would be much appreciated. 
PS: I have the solution at hand ( involving a product of $\sinh$ and $\sin$ ) but I am stuck on how to obtain it. 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The solution which satisfies the boundary condition
$\ds{\,\mrm{u}\pars{0,y} = 0}$ is given by
\begin{align}
\mrm{u}\pars{x,y} & = \int_{-\infty}^{\infty}\mrm{A}\pars{k,y}\sinh\pars{kx}
\,\dd k\label{1}\tag{1}
\end{align}
where $\ds{\,\mrm{A}\pars{k,y}}$ satisfies
$\ds{k^{2}\,\mrm{A}\pars{k,y} + \partiald[2]{\mrm{A}\pars{k,y}}{y} = 0}$. Expression \eqref{1} is reduced to
\begin{align}
\mrm{u}\pars{x,y} & = \int_{-\infty}^{\infty}
\bracks{\mrm{a}\pars{k}\sin\pars{ky} + \mrm{b}\pars{k}\cos\pars{ky}}\sinh\pars{kx}
\,\dd k\label{2}\tag{2}
\\[5mm]
\mbox{and}\quad \left.\partiald{\mrm{u}\pars{x,y}}{x}\right\vert_{\ x\ = 0} & = {\sin\pars{ny} \over n} = \int_{-\infty}^{\infty}
\bracks{k\,\mrm{a}\pars{k}\sin\pars{ky} + k\,\mrm{b}\pars{k}\cos\pars{ky}}
\,\dd k\label{3}\tag{3}
\end{align}
\eqref{3} is satisfied with
$\ds{\,\mrm{a}\pars{k} = {\delta\pars{k - n} \over kn}}$ and $\ds{\,\mrm{b}\pars{k} = 0}$:
$$\bbox[#ffe,10px,border:2px dotted navy]{\ds{%
\mrm{u}\pars{x,y} =
{\sin\pars{ny}\sinh\pars{nx} \over n^{2}}}}
$$
