Generalized solution of Laplace equation with unknown boundary conditions Background: For example, the 2D Laplace's equation:
$$
\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} =0
$$
, when solved by separation of variable $V=X(x)Y(y)$, has the solution of the form:
$$
V_k(x,y)=(Ae^{kx}+Be^{-kx}) + (C\sin ky+D\cos ky) \tag{1}
$$
Usually, a set of (periodic) boundary conditions are given, which might eliminate some terms in $V_k$, and the general solution is the discrete sum of all the eigenvectors:
$$
V(x,y)=\sum_{k=-\infty}^{\infty}V_k(x,y) \tag{2}
$$
, where $k$'s have a specific form depending on the boudary conditions given.
Situation: When the boundary conditions are not given, $V_k$ in (1) is of the most general form and $k$ can take any possible values in $(-\infty, \infty)$, which means the sum in (1) must be changed into an integral (correct me if I am wrong!).
Question: When the boundary conditions are not yet given, $k$ can be continuous. Thus, what is the generalized version of the sum in (2)?
Attempt: All I have tried so far is coming up with the (obvious?) following integral:
$$
V(x,y)=\int_{-\infty}^{\infty}[(A(k)e^{kx}+B(k)e^{-kx}) + (C(k)\sin ky+D(k)\cos ky)]dk
$$
However, I have a feeling that this integral is not the correct answer.
 A: $$
\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} =0 \tag 1
$$
You say that you solved by separation of variable $V=X(x)Y(y)$ and found the solution on the form:
$$V_k(x,y)=(Ae^{kx}+Be^{-kx}) + (C\sin ky+D\cos ky)$$
This is not correct. Put it into Eq.$(1)$ and observe that the equality is not satisfied.
Unfortunately the detail of the calculus is not shown. So one cannot check it and see where is the mistake or deficiency.
For particular solutions on the form $\quad V_k=X(x)Y(y)\quad$ we have $\quad X''Y+Y''X=0\quad$ which implies :
$$\frac{X''}{X}=-\frac{Y''}{Y}=\lambda^2 \qquad \begin{cases}
X=e^{\pm\lambda x}\\
Y=e^{\pm i\lambda y}
\end{cases}$$
where $\lambda$ can be complex. So the functions $X(x)$ and $Y(y)$ can be exponential or sinusoidal or combination of both.  In order to simplify the writing, let the solution be expressed on exponential form without forgetting that the exponential can be replaced by sinusoidal or product of exponential and sinusoidal functions.
$$V_\lambda(x,y)=e^{\lambda(x\pm iy)}$$
Any linear combinations of $V_\lambda(x,y)$ with different $\lambda$ are solutions of Eq.$(1)$. For example :
$$V(x,y)=\sum_\lambda A_\lambda e^{\lambda(x+ iy)}+\sum_\lambda B_\lambda e^{\lambda(x- iy)}$$
where the coefficients $A_\lambda$ and $B_\lambda$ are arbitrary constants.
A more general solution than a discret sum is expressed on integral form :
$$V(x,y)=\int f(\lambda) e^{\lambda(x+ iy)}d\lambda+\int g(\lambda) e^{\lambda(x- iy)}d\lambda$$
$f(\lambda)$ and $g(\lambda)$ are arbitrary function insofar the integral be convergent.
We observe that the first integral is a function of $(x+iy)$ and the second is a function of $(x-iy)$. Thus the above solution can be written on the form :
$$V(x,y)=F(x+iy)+G(x-iy)$$
$F$ and $G$ are arbitrary functions to be determined according to some boundary conditions.
