Can Laplace's equation be solved using Fourier transform instead of Fourier series? Sorry for the long text, but I am unable to make my question more compact.
Any periodic function can be Fourier expanded. Usually, they say in mathematical physics books, if the function is not periodic we use Fourier transform which is more general than Fourier series expansion. 
If Fourier transform is more general, cannot we use it to expand a periodic functions as well? Why periodic functions in textbooks are only Fourier expanded but not Fourier transformed?
More specifically, the boundary value problems that we solve in electromagnetism (like in chapter 3 of Griffiths) in which for example some potential is specified on the boundary of some region and we want to find the potential inside that region, this problem is usually solved by separation of variable then eventually applying Fourier series expansion to fit the boundary conditions. Those problems are never solved using Fourier transform, why is that? is it because that in Fourier series expansion one has control on truncating the series to whatever accuracy one wants whereas for Fourier transform one cannot do that? or is it an issue of convergence? 
If both are viable there must be some criteria on using one over the other!
If one can point out a reference in which Laplace's equation is solved once with Fourier series and once with Fourier transform that will be greatly appreciated. 
 A: The Fourier transform of a periodic function is a delta function at every integer position with coefficient equal to the corresponding Fourier series value. You can show this by multiplying the function by a very wide Gaussian and taking the limit. The mathematical theory is made rigorous in the subject of tempered distributions.
A: For the Fourier transform of a function to exist, its absolute value must be integrable, $\int\limits_{-\infty}^\infty |f(x)|\mathrm{d}x<\infty$. The absolute value of a periodic function is not integrable on an infinite domain, so no Fourier transform. [To enjoy the full power of Fourier analysis, the function should be square integrable, $\int\limits_{-\infty}^\infty |f(x)|^2\mathrm{d}x<\infty$.]
For the Fourier expansion of a periodic function, the function has to be integrable on the finite domain of one period of the function, $\int\limits_0^L |f(x)|\mathrm{d}x<\infty$, instead of on the whole real line, which many or most of the periodic functions one meets in electromagnetism problems will be.
So, the difference between a Fourier transform and a Fourier expansion of a periodic function is that the integration is on an infinite domain, respectively a finite domain.
Fourier transforms/expansions are well suited to rectilinear coordinate systems, but they are generally less well-suited to problems in which the boundary conditions pick out curved coordinate systems. Fourier analysis is nonetheless often usable as a first approximation, as when electromagnetic fields are directed along a curved wave-guide.
A: You probably can answer this question yourself. You know that any periodic function can be expanded in a Fourier series. If you Fourier transform said series, what do you get?
Hint: 
\begin{equation}
\int e^{inx}e^{-ipx}dx = 2\pi\delta(p-n)
\end{equation}
