Can a non-periodic function have a Fourier series? Consider two periodic functions. Assume their sum is not periodic.
The periodic functions can be represented by a Fourier series. If you add up the Fourier series, you get a series that represents their sum. But their sum is not periodic, yet you have described it using a Fourier series.
I thought that non-periodic functions can't be represented by a Fourier series. Why isn't this a contradiction?

In order to find its Fourier series, a periodic function with period $R$ should be thought of as a function defined on a circle of circumference $R$, call it $S^1_R$. The Fourier series of the function is then its representation in the basis of $L^2(S^1_R)$ given by orthonormal eigenfunctions of the Laplace operator. 
If two functions have incommensurate periods, then their sum is nonperiodic, does not descend to a circle of any circumference, and therefore does not have a Fourier series.
As functions on $\mathbb{R}$, if they are sufficiently nice, the Fourier transform gives an analogous decomposition, but because there are so many more eigenfunctions of the Laplace operator on $\mathbb{R}$, the sum is an integral. Compare:
Let $e_\omega(t) = e^{2\pi i\omega t}$ for $\omega$ real.
$$ \mbox{periodic ($\omega$ is an integral multiple of a base frequency): }\\ f(t) = \sum_\omega\langle f,e_\omega(t)\rangle e_\omega$$
$$ \mbox{nonperiodic ($\omega$ ranges over $\mathbb{R}$): }\\ f(t) = \int\langle f,e_\omega(t)\rangle e_\omega\ d\omega$$
where
$$\langle f,e_\omega\rangle = \int f(x)\overline{e_\omega(x)}\ dx$$
You will recognize that if we integrate over the circle, $\langle f,e_\omega\rangle$ gives the series of Fourier coefficients and if we integrate over $\mathbb{R}$, it is the Fourier transform.
