If you're interested in $L^2$ convergence, then you could use a Fourier Series of exponentials, a Fourier cosine series, a Fourier sin series, or a Fourier transform, which is an integral "sum" of terms. All of these will converge in $L^2$. You can also use various series where the frequencies are not harmonic. This is because $Lf=-\frac{d^2}{dx^2}f$ is self-adjoint on the twice absolutely continuous functions on $[a,b]$ with any of the following endpoint conditions
$$
\cos(\alpha)f(a)+\sin(\alpha)f'(a)=0 \\
\cos(\beta)f(b)+\sin(\beta)f'(b) = 0.
$$
These conditions yield different sets of eigenfunctions for $L$, and each such set is a complete orthogonal set of functions for $L^2[a,b]$.
As a simple example, let $\alpha=0$ $\beta=\pi/2$ so that the conditions above become
$$
f(a)=0,\;\; f'(b)=0.
$$
For normalization, choose $f'(a)=1$. The solutions satisfying the conditions at $a$ have the form
$$
f_{\lambda}(x)= \frac{\sin(\sqrt{\lambda}(x-a))}{\sqrt{\lambda}}
$$
These become eigenfunctions of the problem if $f_{\lambda}'(b)=0$, which gives the eigenvalue equation
$$
\cos(\sqrt{\lambda}(b-a))=0
$$
The eigenvalues $\lambda$ must satisfy
$$
\sqrt{\lambda}(b-a)=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\cdots .
$$
The eigenfunctions are, up to multiplicative constants, the functions $\sin(\lambda (x-a))$ where $\lambda$ values are chosen so that $\cos(\lambda(b-a))=0$. If you choose mixed conditions where the $\alpha,\beta$ are not simple, then you end up with a variety of non-trivial sequences of eigenvalues. The corresponding eigenfunctions form a complete orthogonal basis of $L^2[a,b]$.
