Given a periodic signal $\tilde{f}(t)$ of period $T_0$, a Fourier coefficient $F_k$ is the projection $\tilde{f}(t)$ along the basis function $e^{jn\omega_0 t}$. The reason why the Fourier series coefficients completely characterizes the original signal $\tilde{f}(t)$ is because the set of basis functions $\{e^{jn\omega_0 t}: n \in \mathbb{Z}\}$ is orthogonal, i.e. $$ \int_{T_0} e^{jn\omega_0 t} e^{-jm\omega_0 t} dt = \begin{cases}T_0, &n=m \\ 0, & n\neq m\end{cases} $$
Assuming that the above is correct, I am trying to understand why the Fourier transform also completely characterizes an aperiodic function $f(t)$, where the basis is now $\{e^{j\omega t}: \omega \in \mathbb{R}\}$. Now, the orthogonality principle requires solving the integral $$ \int_\mathbb{R} e^{j\omega_1 t} e^{-j\omega_2 t} dt. $$ However, this integral does not solve to the orthogonality results that I would like. Is my approach valid, or is there a flaw in my reasoning somewhere?
