I know that $\{e^{inx}\}_{n\in\mathbb N}$ is a basis of $L^2(\mathbb S^1)$ where $\mathbb S^1=\mathbb R/\mathbb Z$. Using this result, and the fact that $$\left<f ,g \right>=\int_0^1 f(x)e^{-inx}\mathrm d x,$$
is a scalar product over $L^2(\mathbb S^1)$, we can wrrite any function $$f:\mathbb S^1\longrightarrow \mathbb R$$ as
$$f(x)=\sum_{n\in\mathbb N}\int_0^1f(x)e^{-iny}\mathrm d ye^{inx},$$
or, as usually denoted, by setting $$c_n=\int_0^1f(x)e^{-inx}\mathrm d x,$$
we write $$f(x)=\sum_{n\in\mathbb Z}c_ne^{inx}=:Sf(x),$$
which is called it's Fourier Series.
Question : Now, does $\{e^{i\alpha x}\}_{\alpha \in \mathbb R}$ a basis of $L^2(\mathbb R)$ ?
If yes, then (without rigor), for $f:\mathbb R\longrightarrow \mathbb R$, we could write
$$f(x)=\int_{-\infty }^\infty \int_{-\infty }^\infty f(y)e^{-i \alpha y}\mathrm d ye^{i\alpha x}\mathrm d \alpha ,$$
what is in fact exactly the inversion of Fourier transform, i.e. $$f(x)=\int_{-\infty }^\infty \hat f(\alpha )e^{i\alpha x}\mathrm d \alpha .$$
To me, if my conjecture that $L^2(\mathbb R)=span\{e^{i\alpha x}\}_{\alpha \in\mathbb R}$, then this formula would make totally sense (as replacing $$\sum_{\alpha \in\mathbb R}\hat f(\alpha )e^{i\alpha x}\quad \text{by}\quad \int_{\mathbb R}\hat f(\alpha )e^{i\alpha x}\mathrm d \alpha,$$
since an integral "can be seen" as a continuous sum.) Of coure that to have the existence of the Fourier inverse, we need that $f$ is Schwartz, but as I said, I ask the question without rigor; in other words, we suppose that we have all good conditions for that things exist.
Best Answer
The functions $e^{i\alpha x}$ are not in $L^2$, which means they cannot be a basis. However, there is a general principle associated with Sturm-Liouville theory that can help give you an approximation to what you want. Integrals over any small interval of the parameter $\alpha$ are in the space, and, for such intervals $I$, $J$, you have orthogonality: $$ \left\langle \int_{I} e^{i\alpha x}d\alpha,\int_{J} e^{i\alpha x}d\alpha\right\rangle =0, \;\; \mbox{ if $I\cap J$ is of measure $0$ }. $$ Furthermore, if $|I|$ is the length of the interval $I$, then $$ \left\|\frac{1}{\sqrt{2\pi|I|}}\int_{I}e^{i\alpha x}d\alpha\right\|^2=1. $$ So you can divide up the real axis into very small disjoint intervals $I_n$, and you end up with an orthonormal set $$ \left\{\frac{1}{\sqrt{2\pi|I_n|}}\int_{I_n}e^{i\alpha x}d\alpha\right\}_{n=-\infty}^{\infty} $$ If you let $e_n$ denote the integral over $I_n$, and you attempt to expand in this basis $\{ e_n \}$, you end up with what looks like a type of Riemann integral approximation for the Fourier integral inversion problem. $$ \sum_{n=-\infty}^{\infty}\langle f,e_n\rangle e_n = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\left(\frac{1}{|I_n|}\int_{I_n}\hat{f}(\alpha)d\alpha\right)\int_{I_n}e^{i\alpha t}dt $$ The term in parentheses is the integral average of the Fourier transform over $I_n$. So this is very much like a Riemann approximation of the inversion integral applied to the Fourier transform. And it makes Mathematical sense. If the function $f$ is constant on each of intervals $I_n$, then (ignoring values at endpoints) the above is a correct inversion integral to give you back $f$.