Fourier's Trick

Let $ \{f_n\} $ be an orthonormal basis, that is

$$ \langle f_n, f_m \rangle = \delta_{nm} = \begin{cases} 1 & n = m \\ 0 & n \neq m \end{cases} $$

Then any function can be written as a linear combination of $ \{f_n\} $

$$ g(x) = \sum_{n=1}^{\infty} c_n f_n(x) $$ $$ \int_{-\infty}^{\infty} dx \, f_m^*(x)g(x) = \int_{-\infty}^{\infty} dx \, f_m^*(x) \left( \sum_{n=1}^{\infty} c_n f_n(x) \right) $$ $$ = \sum_{n=1}^{\infty} c_n \int_{-\infty}^{\infty} dx \, f_m^*(x)f_n(x) $$

Using orthonormality,

$$ \int_{-\infty}^{\infty} dx \, f_m^*(x)f_n(x) = \delta_{mn} \quad \text{where} \quad \delta_{mn} = \begin{cases} 1 & n = m \\ 0 & n \neq m \end{cases} $$

so this becomes

$$ \sum_{n=1}^{\infty} c_n \delta_{mn} = c_m $$

The Kronecker delta $ \delta_{mn} $ kills every term except the one for which $ n = m $. Therefore, the $ n $-th coefficient in the expansion of $ f(x) $ is

$$ c_n = \int_{-\infty}^{\infty} dx \, f_n^*(x)f(x) $$