Fourier's Trick

$$ \psi(x,0) = \sum_{n=1}^{\infty} c_n \psi_n(x) $$ $$ \int_{-\infty}^{\infty} dx \, \psi_m^*(x)\psi(x,0) = \int_{-\infty}^{\infty} dx \, \psi_m^*(x) \left( \sum_{n=1}^{\infty} c_n \psi_n(x) \right) $$ $$ \int_{-\infty}^{\infty} dx \, \psi_m^*(x)\psi(x,0) = \sum_{n=1}^{\infty} c_n \int_{-\infty}^{\infty} dx \, \psi_m^*(x)\psi_n(x) $$

Using orthonormality,

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

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 $ \psi(x,0) $ is

$$ c_n = \int_{-\infty}^{\infty} dx \, \psi_n^*(x)\psi(x,0) $$