The Fourier transform of an aperiodic discrete signal x[n] is given by:
\[X({\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}\]where $j$ is the imaginary unit, $\omega$ is the angular frequency, and $X(e^{j\omega})$ is the Fourier transform of $x[n]$.
The inverse Fourier transform of $X(e^{j\omega})$ can be used to synthesize the original aperiodic discrete signal $x[n]$:
\[x[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi} X({j\omega}) e^{j\omega n} d\omega\]where $X(e^{j\omega})$ is the Fourier transform of $x[n]$, $j$ is the imaginary unit, $\omega$ is the angular frequency, and $x[n]$ is the original aperiodic discrete signal.
substitute the expression for the Fourier transform of $x[n]$ from the analysis equation into the synthesis equation:
\[\begin{aligned} x[n] &= \frac{1}{2\pi}\int_{-\pi}^{\pi} X(e^{j\omega}) e^{j\omega n} d\omega \ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\sum_{k=-\infty}^{\infty} x[k] e^{-j\omega k}\right) e^{j\omega n} d\omega \ &= \frac{1}{2\pi}\sum_{k=-\infty}^{\infty} x[k] \int_{-\pi}^{\pi} e^{j\omega (n-k)} d\omega \ &= \frac{1}{2\pi}\sum_{k=-\infty}^{\infty} x[k] 2\pi \delta_{nk} \ &= \sum_{k=-\infty}^{\infty} x[k] \delta_{nk} \ &= x[n] \end{aligned}\]
where $\delta_{nk}$ is the Kronecker delta function.
