# Inverse Discrete Time Fourier Transform of $1$

$$\textrm{DTFT}(\delta[n]) =1$$, but $$\textrm{IDTFT(1)} = \frac{\sin(\pi n)}{\pi n}$$. Why it is not equal to the unit impulse $$\delta[n]$$?

The IDTFT of $$X(e^{j\omega})=1$$ is indeed
$$x[n]=\frac{\sin(n\pi)}{n\pi}\tag{1}$$
Now, what happens for indices $$n\neq 0$$? As it turns out, you can safely rewrite $$(1)$$ as
$$x[n]=\delta[n]\tag{2}$$
where $$\delta[n]$$ is the discrete-time unit impulse. (HINT: think about where the zeros of $$\sin(x)$$ are).