# Fourier transform accumulation property

It is known that Fourier Transform has an "accumulation property". If we define:

$$X(\omega) = F[x(n)]$$ and $$y(k) = \sum\limits_{n=-\infty}^k x(n)$$ i can write: $$y(k)-y(k-1)=\sum\limits_{n=-\infty}^k x(n)-\sum\limits_{n=-\infty}^{k-1} x(n)=x(k)$$ Where both $x(n)$ and $y(n)$ are discrete functions. Now, if i'm not mistaken, the accumulation property states that the Fourier Transform of $y(n)$ can be written as $$\frac{X(\omega)}{(1-e^{-i\omega})}+\pi \sum\limits_{n=-\infty}^{+\infty} X(2\pi n) \delta(\omega-2\pi n)$$

Now, I understand the second term exists to yield the result of $Y(\omega)$ when $x(n)$, but I don't understand how this writing works, because when $\omega=0$ the first term goes to infinity anyway.

My question is, shouldn't the definition be:

$$\begin{cases} \frac{X(\omega)}{(1-e^{-i\omega})}, \omega\neq0 \\ \pi \sum\limits_{n=-\infty}^{+\infty} X(2\pi n) \delta(\omega-2\pi n), \omega=0 \end{cases}$$

because i think the other expression will always be infinite when $\omega=2 \pi n$.

Consider the discrete fourier transform of the step function $u(t)$
$$\frac{1}{1-e^{i\omega}}+\pi\sum\limits_{n=-\infty}^{+\infty} \delta(\omega-2\pi n)$$
$$\frac{1}{i\omega}+\pi\delta(\omega)$$