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Define $y[n]:=\displaystyle\sum_{m=-\infty}^{n}x[m]$. The DTFT is found as follows: \begin{align*} y[n]&=\sum_{m=-\infty}^{n}x[m]\\ \\ &=\sum_{m=-\infty}^{n-1}x[m]+x[n]\\ \\ &=y[n-1]+x[n]\\ \\ \implies x[n]&=y[n]-y[n-1] \end{align*} Therefore, applying DTFT : \begin{align*} \mathcal{F}\{x\}[n]=X(e^{j\omega})&=\mathcal{F}\{y[n]-y[n-1]\}\\ \\ &=(1-e^{-j\omega})Y(e^{j\omega}) \end{align*} Hence : $$ Y(e^{j\omega})=\frac{1}{1-e^{-j\omega}}X(e^{j\omega}) $$ However, the correct answer is : $$ \frac{1}{1-e^{-j\omega}}X(e^{j\omega})+\color{red}{\pi X(e^{j0})\sum_{k=-\infty}^{\infty}\delta(\omega-2k\pi)} $$ My question is where did the DC term come from and why didn't it work in my derivation?

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    $\begingroup$ Dividing by $1-e^{-j\omega}$ is not valid for $\omega=0$ (and integer multiples of $2\pi$). $\endgroup$ – Matt L. Apr 9 at 9:50
  • $\begingroup$ You are right sir, but how did they fix this when they used the impulse train? @MattL. $\endgroup$ – deerclaysup Apr 9 at 9:52
  • $\begingroup$ I might write up an answer when I have the time to do so, but in the meantime take a look at the answers to this question. $\endgroup$ – Matt L. Apr 9 at 9:58
  • $\begingroup$ ... and this one. $\endgroup$ – Matt L. Apr 9 at 9:59
  • $\begingroup$ Alright will do, thank you very much sir! $\endgroup$ – deerclaysup Apr 9 at 10:00
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Note that the equation

$$x[n]=y[n]-y[n-1]\tag{1}$$

doesn't uniquely determine the sequence $y[n]$. If some $y[n]$ satisfies $(1)$, so does $y[n]+c$ with some constant $c$. Consequently, Eq. $(1)$ determines $y[n]$ only up to a constant, which corresponds to a DC term in the frequency domain.

Hence, the equation

$$Y(e^{j\omega})=\frac{X(e^{j\omega})}{1-e^{-j\omega}}\tag{2}$$

is only valid for $\omega\neq 0$, or, more generally, for $\omega\neq 2\pi k$, $k\in\mathbb{Z}$.

The most straightforward way to derive the DTFT of

$$y[n]=\sum_{m=-\infty}^nx[m]\tag{3}$$

is to realize that $y[n]$ is the convolution of $x[n]$ with the unit step sequence $u[n]$, and, consequently, the DTFT of $y[n]$ is given by

$$Y(e^{j\omega})=X(e^{j\omega})U(e^{j\omega})\tag{4}$$

where $U(e^{j\omega})$ is the DTFT of the unit step $u[n]$:

$$U(e^{j\omega})=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{5}$$

Several ways to derive the result $(5)$ are discussed in the answers to this and this question.

Note that in $(5)$ I use $\delta(\omega)$ instead of $\sum_k\delta(\omega-2\pi k)$ to avoid cluttered notation. The fact that the DTFT is always $2\pi$-periodic is understood.

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  • $\begingroup$ Thank you very much sir now I have fully understood it. $\endgroup$ – deerclaysup Apr 9 at 11:22
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    $\begingroup$ "The fact that the DTFT is always $2\pi$-periodic is understood". I have also encountered more mathematically rigorous folks who insist (to the point of engaging in flame wars) that the DTFT is only defined on some interval $\omega \in [\omega_0, \omega_0 + 2 \pi)$ or $\omega \in (\omega_0, \omega_0 + 2 \pi]$. Either definition works equally well for casual engineering use. $\endgroup$ – TimWescott Apr 9 at 14:52

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