How to calculate the DTFT of $1$? The sequence $x[n] = 1$ is not absolutely summable, so one can not compute the DTFT by using the definition

$$X(\Omega) \space = \sum \limits_{n=-\infty}^{\infty}x[n]e^{-j\Omega n}$$

Can anyone point me to a derivation of DTFT of $1$ from the first principles?

The derivations I came across used the fact that DTFT of $e^{j\Omega_0n}$ is

$$2\pi\sum \limits_{k=-\infty}^{\infty} \delta(\Omega-\Omega_0-2k\pi)$$

which again brings me to the question: how?

  • $\begingroup$ Rodrigo de Azevedo, thanks for the edit! Yes, I'm aware that it should be an impulse train and computing the IDFT will prove it. My quest however is to compute the DTFT and get the result as the impulse train. I want to deduce it mathematically without guessing its DTFT and then taking IDFT to prove it. $\endgroup$
    – Navin
    Mar 24 '19 at 10:57
  • 1
    $\begingroup$ The DTFT of the complex exponential is not the result of a computation (as the DTFT series does not converge, of course). It is just a convenient definition, which we take for granted. It is intuitively justified because it gives the expected result when applying the IDTFT formula. Other than that, there is nothing more to say. $\endgroup$
    – Stelios
    Mar 24 '19 at 12:21

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