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How to integrate this function:

$$x[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi}\delta(\omega-\omega_{0}) e^{jn\omega}d\omega$$

Where:

$$\delta[\omega-\omega_0] = \begin{cases}1&n=\omega_o\\0&n\neq\omega_o\end{cases}$$

The book says the answer is suppose to be:

$$x[n] = \frac{1}{2\pi}e^{jn\omega_o} $$

Here's what i'm trying to do:

$$x[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi}\delta(\omega-\omega_{0}) e^{jn\omega}d\omega$$

$$x[n] = \frac{1}{2\pi} e^{jn\omega_o} \int_{-\pi}^{\pi}d\omega$$

$$x[n] = \frac{1}{2\pi} e^{jn\omega_o} (\pi - (-\pi))$$

$$x[n] = \frac{1}{2\pi} e^{jn\omega_o} (2\pi)$$

$$x[n] = e^{jn\omega_o} $$

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You've made a few mistakes, such as giving an $n$-dependent definition of $\delta (\omega-\omega_0)$ instead of an $\omega$-dependent one, and I'm sure your book wants you to work with Dirac deltas, whereas you've used a Kronecker delta. If $\omega_0\in [a,\,b]$, $\int_a^b\delta(\omega-\omega_0)f(\omega)d\omega=f(\omega_0)$ (otherwise the result is $0$); that defines the Dirac delta, as well as making the problem trivial. If the given formula were really meant to use a Kronecker delta, the integrand would be function-valued and vanish except at $\omega=\omega_0$, and the integral would be $0$. Note that the Dirac delta is a measure, not a true function.

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  • $\begingroup$ That's a good point. I'm mixing up dirac with unit step function. its really: $$\delta[\omega-\omega_0] = \begin{cases}\infty&n=\omega_o\\0&n\neq\omega_o\end{cases}$$ $\endgroup$ – Bill Moore Nov 22 '18 at 16:28
  • $\begingroup$ Its a derivation for DTFT property... I guess its confusing because they use dirac delta in n-domain, and kronecker delta in the frequency domain. $$\delta(\omega-\omega_0) = \begin{cases}\infty&n=\omega_o\\0&n\neq\omega_o\end{cases}$$ $$\delta[n-n_o] = \begin{cases}1&n=n_o\\0&n\neq\n_o\end{cases}$$ $\endgroup$ – Bill Moore Nov 22 '18 at 16:37
  • $\begingroup$ sorry reverse that... dirac delta in frequence domain and kronecker delta in n-domain $\endgroup$ – Bill Moore Nov 22 '18 at 16:43
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Remember the translation property of the Dirac Delta. If $a < x_0 < b$

$$ \int_{a}^b {\rm d}x~ f(x) \delta(x - x_0) = f(x_0) $$

If you apply this to your problem you get

$$ \frac{1}{2\pi} \int_{-\pi}^\pi {\rm d}\omega ~ \delta (\omega - \omega_0) e^{j n \omega} = \frac{1}{2\pi} e^{j n \omega_0} $$

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