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The DTFT is given by:

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$$

The IDTFT is given by:

$$x[n]=\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\omega})e^{j\omega n}d\omega$$

I have been able to show by substitution of the DTFT into the IDTFT that the transform and a subsequent inverse transform return $x[n]$:

$$\begin{align} x[n]&=\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\omega})e^{j\omega n}d\omega\\ &=\frac{1}{2\pi}\int_{0}^{2\pi} \left( \sum_{k=-\infty}^{\infty}x[k]e^{-j\omega k} \right)e^{j\omega n}d\omega\\ \end{align}$$

Swap the order of integration and summation:

$$x[n]=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\int_{0}^{2\pi}x[k]e^{j\omega (n-k)}d\omega$$

Argue that given $e^{j\omega (n-k)}$ is an odd function, it will only evaluate to anything other than 0 when $k=n$:

$$\begin{align} x[n]&=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\int_{0}^{2\pi}x[k]e^{j\omega (n-k)} d\omega \ \delta[n-k]\\ &=\frac{1}{2\pi}\int_{0}^{2\pi}x[n]d\omega \\ &=\frac{2\pi}{2\pi}x[n]\\ \end{align}$$

However, I have been unable to show the dual case: that the inverse transform (IDTFT) substituted into the forward transform (DTFT) gives $X(e^{j\omega})$. How can we show this?

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    $\begingroup$ i hope you don't mind that i "cleaned" some of the notational convention a little. i added a Kroenecker delta $\delta[n-k]$ to it. $\endgroup$ – robert bristow-johnson Feb 12 '19 at 0:49
  • $\begingroup$ Thanks, if it makes it clearer then that’s a good thing! $\endgroup$ – Requiens Feb 12 '19 at 1:13
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    $\begingroup$ Note that all integrals should be over one period, not over two. I've changed the formulas accordingly. $\endgroup$ – Matt L. Feb 12 '19 at 8:07
  • $\begingroup$ oh yeah, that's right. $\endgroup$ – robert bristow-johnson Feb 12 '19 at 10:46
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$$\begin{align}X(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\&=\sum_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{0}^{2\pi}X(e^{j\Omega})e^{jn\Omega}d\Omega\right]\;e^{-jn\omega}\\&=\int_{0}^{2\pi}X(e^{j\Omega})\left[\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{jn(\Omega-\omega)}\right]d\Omega\\&=\int_{0}^{2\pi}X(e^{j\Omega})\delta(\Omega-\omega)d\Omega\\&=X(e^{j\omega})\end{align}$$

where I've used the identity

$$\delta(\omega)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{jn\omega}$$

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  • $\begingroup$ This is perfect! I think my own confusion arose when using the same $\omega$ inside and outside the integral so became confused... using the different notations cleared up the misunderstanding. $\endgroup$ – Requiens Feb 12 '19 at 12:31
  • $\begingroup$ @Resquiens: Yes, you have to be careful with variables inside and outside the integral, and you need to recognize the identity I used for the Dirac impulse. $\endgroup$ – Matt L. Feb 12 '19 at 12:32

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