In short : is the DTFT a reversible operation ? That is to say, is the inverse DTFT of the DTFT of a continuous signal x, equal to x ?
In less short : I tried to demonstrate that this is the case, but can't complete the demonstration. Considering the following DTFT definition of a signal $x:t \to x(t)$ sampled with a period $T_e$ :
$$x_e(t) = \sum_{n=-\infty}^{+\infty}x(nT_e)\delta(t - nT_e)$$
$$\left[DTFT(x)\right](f) = FT(x_e)(f) = X_e(f) = \sum_{n=-\infty}^{+\infty}x(nTe) e^{-2i\pi f nT_e}$$
The Inverse DTFT would be (I guess):
$$\left[IDTFT(X_e)\right](t) = T_e \int_{-F_e/2}^{F_e/2}X(f)e^{2i\pi ft}df$$
where $F_e=1/T_e$ is the sampling frequency. Mixing up the two I end up with : $$T_e \int_{-F_e/2}^{F_e/2}X_e(f)e^{2i\pi ft}df = T_e \int_{-F_e/2}^{F_e/2} \left[\sum_{n=-\infty}^{+\infty}x(nT_e) e^{-2i\pi f nT_e} \right] e^{2i\pi f t} df= T_e \sum_{n=-\infty}^{+\infty} x(nT_e) \int_{-F_e/2}^{F_e/2} e^{-2i\pi f nT_e} e^{2i\pi ft} df$$ where the last integral resolves to : $$\int_{-F_e/2}^{F_e/2} e^{-2i \pi f (nTe-t)}df = \left[\frac{e^{-2i \pi f (nTe-t)}}{-2i\pi(nTe-t)} \right]_{-Fe/2}^{Fe/2} = \frac{e^{-2i \pi Fe/2 (nTe-t)} - e^{-2i \pi -Fe/2 (nTe-t)}}{-2i\pi(nTe-t)} = \frac{e^{2i \pi Fe/2 (nTe-t)}- e^{-2i \pi Fe/2 (nTe-t)}}{2i\pi(nTe-t)} = \frac{sin(\pi Fe(nTe-t))}{\pi(nTe-t)}$$ letting me with : $$ T_e \sum_{n=-\infty}^{+\infty} x(nT_e) \frac{sin(\pi Fe(nTe-t))}{\pi(nTe-t)} \ne x(t)$$
What am I missing ?
Cheers
Nevermind, I used a wrong definition of the inverse DTFT. With :
$$\int_{-\infty}^{+\infty}X_e(f)e^{2i\pi ft}df $$
we get that the IDTFT of the DTFT is :
$$ \int_{-\infty}^{\infty}X_e(f)e^{2i\pi ft}df = T_e \int_{-\infty}^{\infty} \left[\sum_{n=-\infty}^{+\infty}x(nT_e) e^{-2i\pi f nT_e} \right] e^{2i\pi f t} df= \sum_{n=-\infty}^{+\infty} x(nT_e) \int_{-\infty}^{\infty} e^{-2i\pi f nT_e} e^{2i\pi ft} df = \sum_{n=-\infty}^{+\infty} x(nT_e) \int_{-\infty}^{\infty} e^{2i \pi f (t-nT_e)}df = \sum_{n=-\infty}^{+\infty} x(nT_e) \delta(t-nT_e) = x_e(t) $$