Invertibility of Time-Dependent Fourier Transform I am reading Oppenheim &amp; Schafer's (O&amp;S) Discrete Time Signal Processing (2nd or 3rd edition, does not matter) and I find hard to understand a technicality behind the Time-Dependent Fourier Transform (TDFT from now on). Specifically, I have a question regarding the invertibility of the TDFT.

According to the textbook, the TDFT is given by
$$X[n, \lambda) = \sum_{m=-\infty}^{+\infty}x[n+m]w[m]e^{-j\lambda m}$$
which can be thought as the DTFT of a windowed signal portion around sample $n$. So, by moving one sample at a time, we have "a collection" of DTFTs of windowed signal portions around sample $n$, and the TDFT is a 3D-function (consider magnitude only) with axis $n$ vs axis $\lambda$.

Then, O&amp;S mentions that the TDFT is invertible iff $w[m]$ has at least one nonzero sample. 

If we take the Fourier synthesis equation, we get: $$x[n+m]w[m]=\frac{1}{2\pi}\int_0^{2\pi} X[n,\lambda)e^{j\lambda m}d\lambda, \: \: \: -\infty &lt; m &lt; +\infty$$
which makes perfect sense as it reconstructs the windowed signal around sample $n$ from complex exponentials weighted by the DTFT $X[n, \lambda)$.
Dividing by the window: $$x[n+m] = \frac{1}{2\pi w[m]}\int_0^{2\pi} X[n, \lambda)d\lambda$$ if $w[m] \neq 0$.

Here's my question: where did the $e^{j\lambda m}$ term go??

Isn't this equation supposed to recover the full windowed signal $x[n+m]$? The integral looks as a number to me (summing $X[n,\lambda)$ values for all $\lambda$ at index $n$ of the TDFT).

The same equation is in all O&amp;S versions, so I guess it's not a typo and I am missing something very fundamental here :D

Thanks in advance.
I have the 2nd edition, and there the case $m=0$ is considered, so the equation reads (Eq. $(10.21)$ on p. 716)

$$x[n] = \frac{1}{2\pi w[0]}\int_0^{2\pi} X[n, \lambda)d\lambda$$

(which of course assumes $w[0]\neq 0$). This makes sense because $e^{j\lambda m}$ equals $1$ for $m=0$.