I am reading Oppenheim & Schafer's (O&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&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: $$x[n+m]w[m]=\frac{1}{2\pi}\int_0^{2\pi} X[n,\lambda)e^{j\lambda m}d\lambda, \: \: \: -\infty < m < +\infty$$ 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&S versions, so I guess it's not a typo and I am missing something very fundamental here :D
Thanks in advance.