In one Physics paper several times one encounters a limit of the following form $$\lim_{L\to \infty}\int_0^\pi e^{\pm i\omega L(\frac{\pi}{2}-\tau)}\int_{S^2} \varepsilon(\hat{x})f(\tau,\hat{x})d\tau d^2\hat{x}.$$
The integral over $S^2$ likely plays no role in this issue.
In the paper the authors essentially say that this limit "forces $\tau$ to be in a small neighborhood of $\pi/2$" and therefore what they do in the end is to set the exponential to one and then write the limit as something of the form
$$\lim_{L\to \infty}\int_0^\pi e^{\pm i\omega L(\frac{\pi}{2}-\tau)}\int_{S^2} \varepsilon(\hat{x})f(\tau,\hat{x})d\tau d^2\hat{x}=\int_{\widetilde{\mathcal{I}}^+} \varepsilon(\hat{x})f(\tau,\hat{x})d\tau d^2\hat{x},$$
where ${\widetilde{\cal I}^+}$ denotes the region of $[0,\pi]\times S^2$ comprising the suggested neighborhood of $\pi/2$.
In effect it seems to me that what they are saying is that
$$\lim_{L\to \infty}\int_0^\pi e^{\pm i\omega L(\frac{\pi}{2}-\tau)}f(\tau,\hat{x})d\tau = \int_{\pi/2-\epsilon}^{\pi/2+\epsilon} f(\tau,\hat{x})d\tau,\quad \epsilon\ll 1.$$
Would that really be true? How can it be made precise in case it is true? What is actually the neighborhood? And shouldn't be a limit on the right-hand side? In short, what is the rigorous counterpart of this argument of "throwing away the exponential but still keeping the integral over a small neighborhood of $\pi/2$" to deal with this limit?