Limit of integral containing $e^{iomega L(frac{pi}{2}-tau)}$ as $Lto infty$

In one Physics paper several times one encounters a limit of the following form $$lim_{Lto infty}int_0^pi e^{pm iomega L(frac{pi}{2}-tau)}int_{S^2} varepsilon(hat{x})f(tau,hat{x})dtau d^2hat{x}.$$

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

$$lim_{Lto infty}int_0^pi e^{pm iomega L(frac{pi}{2}-tau)}int_{S^2} varepsilon(hat{x})f(tau,hat{x})dtau d^2hat{x}=int_{widetilde{mathcal{I}}^+} varepsilon(hat{x})f(tau,hat{x})dtau d^2hat{x},$$

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

$$lim_{Lto infty}int_0^pi e^{pm iomega L(frac{pi}{2}-tau)}f(tau,hat{x})dtau = int_{pi/2-epsilon}^{pi/2+epsilon} f(tau,hat{x})dtau,quad epsilonll 1.$$

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?