Regge behavior of the 1342 ordering of the Veneziano amplitude

Question

I'm trying to understand the Regge limit of the following open string diagram,

\(\frac{\Gamma(-1-\alpha' s)\Gamma(-1-\alpha' u)}{\Gamma(-2-\alpha'(s+u))} \)

The other diagrams I understand fine and lead to the usual Regge behavior, \((1+\exp(-i\pi \alpha' t))(\alpha' s)^{\alpha' t}\)

But this diagram is confusing. We take s to infinity along a direction where the imaginary part of s goes to infinity. If we want to be right above the branch cut then we should go in the direction \(s+i\epsilon s\)

This gives \(\frac{(-\alpha' s)^{-\alpha' s}(-\alpha' u)^{-\alpha' u}}{\Gamma(2+\alpha' t)}\)

The phase of this is \(\exp(i\pi \alpha' s)\)

At this point all the old dual resonance model papers say that this is exponentially suppressed, which I understand because the imaginary part of s is large. But what about when we go back to the physical region, where the imaginary part of s is zero? Then this seems to give a nonvanishing result, which modifies the standard Regge answer. Any ideas on why this is wrong?