Unitarity constraints for real soft photons (Weinberg, Section 13.3) Weinberg obtains differential scattering rate for a process with soft photons,
$$\begin{align*}
         d \Gamma_{\beta \alpha}^{\lambda}(\omega_1, ...\omega_N) =  \Gamma^\lambda_{\beta \alpha} A(\alpha \rightarrow \beta)^N \frac{d \omega_1}{\omega_1} ... \frac{d \omega_N}{\omega_N}, 
    \end{align*}\tag{13.3.5}$$
and says:

the unitarity demands that if we use an infrared cut-off for virtual photons as shown by the subscript, we must use the same cut-off for the real photons.

Why is it the case? The $\omega$ factors are the energies for the emitted, real soft photons.
 A: The answer to your question lies in the heart of the Optical Theorem. I will refer you to discussions in P&S (Chapter 7.3), but there are also other textbooks explaining that subject. I will not explain the Optical theorem here, as I think the details in its derivation are not useful for understanding what you are asking, but if you try to study it and something doesn't make sense, you can always leave a comment.
So, the Optical Theorem (for a simple $2\rightarrow2$ process) says that due to unitarity ($SS^{\dagger}=1\Rightarrow -i(T-T^{\dagger})=T^{\dagger}T$ where $T$ is the non trivial part of the S-matrix, defined according to $S=1+iT$), the imaginary part of the process in which there exists a photon loop, will be (up to a factor of 2) equal to the square of the process, corresponding to the Feynman diagram with the photon propagator cut and the two photons formed by that cut being put on shell (the two fermions on the other side of the diagram can either be the initial 2 or the final two fermions), integrated over all the possible momenta the two external photons can have.
Having said that, since the amplitude of the total process will contain a momentum integral because of the closed photon loop, and since the latter amplitude must be equal to the square of the amplitude of the diagram in which the internal photon (loop) propagator is cut, integrated over the possible momenta of the two fermions, it can be inferred that the integral in the LHS (loop integral) will have the same limits with the integral in the RHS (external photon integral), with one of the integrals wrt the external photons in the cut diagram being negated by a $\delta$ function.
I have not provided a fully explainatory solution, but the example of $2\rightarrow2$ fermion scattering with a single photon loop is sufficient to understand how the integral limits are to determined from the argument of unitarity. More general cases surely exist, but they all work out the same. Also, I couldn't exactly describe schematically the details, but I think if you have a look at P&S Chapter 7.3, things will start to make more sense. If there are any questions/concerns, please do let me know in the comments.
