Computing the amplitude squared for $e^-\mu^-\rightarrow e^-\mu^-$ at tree level we get \begin{equation} \frac{1}{4}\sum_\mathrm{spins}|\mathcal{M}(s,t)|^2=2e^4\frac{s^2+u^2}{t^2} \end{equation} which is IR divergent as $t\rightarrow 0$. The KLN theorem tells us that all IR divergences cancel when computing cross sections, so what cancels this divergence?
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The singularity of the differential cross section as $t \rightarrow 0$ has the physical interpretation that an on-shell photon is created and travels an infinite distance before interacting.
For massless electrons t takes the form $t \sim (1-\cos\theta)$, and we see that $t\rightarrow 0$ corresponds to forward scattering . If we then use the amplitude to calculate the total cross section we find $$\sigma \sim \int d(\cos\theta)\frac{1}{\sin(\theta/2)^4}$$ (this is basically just the Rutherford formula), and the cross section is infinite.
To see what this means, recall that classically forward scattering corresponds to an infinite impact parameter. The infinite cross section can then be interpreted as that the charged electron and muon will always scatter, even if they are on the other side of the universe, i.e the EM force is long-ranged
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