 Solution

In the forward direction ${\mathbf{k}}_f={\mathbf{k}}_i$, the first Born amplitude $-C\int V$ is real. Symmetrize the second-order double integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$ before taking its imaginary part. This replaces $e^{-i{\mathbf{k}}_i\cdot \mathbf{d}}$ by $\cos ({\mathbf{k}}_i\cdot \mathbf{d})$, and givesThe preceding total-cross-section integral equals the same real integral multiplied by $4\pi /k$. ThereforeThis is the optical theorem at second Born order. Comparing only the real first-order forward amplitude with a nonzero second-order scattering cross-section would mix perturbative orders. Terms of order $V^3$ require both the next amplitude order and interference in the scattering cross-section.