= Solution
Expanding the modulus squared yields an outgoing-angle factor $e^{-i\mathbf k_f\cdot\mathbf d}$ and an incoming factor $e^{i\mathbf k_i\cdot\mathbf d}$, where $\mathbf d=\mathbf r-\mathbf r'$. Choosing a polar axis along $\mathbf d$ gives
$$
\int d\Omega_f\,e^{-ik\widehat{\mathbf k}_f\cdot\mathbf d}
=2\pi\int_{-1}^1e^{-ikdu}\,du=4\pi\frac{\sin kd}{kd}.
$$
Consequently
$$
\boxed{\sigma^{(1)}_{\rm total}=\frac{m^2}{\pi\hbar^4}
\int d^3r\,d^3r'\,V(\mathbf r)V(\mathbf r')
\frac{\sin kd}{kd}\,e^{i\mathbf k_i\cdot\mathbf d}}.
$$
The apparent $d=0$ singularity is removable: $\sin(kd)/(kd)\to1$. Exchanging $\mathbf r$ and $\mathbf r'$ cancels the imaginary part, while nonnegativity follows directly from its origin as an integral of $|f^{(1)}|^2$. A complex absorbing potential would require $V(\mathbf r')^*$ instead; the lossless <optical theorem> comparison below uses real $V$.
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