$\let\th=\theta \let\dag=\dagger \def\bk{\mathbf k} \def\br{\mathbf r}
\def\st{\sigma_{\mathrm{tot}}} \def\Im{\mathrm{Im}}$
Shortly, because it's a first-order perturbative approximation. Let me recall how optical theorem is proven. There are two ways, equivalent at their root:

- flux conservation
- unitarity of $S$-matrix.

By flux conservation I mean the following. Start from
$$u(\br) = e^{i\bk\cdot\br} + f(\th)\,{e^{ikr} \over r}$$
as the asymptotic form of solution for a scattering problem. We expect
that asymptotically outgoing flux equates ingoing one, i.e. that total asymptotic flux vanishes. Let's write shortly
$$u = u_1 + u_2.$$
Computing flux, which is quadratic in $u$, we shall find three terms:
$$\Phi(u_1) + \Phi(u_2) + \hbox{interference term}.$$

$\Phi(u_1)$ is obviuosly zero. $\Phi(u_2)$ is proportional to $\st$,
and the interference term gives $\Im f(0)$. Result is the optical theorem.

Now note that $f(\th)$ is of the first order in $V(r)$, whereas $\st$
is of second order. A first-order perturbative approach neglects second-order terms, thus giving $\Im f(0) = 0$.

Unitarity of $S$-matrix means $SS^\dag=I$. Define $S = I + T$. Then
$$I = SS^\dag = I + T + T^\dag + TT^\dag.$$
Again $T$ expresses outgoing wave, i.e. scattering amplitude (with an
$i$ multiplier, so that $T+T^\dag$ is $\Im f$). $TT^\dag$ gives $\st$,
and the above argument may be repeated.

This post imported from StackExchange Physics at 2022-08-14 10:22 (UTC), posted by SE-user Elio Fabri