Using the optical theorem to calculate the imaginary part of a loop diagram I'm trying to calculate the imaginary part of this diagram

in $\phi^4$ theory, using the optical theorem, and I'm having trouble. 
All of the examples I can find use the theorem to relate the imaginary part of the total 2-particle to 2-particle forward scattering amplitude to the total cross-section; that's not what I'm trying to do. I see a lot of equations that look like this:
$$ 2 \operatorname{Im} A = \int d\Pi | B |^2 $$
wherein $A$ is the diagram above and $B$ is 

but nobody really discusses such equations or works any examples. Does that mean that I just take the modulus squared of the tree-level diagram: $|i \lambda|^2 = \lambda^2$? That seems too simple, which brings me to the integration over $d\Pi$. What the heck is $d\Pi$? Peskin and Schroeder don't say, but they make vague mention of the phase space of the intermediate particles, so is $d\Pi$ just the differential phase space of the "two" $\phi$s in the loop? If so, how do I go about setting up and evaluating that integral? If not, what is $d\Pi$, and how do I evaluate the integral over it?
 A: d$\Pi$ is, indeed, the differential phase-space. Peskin and Schroeder have an equation of exactly the form above in figure 7.6 on page 235, and although they don't say what $d\Pi$ is there, they do define a similar, but more specific, quantity $d\Pi_n$, the differential phase-space for $n$ particles, in equation 4.80 on page 106:
$$ d\Pi_n = \left ( \prod _f \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f} \right ) (2\pi)^4 \delta^{(4)}\left (P-\sum_f p_f \right) $$
wherein the $\Pi$ on the left is a variable indicating the $n$-body phase space, the $\prod$ on the right is a product symbol, $P$ is the net external 4-momentum, and the subscript $f$ indicates the final state 4-momenta of the $n$ particles.
Noting that
$$\int \frac{d^4 p}{(2\pi)^4} 2\pi\delta(p^2-m^2) = \int\frac{d^3 p}{(2\pi)^3}\frac{1}{2 E_{\vec p}}$$
this prescription for $d\Pi$ yields the same integrals for evaluation of a diagram via the optical theorem as do Cutkosky's cutting rules, confirming that this is the correct $d\Pi$ and not just a coincidence of notation.
That said, there are still serious complications in the evaluation of those integrals. See this related question for details.
