I am considering the $\phi^3$ theory in $d = 3$ dimensions. The propagator is given by the following formula
$$ G(p^2) = \frac{-i}{p^2 + m_0^2 + M^2(p^2) - i\epsilon} $$
where
$$ M^2(p^2) = -\frac{g_0^2}{16\pi}\int_0^1 \frac{dx}{\sqrt{m_0^2 + x(1-x)p^2}} + O(g_0^4) = -\frac{g_0^2}{8\pi \sqrt{p^2}}\tan^{-1}\bigg(\frac{\sqrt{p^2}}{2m_0}\bigg) + O(g_0^4) $$ and $$ m_0^2 \approx m^2 + \frac{g_0^2}{16\pi m_0}\ln3 - O(g_0^4) $$
The propagator can be written in terms of the spectral density $\rho$ as follows:
$$ G(p^2) = \int_0^\infty \frac{d(\mu^2)}{2\pi} \rho(\mu^2) \frac{-i}{p^2 + \mu^2 - i\epsilon} $$
where
$$\rho = 2\pi Z \delta(\mu^2 - m^2) + \sigma(\mu^2)\theta(\mu^2 - 4m^2)$$ Textbooks like Peskin and Schroeder just mention the above form of the spectral density but not mention the exact form of the function $\sigma$ for any QFT and therefore I wanted to see whether I could obtain the function myself for the $\phi^3$ theory. To proceed, I considered the following function:
$$ F(s) = \int_0^\infty \frac{d(\mu^2)}{2\pi} \rho(\mu^2) \frac{i}{s - \mu^2} = G(p^2 = i\epsilon - s) $$
Given the form of the spectral density, I was able to show that
$$ \lim_{\epsilon \to 0}\big[ F(s + i\epsilon) - F(s-i\epsilon)\big] = \sigma(s) $$
whenever $s$ is real and $\ge 4m^2$. Consequently, $s = 4m^2$ is a branch point. My idea was to substitute
$$ F(s) = G(p^2 = i\epsilon - s) = \frac{-i}{ - s + m_0^2 + M^2(i\epsilon - s)} $$
into the limit formula to determine $\sigma$. Evaluating the limit, however, I keep on getting zero for $s \ge 4m^2$
How should I determine $\sigma$?
