How to get the imaginary part from the Källén-Lehmann propagator

During field theory course the Källén-Lehmann propagator was defined as follows:

$$D_F(p^2) = \frac{i}{p^2-m^2+i\epsilon} + \int^{\infty}_{4m^2}ds\rho(s)*\frac{i}{p^2-s+i\epsilon} \tag{1}$$

afterwards, it was stated that by using:

$$\frac{1}{x+i\epsilon} = PV(\frac{1}{x}) - i\pi\delta(x) \tag{2}$$

I could get (with $$\theta$$ being the Heaviside step function):

$$Im[iD_F(p^2)] = \pi\delta(p^2-m^2) + \pi\theta(p^2>4m^2)\rho(p^2) \tag{3}$$

I don't understand how to go from (1) to (3). any help would be much appreciated.

• It's not that difficult: set $x = p^2 - m^2$, then... – Hans Moleman Jul 2 at 15:29

From: $$D_F(p^2) = \frac{i}{p^2-m^2+i\epsilon} + \int^{\infty}_{4m^2}ds\rho(s)\cdot\frac{i}{p^2-s+i\epsilon} \tag{1}$$ you get that: $$iD_F(p^2) = \frac{-1}{p^2-m^2+i\epsilon} - \int^{\infty}_{4m^2}ds\rho(s)\cdot\frac{1}{p^2-s+i\epsilon}$$ For the first bit, set $$x = p^2-m^2$$ and use your identity to show: $$\frac{-1}{p^2-m^2+i\epsilon} = -PV \left ( \frac{1}{p^2-m^2} \right ) + i\pi \delta(p^2-m^2).$$ For the second bit, set $$x = p^2-s$$ and use your identity to show: $$\int^{\infty}_{4m^2}ds\rho(s)\cdot\frac{-1}{p^2-s+i\epsilon} = -\int^{\infty}_{4m^2}ds\rho(s)\cdot \left [ PV \left ( \frac{1}{p^2-s} \right ) -i\pi \delta(p^2-s) \right ] = \\-PV \left ( \frac{1}{p^2-s} \right )\int^{\infty}_{4m^2}ds\rho(s) + \int^{\infty}_{4m^2}ds \rho(s)\,i\pi \delta(p^2-s) = \\ \mathrm{sth\,real} + i\pi\rho(p^2)\theta(p^2>4m^2),$$ where in the last equation I have used $$\int \delta(x-x_0)f(x)dx = f(x_0)$$, and the Heaviside step function to enforce the limits of integration.
$$Im[iD_F(p^2)] = \pi\delta(p^2-m^2) + \pi\theta(p^2>4m^2)\rho(p^2) \tag{3}$$