Binomial expansion of massive spin 0 propagator

[RedX]

I've seen written the following manipulation of the KG Feynman propagator:

$$\frac{1}{p^2-m^2+i\epsilon}=\frac{1}{p^2+i\epsilon} \frac{1}{(1-\frac{m^2}{p^2+i\epsilon})}= \frac{1}{p^2+i\epsilon} (1+\frac{m^2}{p^2+i\epsilon}+\left(\frac{m^2}{p^2+i\epsilon}\right)^2+...)$$

I don't think this can be valid unless $$|p^2+i\epsilon|>m^2$$.

However if you are going to integrate this expression to get the inverse Fourier transform:

$$\int d^4p \frac{e^{ip(x-y)}}{p^2-m^2+i\epsilon}$$

then this integrals picks out the poles which forces $$p^2=m^2-2iE_p\epsilon$$.

However, since the epsilon term comes in with a minus sign, it could potentialy cancel out the +iepsilon term in $$|p^2+i\epsilon|>m^2$$.

So is the binomial expansion valid?

 

[Evalesco]

Yes, the binomial expansion is valid, as long as the absolute value of p^2+i\epsilon is greater than m^2. However, when integrating to calculate the inverse Fourier transform, this condition may not be satisfied and the epsilon term could potentially cancel out the +iepsilon term.