Calculation of one-loop diagram in $\phi^4$ theory

Question

In Folland's book Quantum Field Theory, page 207, he gives the value of the amputated one-loop $\phi^4$ diagram as $$I(p) = \frac{(-i\lambda)^2}{2} \int \frac{-i}{-q^2 + m^2 - i\epsilon} \cdot \frac{-i}{-(q+p)^2 + m^2 - i\epsilon} \frac{d^4q}{(2\pi)^4}. \tag{1}$$ To evaluate this integral, he first Wick rotates by substituting: $$q^0 \rightarrow iq^0 \\ p^0 \rightarrow ip^0$$ which he claims gives $$\frac{(-i\lambda)^2}{2} \int \frac{-i}{(|q|^2 + m^2)(|q+p|^2 +m^2)} \frac{d^4q}{(2\pi)^4}. \tag{2}$$ He then goes on to introduce Feynman parameters and carries on with the computation. I am confused on the numerators of (1) and (2). After carrying out the multiplication of the two fractions in (1), shouldn't the numerator of (2) be $-1$? Also, how is it justified to drop the terms containing $\epsilon$?

Answer 1

I am confused on the numerators of (1) and (2). After carrying out the multiplication of the two fractions in (1), shouldn't the numerator of (2) be $-1$?

Don't forget that you have a $dq_0$ in your measure $d^4q = dq_0 d^3q$.

So you pick up another factor of $i$ when you make the $q_0\to i q_0$ substitution (change of variables), since $dq_0 \to i dq_0$.

Also, how is it justified to drop the terms containing $\epsilon$?

It is justified to set $\epsilon$ to zero because the rest of the denominator can no longer be zero. For example, there is no way for $|q|^2 + m^2$ to be zero since both terms are non-negative, and $m^2$ is positive (since $m\neq 0$).

The $\epsilon$ is there to make the integration meaningful when the denominator can be zero, and it is understood that we want $\epsilon$ to go to zero. So, after the Wick rotation we can just explicitly set $\epsilon$ to zero, since that is the same as the limit as it goes to zero.