# Wick Rotation and sign of the integrand in Weinberg's book

I'm studying from Weinberg's QFT volume 1, chapter 11. I have a problem with equation $$(11.2.7)$$.

Starting from eq. $$(11.2.5)$$

\begin{align} \Pi^{\rho\sigma} (q) = \frac{-ie^2}{(2\pi)^4} \int_0^1dx \int d^4p \, [p^2 + m^2 -i\epsilon + q^2x (1-x)]^{-2} \, \\ \times \, \text{Tr} \{[-i (\not\!{p} + \not\!{q} x) + m]\gamma^\rho[-i(\not\!{p} - \not\!{q}(1-x))+m]\gamma^\sigma \} \tag{11.2.5} \end{align}

And we have

\begin{align} \text{Tr} \{[&-i (\not\!{p} + \not\!{q} x) + m]\gamma^\rho[-i(\not\!{p} - \not\!{q}(1-x))+m]\gamma^\sigma \} \\= 4 [&-(p+qx)^\rho (p-q(1-x))^\sigma + (p+qx)\cdot(p-q(1-x)) \eta^{\rho\sigma}\\ &- (p+qx)^\sigma (p-q(1-x))^\rho + m^2 \eta^{\rho\sigma}] \tag{11.2.6} \end{align}

Looking at the first term of $$(11.2.6)$$ the $$0-0$$ component is, for example proportional to

$$\Pi^{00} (q)\propto -i \int d^4p (-p^0 p^0) \tag{1}$$

Then he performs a Wick Rotation $$p^0= ip^4$$; $$d^4p = i (d^4p)_E$$ and obtains the following result:

\begin{align} \Pi^{\rho\sigma} (q) = \frac{4e^2}{(2\pi)^4}& \int_0^1dx \int (d^4p)_E \, [p^2 + m^2 + q^2x (1-x)]^{-2} \, \\ \times \, [&-(p+qx)^\rho (p-q(1-x))^\sigma + (p+qx)\cdot(p-q(1-x)) \eta^{\rho\sigma}\\ &- (p+qx)^\sigma (p-q(1-x))^\rho + m^2 \eta^{\rho\sigma}] \tag{11.2.7} \end{align}

And he says that $$\eta$$ can be interpreted as a euclidean metric, therefore he has that the term $$4-4$$ is

$$\Pi^{44} (q)\propto \int (d^4p)_E \, (-p^4 p^4) \tag{2}$$

while I obtain that it should be

$$\Pi^{44} (q)\propto -\int (d^4p)_E \, (-p^4 p^4) \tag{3}$$

Let me explain, from $$(1)$$ I have

\begin{align} \Pi^{00} (q)\propto -i \int d^4p (-p^0 p^0) \xrightarrow{\text{p^0=ip^4}} &\Pi^{44} (q)\propto -i^2 \int (d^4p)_E \, (-i^2 p^4 p^4)\\ &= \int (d^4p)_E \, (p^4 p^4) = -\int (d^4p)_E \, (-p^4 p^4) \end{align}

Which is equal to $$(3)$$ and different from what Weinberg has, which is (2).

Since I've seen something similar in another Padmanabhan's book too, there must be something basic that I'm missing here and it can't be a typo in the book, can you please help me and point it out?

EDIT: Is it possible that he's just changing the measure from Minkowskian to Euclidean but not applying the transformation to the integrand?

The main point is that temporal components/indices of the vacuum polarization tensor $$\Pi^{\rho\sigma}(q)$$ are Wick-rotated. In particular $$\Pi^{00}(q)\to-\Pi^{44}(q)$$.