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I'm reading page 193 of section 6.3 of the QFT textbook by Peskin and Schroeder. There are two integrals that we need to evaluate for the calculation in this section. (here, $\Delta>0$)

$$\int\frac{d^4l}{(2\pi)^4}\frac{1}{(l^2-\Delta+i\epsilon)^3}=\int\frac{d^3l}{(2\pi)^3}\int\frac{dl^0}{(2\pi)}\frac{1}{((l^0)^2-|\mathbf{l}|^2-\Delta+i\epsilon)^3}$$

$$\int\frac{d^4l}{(2\pi)^4}\frac{l^2}{(l^2-\Delta+i\epsilon)^3}=\int\frac{d^3l}{(2\pi)^3}\int\frac{dl^0}{(2\pi)}\frac{(l^0)^2-|\mathbf{l}|^2}{((l^0)^2-|\mathbf{l}|^2-\Delta+i\epsilon)^3}$$

They mention a validity condition for performing a wick rotation on an integral; that the integrand should fall off sufficiently rapidly at large $|l^0|$ where $l^0$ is the complex integration variable above. I'm assuming this is so that we can freely close the contour without changing the value of the integral (after that we can rotate the contour).

I think I see how this condition holds for the first integral. If I integrate over an arc in the complex $l^0$ plane, I set $l^0=Re^{i\theta}$ so the integration measure contributes one factor of $R\equiv|l^0|$ and the integrand (as $R\rightarrow\infty$) will behave like $1/R^6$ so overall we have suppression by $1/R^5$ which indeed goes to zero as $R\rightarrow\infty$. The analogous procedure for the second integral yields an overall suppression by $1/R^3$, which again vanishes as $R\rightarrow\infty$, so it seems to me we should still be able to do the Wick rotation, yet P&S mention in that page that the Wick rotation cannot be justified for the second integral (which confused me, I don't see why it is not justified). Now, still looking at the second integral above, if I do the Wick rotation anyway, and then change to a Euclidean integration variable, I get an integral that diverges anyway (easily seen in 4-dim spherical coordinates), but I'm not sure if this is related to my confusion about the validity of the Wick rotation itself or not. Any clarification would be appreciated.

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Well. We should first move to the spherical coordinate, and the integration measure has a $r^3$ factor, so that the second integrand tends to 1/r. As a result, the integral behaves as ~log(R) as R goes to infinity.

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