# Schroeder's Minkowski Space Integral - Concerns about Wick Rotations

In the Appendix of Peskin & Schroeder's "An Introduction to Quantum Field Theory" there is a list of integrals in Minkowski space. Of particular interest to me is the integral (A.44): $$I(\Delta) = \int \frac{d^{d}\ell}{(2\pi)^d} \frac{1}{(\ell^{2} - \Delta)^{n}} = \frac{(-1)^{n} i }{(4\pi)^{d/2}} \frac{\Gamma(n-\tfrac{d}{2})}{\Gamma(n)} \left( \frac{1}{\Delta} \right)^{n-d/2}$$

$\Delta$ doesn't depend on $\ell$, and Peskin uses the metric $(+---)$. Also, $d$ is the dimension of the space we're looking at, obviously. Peskin derives this integral by Wick rotating to Euclidean space such that $\ell^{0} = i \ell^{0}_{\mathrm{E}}$, and $\ell^{2} = - \ell^{2}_{\mathrm{E}}$ (this is shown in more detail in Chapter 6.3).

(Context: I need to look at the above integral for $n=3$. Supposedly this diverges in $d=4$, which is what I am looking at)

• Because of the Wick rotation used, is the above integral exact or does it just work some of the time, or as an approximation? I always thought that it was exact, but recently I had a discussion with someone who said this is not true and a Wick rotation doesn't necessarily yield the exact result. I am confused and would like some clarification on this.

• Secondly, Schroeder uses the $(+---)$ metric (aka the wrong metric for those East Coast fans like myself). This means that $\ell^{2} = (\ell^{0})^2 - ||\boldsymbol{\ell}||^{2}$. I am wondering, does the above integral change in its result if we use the metric $(-+++)$? Mostly I am anticipating that an extra minus sign floats around somewhere since now we'd have $\ell^{2} = -(\ell^{0})^2 + ||\boldsymbol{\ell}^{2}||$.

• Be wary when using equations from P&S as editions have errors detailed on their site. I spent hours questioning a result only to find it was because of an error in P&S. In this case, the integral is perfectly fine. – JamalS Mar 10 '17 at 23:00
• Really appreciate the heads up! I never knew that P&S has a website with errata, I'll have to check that out – QuantumEyedea Mar 11 '17 at 1:51

1. The Euclidean integral $$I_E(\Delta) ~:=~\int \! \frac{d^d\ell_E}{(2\pi)^d} \frac{1}{(\Delta+\ell_E^2)^n}~\stackrel{(A.44)}{=}~\frac{1}{(4\pi)^{d/2}}\frac{\Gamma\left(n-\frac{d}{2}\right)}{\Gamma(n)} \Delta^{\frac{d}{2}-n}\tag{E}$$ is integrable iff $$n>\frac{d}{2}.$$ See also the superficial degree of divergence.
2. For $n<\frac{d}{2}$ the Euclidean integral $I_E(\Delta)$ is declared in dimensional regularization to be equal to the rhs. of eq. (E) via analytic continuation.
3. The Minkowski integral in the $(\mp,\pm,\ldots,\pm)$ convention is defined via Wick rotation $$\ell_M^0~=~i\ell_E^0 \tag{A.43}$$ to be given by the Euclidean integral $$I_M(\Delta) ~:=~\int \! \frac{d^d\ell_M}{(2\pi)^d} \frac{1}{(\Delta\pm\ell_M^2)^n} ~:=~i I_E(\Delta),\tag{M}$$ respectively.
• In 3, you're saying that the integral has the same result regardless of which $(\pm,\mp,\ldots,\mp)$ is used? – QuantumEyedea Mar 11 '17 at 1:50