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)

I have two questions about this integral:

  • 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 $\ell^{2} = -(\ell^{0})^2 + ||\boldsymbol{\ell}^{2}||$.

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    $\begingroup$ 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. $\endgroup$ – JamalS Mar 10 '17 at 23:00
  • $\begingroup$ Really appreciate the heads up! I never knew that P&S has a website with errata, I'll have to check that out $\endgroup$ – 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.

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  • $\begingroup$ In 3, you're saying that the integral has the same result regardless of which $(\pm,\mp,\ldots,\mp)$ is used? $\endgroup$ – QuantumEyedea Mar 11 '17 at 1:50
  • $\begingroup$ Yes, for the way that I've defined it. $\endgroup$ – Qmechanic Mar 11 '17 at 6:23

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