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Regularization is a small deformation of the theory in a scale that is far from your concerns (see this). By the Renormalization Group we understand that at IR scale, some UV deformations are irrelevant. The analytical continuation like $n^{-s}$ when $s$ is small affects more large $n$ values than the small $n$. This deformation is an smooth one through all ...


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Thanks to Qmechanic I got convinced that maybe it's better to compute that product from scratch: $$ P_{a,b} = \frac{1}{b}\prod_{n=0}^\infty(an+b) = b^{-1}\exp\left\{\sum_{n=0}^\infty\log(an+b) \right\}; $$ so the problem is to evaluate the infinite sum in term fo Hurwitz zeta function: $$ \zeta(s;z) := \sum_{n=0}^\infty(n+z)^{-s}, $$ and its analytic ...


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In this answer, we give a heuristic explanation for the formula (18) in Ref. 1. Consider two zeta-function regularized infinite products $$\tag{1a} F_a(b)~:=~\prod_{\lambda \in \mathbb{N}+b} a~=~a^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= ...


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Trying to answer my old question myself: The cited piece is talking about an infrared cutoff, which basically amounts to giving the propagating particle a mass which appears squared in a propagator. So I think, what is meant is considering $$\lim_{\Lambda\to 0}\int \frac{d^4p}{p^2+\Lambda}$$ and this $\Lambda$ of course has a mass dimension of 2.


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The first thing that you need to do is put the integral variable $p$ in a more clear way. You can doo this with Feynman Parameters. $$ \frac{1}{AB}=\int_{0}^{1} dx \frac{1}{[A+(B-A)x]^2} $$ I will give you an exercise: find the $A$ and $B$, as well as $\Delta$ and $C$, such that: $$ ...


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(This argument is for a one-dimensional system, but similar arguments can be given in higher dimensions. We work in units with $c = \hbar = e = 1$). Suppose we have some regulator procedure parameterized by a momentum cutoff $\Lambda$. Then, for distance $L$ between two parallel plates, we can expand the regularized energy sum in powers of the cutoff as ...



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