# Why do people rule out zeta regularization for renormalization?

Using zeta regularization one can get a formula for regularizing the integral $\int_{a}^{\infty}x^{m-s}\text dx$ for any $m$.

However, I have not seen anywhere. For example, I do not know why in physics you can use zeta regularization to define infinite products $\prod _{n=0}^{\infty}a_{n}$, but they do not use zeta regularization for the regularization of divergent integrals like $\int_{a}^{\infty}x^{m-s}dx$ for any $m$ even though there is a known formula

$$\begin{array}{l} \int\nolimits_{a}^{\infty }x^{m-s} \text dx =&\frac{m-s}{2} \int\nolimits_{a}^{\infty }x^{m-1-s} \text dx +\zeta (s-m)-\sum\limits_{i=1}^{a}i^{m-s} +a^{m-s} \\ &-\sum\limits_{r=1}^{\infty }\frac{B_{2r} \Gamma (m-s+1)}{(2r)!\Gamma (m-2r+2-s)} (m-2r+1-s)\int\nolimits_{a}^{\infty }x^{m-2r-s} \text dx, \end{array}$$

which I think is better than Pauli-Villars or dimensional regularization. Why do people ignore it? Is there a plot against zeta regularization?

• This question does not seem to have any physics content. People don't rule out zeta function regularization. It is often used in one-loop calculations and whole books have been written about it. However, it is awkward to use at higher loops which prevents it's mainstream use in large scale calculations. – Simon Oct 29 '13 at 0:27