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I only recently heard about the concept of $\zeta$-regularization, which allows the evaluation of things like

$$\infty != \prod_{k=1}^\infty k = \sqrt{2\pi}$$

and

$$\infty \# = \prod_{k=1}^\infty p_k = 4\pi^2$$

where $n\#$ is a primorial, and $p_k$ is the $k$-th prime. (The expression for the infinite product of primes is proven here.) That got me wondering if, given a sequence of positive integers $m_k$ (e.g. the Fibonacci numbers or the central binomial coefficients), it is always possible to evaluate the infinite product

$$\prod_{k=1}^\infty m_k$$

in the $\zeta$-regularized sense. It would seem that this would require studying the convergence and the possibility of analytically continuing the corresponding Dirichlet series, but I am not too well-versed at these things. If such a regularization is not always possible, what restrictions should be imposed on the $m_k$ for a regularized product to exist?

I'd love to read up on references for this subject. Thank you!

Timmy Turner
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  • How did you get $\prod k =\sqrt{2\pi}$? – Fabian Aug 02 '12 at 09:54
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    @Fabian, if $\zeta^\prime(s)=-\sum\limits_{j=1}^\infty \frac{\log,k}{k^s}$, then consider $\exp(-\zeta^\prime(0))$... – Timmy Turner Aug 02 '12 at 09:57
  • a free online preprint about the second equation is at http://cds.cern.ch/record/630829 (2013-05-13) – Gottfried Helms May 13 '13 at 17:50
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    I don't know if it helps or if it is obvious, but via exponentiation the problem is exactly the same as a classic series. – geodude Aug 24 '13 at 11:31
  • It seems to me that a physical interpretation of regularization would be to "measure" an object as it passes through the entrance to a black hole at a velocity asymptotically approaching the speed of light. – frogfanitw Jan 20 '16 at 05:12
  • @TimmyTurner It very much bothers me that $(-1/2)!=\Gamma(1/2)=\sqrt{2\pi}$, and $\sum_{n=1}^{-1/2}1=-1/2$. I almost feel like some divergent series have a strange "plug in -1/2" type of nature. Think it is question worthy? – Simply Beautiful Art Jul 27 '16 at 00:14
  • It looks like an analog of Ramanujan summation to me: $\infty!=\Pi_{k=1}^\infty k=\exp\Sigma_{k=1}^\infty\log k$. https://en.wikipedia.org/wiki/Ramanujan_summation – Otherwise Oct 22 '16 at 17:20
  • http://math.stackexchange.com/questions/1325169/zeta-regulated-product-solving-without-the-zeta-function?rq=1 My own shot about the subject without any background knowlegde cause i was just as curious. I know i should add some constrains for the methode. – Gerben Nov 28 '16 at 03:20
  • To me, this looks wrong: definitely, $\operatorname{reg}\sum_{k=1}^\infty \ln k=\ln \sqrt{2\pi}$. But exponent of regularized value is not the same as regularized value of exponent! – Anixx Jul 05 '22 at 14:16

2 Answers2

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Given an increasing sequence $0<\lambda_1<\lambda_2<\lambda_3<\ldots$ one defines the regularized infinite product $$ \prod^{\infty}_{n=1}\lambda_n=\exp\left(-\zeta'_{\lambda}(0)\right), $$ where $\zeta_{\lambda}$ is the zeta function associated to the sequence $(\lambda_n)$, $$ \zeta_{\lambda}(s)=\sum^{\infty}_{n=1}\lambda_n^{-s}. $$ (See the paper: E.Munoz Garcia and R.Perez-Marco."The Product over all Primes is $4\pi^2$". http://cds.cern.ch/record/630829/files/sis-2003-264.pdf )

In the paper( https://arxiv.org/ftp/arxiv/papers/0903/0903.4883.pdf ) I have evaluated the $$ \prod_{p-primes}p^{\log p}=\prod_{p-primes}e^{\log^2p}=\exp\left(24\zeta''(0)+12\log^2(2\pi)\right) $$ where $\zeta''(0)$ is the second derivative of Riemann's Zeta function in $0$.

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The claim that $\operatorname{reg} \prod_{k=1}^\infty k = \sqrt{2\pi}$ looks misleading and wrong to me.

Definitely, $\prod_{k=1}^\infty k=\exp \sum_{k=1}^\infty \ln k$.

Also, definitely $\operatorname{reg}\sum_{k=1}^\infty \ln k=\ln \sqrt{2\pi}$

But exponent of the regularized value is not the same as the regularized value of exponent!

More honestly, the value $\sqrt{2\pi}$ corresponds not to the regularized (finite) value of the product, but to its hypermodulus as defined here. Hypermodulus is the exponent of the scalar finite part of the logarithm of the object.

Anixx
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