Zeta regularization of Infinite product

Question

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{-1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Quine, Heydari and Song 1993 stating $P_{1,b}$ as mine but

$$ P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma \left( 1+\frac{b}{a}\right )}. \tag{18} $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

Answer 1

In this answer, we give a heuristic explanation for the formula (18) in Ref. 1.

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 $\Lambda= \mathbb{N}+b$. Here $a\in\mathbb{C}\backslash\{0\}$ and $b\in\mathbb{C}$ are two complex numbers.

2. Notice that if we put $b=0$, we get the well-known zeta function regularization formulas $$\tag{2a} F_a(b\!=\!0)~:=~\prod_{n \in \mathbb{N}} a~=~a^{-\frac{1}{2}} $$ and $$\tag{2b} G(b\!=\!0)~:=~\prod_{n \in \mathbb{N}}n~=~\sqrt{2\pi}. $$

3. Using OP's argument, one would falsely have expected that the infinite product (1a) should be independent of the shift $b\in\mathbb{C}$. Here we will employ another kind of logic. If we shift the half-lattice $\Lambda= \mathbb{N}+b$ by one unit $b\to b-1$, we would expect one more element in the half-lattice, and thereby one more factor in the infinite product. This leads to the following very powerful functional equations/recursion relations, $$\tag{3a} F_a(b\!-\!1)~=~a ~ F_a(b) $$ and $$\tag{3b} G(b\!-\!1)~=~b ~ G(b). $$ Note in particular that eqs. (3a) and (3b) are in fact satisfied by the zeta function regularization (1a) and (1b), respectively!

4. Therefore we naturally arrive at the regularized product formula (18) in Ref. 1, $$\prod_{n \in \mathbb{N}}(an+b) ~=~\prod_{\lambda \in \mathbb{N}+\frac{b}{a}} a\lambda ~=~\left[\prod_{\lambda \in \mathbb{N}+\frac{b}{a}} a\right]\left[\prod_{\lambda \in \mathbb{N}+\frac{b}{a}}\lambda\right]$$ $$\tag{18}~=~F_a\left(\frac{b}{a}\right)G\left(\frac{b}{a}\right) ~=~a^{-\frac{1}{2}-\frac{b}{a}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{b}{a}+1\right)}.$$

References:

1. J.R. Quine, S.H. Heydari and R.Y. Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993) 213; eq. (18).

Answer 2

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 continuation. We define

$$\tag{*} \hat\zeta(s;a,b) := \sum_{n=0}^\infty(an+b)^{-s} = a^{-s}\zeta(s,b/a). $$

we note that

$$ \partial_s\hat\zeta(0;a,b) = -\sum_{n=0}^\infty\log(an+b), $$

so that we will have

$$ P_{a,b} = b^{-1}\exp\left\{-\partial_s\hat\zeta(0;a,b)\right\}. $$

But the derivative of $\hat\zeta$ can be evaluated using (*):

$$ \partial_s\hat\zeta(0;a,b) = \left(\frac{1}{2}-\frac{b}{a}\right)\log a -\frac{1}{2}\log 2\pi + \log\Gamma\left(\frac{b}{a}\right); $$

where we used the well-know identities

$$ \zeta(0;z) = \frac{1}{2}-z, \qquad\qquad\qquad \partial_s\zeta(0;z) = \log\Gamma(z) - \frac{1}{2}\log 2\pi. $$

Combining all together we arrive at the conclusion:

$$ P_{a,b} = a^{-\frac{1}{2}-\frac{b}{a}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{b}{a}+1\right)}. $$

Answer 3

I will complement the current answer by pointing out what the incorrect step was in the question. I was going to write it as a comment but it was too long.

The mistake in the computation is the erroneous equation $$\prod_{n=1}^\infty a\left(n+\frac{b}{a}\right)=\prod_{n=1}^\infty a\prod_{n=1}^\infty \left(n+\frac{b}{a}\right).$$ This is not correct in general because in infinite dimensions we do not have $\det(AB)=\det(A)\det(B)$. In fact, the right hand side of the equation is ill-defined. Even if one were to accept that $\prod_{n=1}^\infty a=1/\sqrt{a}$, which is not true, even within the realm of $\zeta$-function regularization, the right hand side would yield $$a^{-1/2}\prod_{n=1}^\infty \left(n+\frac{b}{a}\right).$$ But the correct thing is what is obtained by using (1) in J. R. Quine, S. H. Heydari, and R. Y. Song, “Zeta regularized products,” Trans. Amer. Math. Soc., vol. 338, no. 1, pp. 213–231, 1993, doi: 10.1090/S0002-9947-1993-1100699-1 $$\prod_{n=1}^\infty a\left(n+\frac{b}{a}\right)=a^{Z(0)}\prod_{n=1}^\infty \left(n+\frac{b}{a}\right).$$ In other words, taking out constants from infinite products depends on the rest of the elements in the product. In this case, we simply have $$Z(s)=\zeta(s,b/a)-\left(\frac{b}{a}\right)^{-s}\rightarrow Z(0)=\frac{1}{2}-\frac{b}{a}-1=-\frac{1}{2}-\frac{b}{a}$$ at $s=0$. We conclude the correct equation $$\prod_{n=1}^\infty a\left(n+\frac{b}{a}\right)=a^{-\frac{1}{2}-\frac{b}{a}}P_{1,b/a}.$$