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I want to compute the following identity

$ \prod_{n=-\infty}^{\infty}(n+a) = a \prod_{n=1}^{\infty} (-n^2)(1- \frac{a^2}{n^2}) = 2 i \sin(\pi a) $

It seems strange this identity holds to me. Can anyone gives some explict procedure of this identity?


I found that this procedure is computed in the $\zeta$-function regularization which states \begin{align} \prod_{n=1}^{\infty} a = a^{\zeta(0)} = a^{-\frac{1}{2}}, \quad \prod_{n=-\infty}^{\infty} a = a^{2 \zeta(0) +1} =1, \quad \prod_{n=1}^{\infty} n^{\alpha} = e^{-\alpha \zeta'(0)} = (2\pi)^{\frac{\alpha}{2}} \end{align}


Using $\frac{\sin(\pi a)}{\pi a} = \prod_{n=1}^{\infty} (1-\frac{a^2}{n^2})$ and from $\zeta$-reguralization $\prod_{n=1}^{\infty}n^2 = 2\pi$

i obtain \begin{align} a\prod_{n=1}^{\infty} (n^2) (1-\frac{a^2}{n^2})=2 \sin(\pi a) \end{align}

But i don't know where factor $i$ comes from. i guess from ($-n^2$) the i factors turns out, but i am not sure.

Also I want to know how to obtain

\begin{align} \prod_{n=-\infty}^{\infty}(n+a) = a \prod_{n=1}^{\infty} (-n^2)(1- \frac{a^2}{n^2}) \end{align}

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  • $\begingroup$ Possibly math.stackexchange.com/questions/447011/… is what you are looking for? $\endgroup$ – sciona Jan 22 '15 at 6:18
  • $\begingroup$ @sciona, it seems similar process, but some factors are different.$i.e$ additional $(-n^2)$ is mulitplyied above $\endgroup$ – phy_math Jan 22 '15 at 6:19
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Your identity isn't right. We can compute the zeta regularized value for the two infinite products in your identity and in general,

$$\prod_{n=-\infty}^\infty (n+a) \quad\ne\quad a \prod_{n=1}^\infty (-n^2) \prod_{n=1}^\infty\left(1 - \frac{a^2}{n^2}\right) \quad\ne\quad 2 i\sin(\pi a)$$

In certain sense, the $i$ in the last term comes from $$\prod_{n=1}^\infty(-1)^n = (-1)^{-\frac12} = \pm i$$ In the RHS, which sign of $i$ you get depends on how you interpret $-n$ when you throw it to the machine of zeta function regularization. It in turn depends on the imaginary part of $a$.
Another thing is the factorization

$$\prod_{n=-\infty}^\infty (n+a) \stackrel{?}{=} a \prod_{n=1}^\infty (-n^2) \prod_{n=1}^\infty\left(1 - \frac{a^2}{n^2}\right)\tag{*1}$$

is not $100\%$ legal. When you juggle infinite terms like this, it misses another phase factor which depends on $a$. Instead of $\displaystyle\;\prod_{n=1}^\infty(-n^2)\;$, let us just compute the zeta regularized value of LHS of $(*1)$ directly:

$$\prod_{n=-\infty}^\infty (n+a) = a\prod_{n=1}^{\infty}(-n+a)(n+a)\tag{*2}$$

Let's consider the case $\Im \alpha > 0$. Notice

$$0 < \arg(x + a) < \pi,\quad\forall x \in \mathbb{R}$$

We need to interpret $-n + a$ as $e^{\pi i}(n - a)$ when we throw it to the machine.
The zeta function corresponding to the infinite product $(*2)$ is given by:

$$Z(s) \stackrel{def}{=} a^{-s} + \sum_{n=1}^\infty \left[ \frac{1}{ e^{\pi i s}(n - a)^s} + \frac{1}{(n + a)^s}\right] = a^{-s} + e^{-\pi i s} \zeta(s,1-a) + \zeta(s,1+a) $$ where $\displaystyle\;\zeta(s,t) = \sum_{n=0}^\infty \frac{1}{(n+t)^s}\;$ is the Hurwitz zeta function.

By definition, the zeta regularized value of the product $(*2)$ is

$$\left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta \stackrel{def}{=} e^{-Z'(0)} = a \exp\left[\pi i \zeta(0,1-a) - \zeta'(0,1-a) - \zeta'(0,1+a)\right]$$

Recall following relations for Hurwitz zeta function and Gamma function:

$$\begin{align} \zeta(0,t) &= \frac12 - t\\ \left.\frac{\partial}{\partial s}\zeta(s,t)\right|_{s=0} &= \log\Gamma(t) - \frac12\log(2\pi)\\ \Gamma(1+a) &= a \Gamma(a)\\ \Gamma(a)\Gamma(1-a) &= \frac{\pi}{\sin\pi a} \end{align}$$

We find when $\Im a > 0$, $$ \left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta = \frac{2\pi a e^{(a - \frac12) \pi i}}{\Gamma(1-a)\Gamma(1+a)} = 2e^{(a - \frac12) \pi i}\sin\pi a $$ If one repeat the argument for $\Im a < 0$, we need to interpret $-n + a$ as $e^{-\pi i}(n - a)$ and we will pick out a different phase factor $e^{-(a - \frac12) \pi i}$. Combine these, we have:

$$\left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta = 2e^{\epsilon(a - \frac12) \pi i}\sin(\pi a) \quad\text{ where }\quad \epsilon = \begin{cases} +1,& \Im a > 0\\ ???,& \Im a = 0\\ -1,& \Im a < 0 \end{cases}$$

As you can see, this is in general different from the last term $2i\sin(\pi a)$ in your expression.

If we do the same thing to the product $\displaystyle\;\prod_{n=1}^\infty(-n^2)\;$ in RHS of $(*1)$, we find

$$\left[ \prod_{n=1}^\infty (-n^2) \right]_\zeta = 2\pi e^{-\epsilon\frac{\pi}{2} i} \quad\text{ when }\quad "-n" \text{ is interpreted as } e^{\epsilon \pi i} n. $$ Together with the identity $$\prod_{n=1}^\infty \left(1 - \frac{a^2}{n^2}\right) = \frac{\sin\pi a}{\pi a}$$ The RHS of $(*1)$ becomes $2 e^{-\epsilon\frac{\pi}{2} i}\sin(\pi a)$ and differs from LHS of $(*1)$ by a phase factor $e^{\epsilon a \pi i}$.

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  • $\begingroup$ Wow, great!. Can you recommend any relevant materials on zeta regularization? $\endgroup$ – phy_math Jan 22 '15 at 18:06
  • $\begingroup$ @phy_math I don't really know this stuff. I just look-up the definition and carry out the computation. The definition itself isn't that complicated. In any event, I learn this stuff from following paper: + J.R.Quine, S.H.Hyedari and R.Y.Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993), 213-231 $\endgroup$ – achille hui Jan 22 '15 at 22:30
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I do not think that the identity is corret because if you use the infinite product expansion of $\sin(\cdot)$, you get $$\sin (\pi a)=a\prod_{n=1}^\infty\left(1-\frac{a^2}{n^2}\right)$$ This is due to Euler. If your identity is correct then $2i=\prod_{n=1}^\infty (-n^2)$, which I am not sure about.

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  • $\begingroup$ I found that this procedure is computed in the $\zeta$-function regularization. $\endgroup$ – phy_math Jan 22 '15 at 6:29
  • $\begingroup$ Ok, I do not know about that, so maybe my last statement is not correct. I am editing that. $\endgroup$ – Samrat Mukhopadhyay Jan 22 '15 at 6:30

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