Question on $\Pi_{n=1}^\infty\left(1-\frac{x^a}{\pi^an^a}\right)$ and the Riemann Zeta function

Question

Question on what is

$$\Pi_{n=1}^\infty\left(1-\frac{x^a}{\pi^an^a}\right)$$

I deduced that for any $a=2,4,6,\dots$, the above product is simplify-able into a product of sines,

$$\Pi_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right)=\frac{\sin(x)}x$$

$$\Pi_{n=1}^\infty\left(1-\frac{x^4}{\pi^2n^2}\right)=\frac{\sin(x)\sin(xi)}{x^2}$$

And trying to put this into a general form, I got

$$f(x):=\frac{\sin(x)}{x}$$

$$\implies f(x)=f(-x)$$

$$f(x)f(ix)=\Pi_{n=1}^\infty\left(1-\frac{x^4}{\pi^4n^4}\right)=\sqrt{f(xe^{\frac{\pi i}{2}})f(xe^{\pi i})f(xe^{\frac{3\pi i}{2}})f(xe^{2\pi i})}$$

Using $(a+b)(a-b)=a^2-b^2$, we can get something along the lines of

$$\Pi_{n=1}^\infty\left(1-\frac{x^{2^a}}{\pi^{2^a}n^{2^a}}\right)=\sqrt{f(xe^{\frac{1\pi i}{2^a}})f(xe^{\frac{2\pi i}{2^a}})\dots f(xe^{\frac{2^a\pi i}{2^a}})}=\Pi_{n=1}^{2^a}\sqrt{f(xe^{\frac{n\pi i}{2^a}})}$$

$$\implies\Pi_{n=1}^\infty\left(1-\frac{x^a}{\pi^an^a}\right)=\Pi_{n=1}^a\sqrt{f(xe^{\frac{n\pi i}a})}$$

I was wondering if this is correct and if it were possible to... take the Taylor Series of the RHS? Specifically for $a=3$, as this would result in relations to the Riemann Zeta function.

Answer 1

You are correct that for even $a$ you can write the infinite product as finite product over $\text{sinc}$ functions. This result does not extend to odd $a$ so your final equation is not true. I will here add a simple derivation of the identity for even $a$ and some notes about a related identity involving the $\Gamma$ function.

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If $\omega$ is a $n$'th root of unity then $1 - z^{n} = \prod_{k=0}^{n-1} (1 - z\omega^{k})$ since both sides of the equation are polynomials that agree at (the non-root) $z=0$ and have the same $n$ roots. Taking the product of this identity for $z = \frac{x^2}{\pi^2 m^2}$ over $m=1,2,3\ldots,N$ and then taking $N\to\infty$ we get

$$\prod_{m=0}^\infty\left(1 - \frac{x^{2n}}{\pi^{2n}m^{2n}}\right) = \prod_{k=0}^{n-1} \left[ \prod_{m=0}^\infty\left(1 - \frac{x^2 e^{\frac{2\pi i k}{n}}}{\pi^{2}m^2}\right)\right] = \prod_{k=0}^{n-1} \text{sinc}\left(e^{\frac{\pi i k}{n}}x\right)\tag{1}$$

for all $n\in\mathbb{N}$. The case you are interested in corresponds to $2n = 3$ or $n = \frac{3}{2}$ which is not covered by this derivation and neither does the identity above hold true for non-integer $n$.

There is a similar identity that allows for odd exponents in the infinite product, though involving more complicated functions, namely:

$$\prod_{m=0}^\infty\left(1 - \frac{x^{n}}{\pi^{n}m^{n}}\right) = \prod_{k=0}^{n-1} \frac{1}{\Gamma\left(1 - \frac{(-1)^ke^{\frac{\pi i k}{n}}x}{\pi}\right)}\tag{2}$$

where $\Gamma$ is the gamma-function. When $n$ is even Euler's reflection formula, $\frac{1}{\Gamma(1+z)\Gamma(1-z)} = \text{sinc}(z)$, can be applied to simplify the expression above into the form $(1)$ (combind the $k$'th term and the $2n-k$'th term in the product above). The disadvantage of this formula in the study of the infinite product is that the $\Gamma$ function for imaginary aguments is not very pleasant to work with.

Answer 2

Well, if you are seeking the relationship of this general product to the Riemann Zeta, there is an easy way to show it:

$$P(a,x)=\prod_{n=1}^\infty \left(1-\frac{x^a}{\pi^a n^a} \right)$$

$$\ln P(a,x)=\sum_{n=1}^\infty \ln \left(1-\frac{x^a}{\pi^a n^a} \right)=-\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{x^{ak}}{k~\pi^{ak} n^{ak}}$$

Note that $|x| \leq \pi$ for the latter series to converge. By the main definition of the zeta function, we have:

$$\ln P(a,x)=-\sum_{k=1}^\infty \frac{x^{ak}}{k~\pi^{ak}} \zeta (ak)$$