Is there a closed-form expression for $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$?

Question

I would like to ask if for $|x|<1$, we can express the product $\prod_{n=1}^{\infty}(1-\frac{x}{n^3})$ as a function $f(x)$. I tried to use Weierstrass factorization theorem, but without much success.

I would really appreciate a reference or a solution.

Answer 1

As already noticed in comments, the expression can be obtained from the infinite products for $\Gamma$ (either Euler's one, or Weierstrass's one): $$\Gamma(1+z)=\prod_{n=1}^\infty\frac{(1+1/n)^z}{1+z/n}=e^{-\gamma z}\prod_{n=1}^\infty\frac{e^{z/n}}{1+z/n},$$ and the "algebraic" $1-x/n^3=(1-x^{1/3}/n)(1-e^{2\pi i/3}x^{1/3}/n)(1-e^{-2\pi i/3}x^{1/3}/n)$, giving $$\prod_{n=1}^\infty\left(1-\frac{x}{n^3}\right)=\Big(\Gamma(1-x^{1/3})\Gamma(1-e^{2\pi i/3}x^{1/3})\Gamma(1-e^{-2\pi i/3}x^{1/3})\Big)^{-1}.$$ This easily applies to more general "rational infinite products", as outlined here.

Answer 2

Comment:

The bound of this product can be found usig Weierstrassn inequality:

If $a_1, a_2, a_3, \ldots,a_n$ are real positive integers less than unity, and:

$S_n=(a_1+a_2+a_3+ \cdots+a_n)<1$

then:

$1-S_n<(1-a_1)(1-a_2)(1-a_3) \cdots (1-a_n)<\frac 1 {1+S_n}$

Where we can let:

$a_n=\frac x {n^3}$