$\newcommand{\sinc}{\operatorname{sinc}}$ Throughout, let $m,k$ be positive integers, $x>0$ a real number, and denote $\sinc(z)=\sin(z)/z$ with $\sinc(0)=1$.

A famous result of Euler gives $\sinc(x)$ as an infinite product: $$ \prod_{k=1}^{\infty}\cos\left( 2^{-k}x\right)=\sinc(x) $$Less well-known (although mentioned in Mathematica's documentation for Product) but in a similar vein is $$ \prod_{k=1}^{\infty}1-\frac{4}{3}\sin^2(3^{-k}x) = \sinc(x) $$In fact, each of these are special cases of a more general formula that I found (unfortunately after looking at this question): $$ \prod_{k=1}^{\infty} \frac{1}{m} \csc(m^{-k} x)\sin(m^{1-k} x) = \sinc(x) $$The product term reduces to nice sums depending on the parity of $m$. Further, for even $m$ the product telescopes by double-angle and the result is immediate. Note that the right-hand side does not depend on $m$.

My question: what would we get if we switched the roles of $m,k$ in the product? That is, what is the nature of $$ \prod_{m=1}^{\infty} \frac{1}{m} \csc(m^{-k} x)\sin(m^{1-k} x) =S_k(x) $$

The cases $k=1$ and $k\ge 2$ need to be treated separately. For $k=1$, the product term is $\frac{1}{m}\sin (x) \csc \left(\frac{x}{m}\right)$, which approaches $\sinc(x)$ as $m\to \infty$. Thus $S_1(x)=\delta_0(x)$, as for $x\ne 0$ the product diverges to zero. For $k\ge 2$, note that $$ \lim_{m\to\infty} m^{2 k-2} \log\left(\frac{\sin \left(x m^{1-k}\right) \csc \left(x m^{-k}\right)}{m}\right) = \frac{-x^2}{6}, $$implying the product converges by comparison with the corresponding series $\sum_{m\ge 1} m^{2-2k}$. Below are pictures of estimates for the $100^{th}$ partial products, for $k=2,\ldots, 6$ and $-6\pi\le x\le 6\pi$.

enter image description here

Experience has taught me that a closed-form is unlikely but I would nevertheless like to know how $S_k(x)$ depends on $k$, in particular if they are a family of sinc functions as well.

  • $\begingroup$ Probably there is no general closed form, unless the function you made up counts. I tried the logarithm product to sum of logarithms trick with no luck. $\endgroup$ Jun 6, 2021 at 17:50
  • $\begingroup$ If you accept to make me a favor, could you plot the functions I gave in my answer to compare with your. Being blind I am unable to produce decent plots. Thanks and cheers. $\endgroup$ Jan 16, 2022 at 9:39

1 Answer 1


Using very truncated series $$a_m=\frac1m{\sin \left(x m^{1-k}\right) \csc \left(x m^{-k}\right)}$$ $$a_m=1-\frac{1}{6} \left(m^2-1\right) m^{-2 k}x^2+\frac{1}{360} \left(3 m^4-10 m^2+7\right) m^{-4 k}x^4+O\left(x^6\right)$$

Taking logarithms and expanding again $$\log(a_m)=-\frac{1}{6} \left(m^2-1\right) m^{-2 k}x^2-\frac{1}{180} \left(m^4-1\right) m^{-4 k}x^4+O\left(x^6\right)$$ $$\sum_{m=1}^\infty \log(a_m)=-\frac{\zeta (2 k-2)-\zeta (2k)}{6} x^2 -\frac{\zeta (4 k-4)-\zeta (4 k)}{180} x^4 +O\left(x^6\right)$$ $$\color{red}{\prod_{m=1}^\infty a_m \sim \exp\Bigg[-\frac{\zeta (2 k-2)-\zeta (2k)}{6} x^2 -\frac{\zeta (4 k-4)-\zeta (4 k)}{180} x^4+\cdots \Bigg]}$$ This produces quite well the shape of your curves which in fact are gaussian.


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