I have been playing with the following function:
$$
f(x)=\frac{\pi x (1-x^2)}{\sin\pi x}\prod\_{k=2}^\infty \frac{\sin(\pi x/k)}{\pi x/k}
$$
It is hard to get correct numerical values. I'll start with the basic. Can you confirm that (1) we have absolute convergence, (2) the limit exists if $x$ is an integer, (3) if $x$ is a composite number then $f(x)=0$, and if $x$ is prime, $f(x)\neq 0$?


For simplicity, let's consider the absolute value of $f(x)$. Now the interesting part. If $x$ is not too close to a composite number, it sounds like
$$
|f(x)|\sim \exp(-\lambda x)
$$
for some $\lambda >0$, possibly $\lambda\approx 4.5$. Is there an asymptotic formula that can be easily derived? Since $f(x)=0$ if $x$ is composite, that formula would be valid only for some values of $x$, for instance if the fractional part is within some range, or if $x$ is prime, which seems to be where the formula works best.


If this was true, you could approximately compute the number of primes $<n$ as
$$\pi(n)\approx \sum\_{k=2}^n e^{\lambda k} |f(k)|.
$$
The formula is useless for computational purposes, but I am wondering if it might have some theoretical interest. Anyway, my question is this: can you get some asymptotic formula as $x\rightarrow\infty$ depending on the fractional part of $x$ or if $x$ is prime? Mine might not be correct. Even better, what is the value of $\lambda$ assuming it ever exists?



Curious infinite product, convergence, connection to prime numbers
I believe that the doubly infinite product $\prod \_{k=1} ^\infty \prod\_{m=1} ^\infty \left( 1 - \frac {x^2} {k^2 m^2} \right)$ is absolutely convergent and equals $f(x) \frac {\sin^2 (\pi x)} {(\pi x)^2 (1 - x^2)}$. Is that wrong? I am just using $\frac {\sin(\pi x)} {\pi x} = \prod\_{m=1} ^\infty \left(1 - \frac {x^2} {m^2} \right)$ here.


