Okay, we know that 


$$ \frac{sin(x)}{x} = \prod\_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\cdot\pi^2}\Big) $$ .


Is there some known (trigonometric(?)) function that is equal to the following infinite product? 


$$ \prod\_{n=1}^{\infty} \Big(1-\frac{x}{n\cdot\pi}\Big) $$


I'd be happy as well if someone could provide me with a function that is equal to a similar divergent infinite product (a function, for example, that is equal to 'my' inifite product, only $\pi=1$, or $x=x^2$, or something in that direction). 


Thanks in advance,


Max Muller



Closed form of divergent infinite product?
It's a **divergent** infinite product. You might as well ask for the sum of
$$\sum\_{n=1}^\infty\frac{x}{n\pi}.$$
You can "cure" the divergence by multipliying each term by a suitable factor, so
for instance
$$f(x)=\prod\_{n=1}^\infty e^{x/n\pi}\left(1-\frac{x}{n\pi}\right)$$
does converge (as the $n$-th term is like $\exp(x^2/2n^2\pi^2)$). You can
express this in terms of the gamma function which satisfies
$$\frac1{\Gamma(x)}=x e^{\gamma x}\prod\_{n=1}^\infty
e^{-x/n}\left(1+\frac{x}{n}\right).$$
By using the identity
$$f(x)f(-x)=\prod\_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)$$
one can deduce the identity
$$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$


