Closed form of divergent infinite product? 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
 A: 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
$$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$
A: I would suggest the development of the Gamma function
$$1/\Gamma(z) = z e^{\gamma z}\ \Pi_{n=1}^\infty\ (1+{z\over n})\  e^{-{z\over n}}$$
A: Take a look at the first dozen pages of Andrews and Askey, which you can read online - http://books.google.com/books?id=nMm13WXpLt8C&lpg=PP1&dq=andrews%20askey&pg=PA1#v=onepage&q&f=false
Already on page 3, they give the product representation of 1/Gamma, which is essentially your function, modified to make it convergent.
On page 10, they treat the reflection formula, which shows that 1/Gamma is "half of the sine function", i.e it contributes the zeros on the negative x axis.
