# 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

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Your question is related to the Gamma function (en.wikipedia.org/wiki/Gamma_function) at $-1$; but the product is meaningless, the Gamma function has a singularity there, and this all has been known for two centuries. – Charles Matthews Jun 21 '10 at 20:17
Ok, but isn't it pretty 'obvious' that the gamma function has a singularity there, as it's 'almost' equal to my divergent series, which goes into infinity for whatever x. It isn't that 'bad' that the function has a singularity there. – Max Muller Jun 21 '10 at 20:29
It's equal to zero because the sum of x/n pi is infinite. Use that 1 - y <= e^{-y} for positive y to find a proof. Also if you plug in x = pi into your product and multiply out a few terms it will be clear what's going on. This isn't really a mathoverflow kind of question IMO. – Michael Greenblatt Jun 21 '10 at 20:34
Yeh, I'm sorry, I was afraid of that already, but I I couldn't look it up somewhere easily... Sorry. – Max Muller Jun 21 '10 at 20:38

## 3 Answers

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 $$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$

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I should add that this is an example of a Weierstrass product en.wikipedia.org/wiki/Weierstrass_product which can be used to construct entire functions with any admissble set of zeros. – Robin Chapman Jun 21 '10 at 20:37
Ok, thanks, this is very useful, mister Chapman! – Max Muller Jun 21 '10 at 20:39

I would suggest the development of the Gamma function

$$1/\Gamma(z) = z e^{\gamma z}\ \Pi_{n=1}^\infty\ (1+{z\over n})\ e^{-{z\over n}}$$

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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.

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Thanks, castal, I always like references to more useful information. – Max Muller Jun 21 '10 at 21:48