# About infinite products and Euler Gamma functions [closed]

I am interested in knowing how to calculate infinite products like (or reading any reference about it):

$$\prod_{j=1}^{\infty}\left( 1-\left( \frac{x}{a+j\pi} \right) ^2 \right)$$

Inserting it into a Mathematica worksheet (Wolfram research), it returns the following beautiful formula:

$$\frac{\pi^2\Gamma(\frac{\pi+a}{\pi})^2}{\Gamma(\frac{a-x}{\pi})\Gamma(\frac{a+x}{\pi})}$$

where $$\Gamma(x)$$ is the Euler's Gamma function, and $$x$$ and $$a$$ are positive real numbers.

Gustavo

## closed as unclear what you're asking by user44191, Sean Lawton, Stefan Waldmann, Jan-Christoph Schlage-Puchta, David HandelmanMar 16 at 23:53

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• Improve your $\LaTeX$ formulas. – user64494 Mar 13 at 19:00
• Weierstrass products – reuns Mar 13 at 19:08
• According to Maple, you left out a factor $a^2 - x^2$ in the denominator. – Robert Israel Mar 13 at 23:11
• In general there is no method to give closed formulae for infinite sums or products. The reason is simply that most expressions involving limits do not have any easier expression. So in a way every closed formula is a lucky accident. – Jan-Christoph Schlage-Puchta Mar 15 at 18:49

The Weierstrass product identity $$\frac{1}{\Gamma(z)} = e^{\gamma z} z \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right) e^{-z/n}$$ implies $$\frac{\Gamma(s)^2}{\Gamma(s-z) \Gamma(s+z)} = \frac{s^2-z^2}{s^2} \prod_{n=1}^\infty \left(1 - \frac{z^2}{(n+s)^2}\right)$$ (valid wherever you don't run into a division by $$0$$ or a pole of $$\Gamma$$). You're essentially looking at the case $$s = a/\pi$$, $$z = x/\pi$$.