-2
$\begingroup$

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.

Thanks in advance,

Gustavo

$\endgroup$
4
  • 2
    $\begingroup$ Improve your $\LaTeX$ formulas. $\endgroup$
    – user64494
    Mar 13, 2019 at 19:00
  • 3
    $\begingroup$ Weierstrass products $\endgroup$
    – reuns
    Mar 13, 2019 at 19:08
  • $\begingroup$ According to Maple, you left out a factor $a^2 - x^2$ in the denominator. $\endgroup$ Mar 13, 2019 at 23:11
  • $\begingroup$ 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. $\endgroup$ Mar 15, 2019 at 18:49

1 Answer 1

3
$\begingroup$

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 $s = a/\pi$, $z = x/\pi$.

$\endgroup$
2

Not the answer you're looking for? Browse other questions tagged or ask your own question.