So I am directed to use the Weierstrass formula for the Gamma function and then follow this by the reflection formula, this will allow me to find an expression for $f(z)$ so that

$$ sin(z\pi) = f(z)\prod^{\infty}_{n=1} \Bigr(1- \frac{z^2}{n^2}\Bigr) $$

is valid for any number $ z\not\in \mathbb{Z}$.

So the Weierstrass formula is

$$\frac{1}{\Gamma(z)} = ze^{z\gamma} \prod^{+\infty}_{n=1} \Bigr(\Bigr( 1+\frac{z}{n}\Bigr) exp\Bigr(-\frac{z}{n}\Bigr) $$

where $\gamma$ is Euler's constant.

Also the reflection formula is

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

In all honesty I have no idea what I'm supposed to do or where to begin, generally I post my own attempts but for this question I am not even sure what to do.

Any help, hints, directions are greatly appreciated!


The reflection formula is $\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ (the $\Gamma$ function does not vanish anywhere), implying $\frac{\sin(\pi z)}{\pi z}=\frac{1}{\Gamma(1+z)\,\Gamma(1-z)}$. If you know that $$ \frac{1}{\Gamma(1+z)}=e^{\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right) e^{-z/n} \tag{1}$$ by replacing $z$ with $-z$ you also know that $$ \frac{1}{\Gamma(1-z)}=e^{-\gamma z}\prod_{n\geq 1}\left(1-\frac{z}{n}\right) e^{z/n} \tag{2}$$ and by multiplying $(1)$ and $(2)$ $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{3} $$ i.e. $f(z)=\pi z$.

  • $\begingroup$ Thank you ! This makes a lot of sense.. I will edit my mistake with the reflection formula :) $\endgroup$ – Evan Jan 29 '18 at 16:26

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