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Could someone confirm me the validity of the following formula:

$$\zeta\left(z\right)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\left(\frac{e^{-\gamma \left(1-z\right)}}{1-z}\prod_{n=1}^{\infty}\frac{e^{\frac{1-z}{n}}}{1+\frac{1-z}{n}}\right)\zeta\left(1-z\right)$$

for the Riemann Zeta function accross the entire complex plane ?

EDIT: another non-related question: is the notation without parenthesis ambiguous for a mathematician? $$\zeta\left(z\right)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\frac{e^{-\gamma \left(1-z\right)}}{1-z}\prod_{n=1}^{\infty}\frac{e^{\frac{1-z}{n}}}{1+\frac{1-z}{n}}\zeta\left(1-z\right)$$

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  • $\begingroup$ The notation without parentheses is at least undesirable, because of the care that must be taken to ensure that a product converges. Since the $\zeta$ term doesn't involve the index $n$, it would be best for the reader if it were moved outside of the product, to the left. It is not strictly ambiguous. $\endgroup$ – Eric Tressler Jul 30 '13 at 0:15
  • $\begingroup$ Yes, it is true for entire complex plane. The term inside parenthesis is usually called $\Gamma(1-z)$. The functional equation is valid through out the complex plane in the sense that one has to take care of the poles of $\Gamma(1-z)$, which are neutralized by 'trivial zeros' of $\zeta(1-z)$ at negative odd integers. $\endgroup$ – Kunnysan Jul 30 '13 at 7:02

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