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I want to derive Euler's infinite product formula

$$\displaystyle \sin(\pi z) = \pi z \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right)$$

by using Euler's reflection equation $\Gamma(z)\Gamma(1-z) \sin(\pi z) = \pi$ and the definition of $\Gamma(z)$ as an infinite product, namely

$$\displaystyle \Gamma(z) := \frac{1}{z} \prod_{k=1}^\infty \frac{(1+\frac{1}{k})^z}{1+\frac{z}{k}}.$$

To be precise, I obtain that

$$\sin(\pi z) = \pi z(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right)$$

hence I wish to prove

$$(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right) = \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right).$$

I multiplied things out and got it to the form

$$(1-z) \prod_{k=1}^\infty \frac{1 + \frac{1}{k} + \frac{z(1-z)}{k^2}}{1 + \frac{1}{k}} = (1-z) \prod_{k=1}^\infty \left( 1 + \frac{\frac{z(1-z)}{k}}{1+\frac{1}{k}}\right)$$ however the $(1-z)$ factor out front is giving me some trouble; I'm not sure how to proceed.

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  • $\begingroup$ Something looks fishy: Your final product appears to have its zeros in the wrong places. $\endgroup$ – Harald Hanche-Olsen Jun 18 '12 at 12:30
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Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:

$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$

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When k is equal to z it will be equal to 1, 1-1=0. A product series with a zero in it is 0. Since sin(z*pi) is always zero, both sides are equal. But I'm not sure what that is supposed to mean. If I did the equation in two parts leaving out z=k, the part before the z, and the part after, I will get a completely different answer than zero. Is this thing a hack?

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