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I am familiar with the weierstrass infinite product and eulers form yet I'm clueless as to how to derive this infinite product formula below.

$$\Gamma(1+z)=\frac 1{e^{\gamma z}}\sqrt{\frac {\pi z}{\sin\pi z}}\prod\limits_{k=1}^{+\infty}\exp\left(-\frac {\zeta(2k+1)z^{2k+1}}{2k+1}\right)$$

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    $\begingroup$ Simplify the exponential product into the exponential of a series. If you differentiate that series, you’re left with the odd terms of a known generating function, which you can extract. It’s a good idea to try something on MSE, any working or thoughts at all $\endgroup$
    – FShrike
    Aug 7 at 21:42
  • $\begingroup$ Cool thx I'll see how far I get $\endgroup$
    – Richie
    Aug 7 at 21:56
  • $\begingroup$ I took logarithm of both sides $\endgroup$
    – Richie
    Aug 7 at 22:02
  • $\begingroup$ So ur saying I should take derivative of ln gamma $\endgroup$
    – Richie
    Aug 7 at 22:02
  • $\begingroup$ I made a mistake. While $\zeta(2n)$ in the product would have a known evaluation, $\zeta(2n+1)$ is less-known (famously). This will require some other trickery then! $\endgroup$
    – FShrike
    Aug 7 at 22:08

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Weierstrass's $\Gamma(1+z)=z\Gamma(z)=e^{-\gamma z}\prod_{n=1}^\infty(1+z/n)^{-1}e^{z/n}$ gives, for $\color{red}{|z|<1}$, $$\log\Gamma(1+z)=-\gamma z+\sum_{n=1}^\infty\left[\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right] \\=-\gamma z+\sum_{n=1}^\infty\sum_{k=2}^\infty\frac1k\left(-\frac{z}{n}\right)^k=-\gamma z+\sum_{k=2}^\infty\zeta(k)\frac{(-z)^k}{k}.$$

The even part of this series (that is, the sum of terms with even $k$) is then $$\frac12\big(\log\Gamma(1+z)+\log\Gamma(1-z)\big)=\frac12\log\frac{\pi z}{\sin\pi z}$$ by the reflection formula, hence $\log\Gamma(1+z)$ is this quantity plus the odd part: $$\log\Gamma(1+z)=-\gamma z+\frac12\log\frac{\pi z}{\sin\pi z}-\sum_{k=1}^\infty\zeta(2k+1)\frac{z^{2k+1}}{2k+1}.$$

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