全ての素数の積が 4π^2 であることの証明

    \[\zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s} = \prod_{p \in \text{prime}} \left( \sum_{i=0}^\infty \left( \frac{1}{p^s} \right)^i \right) = \prod_{p \in \text{prime}} \frac{1}{1 - \frac{1}{p^s}}\]

より, 両辺に対して対数をとると

    \begin{align*} \log \zeta(s) &= \log \left( \prod_{p \in \text{prime}} \frac{1}{1 - \frac{1}{p^s}} \right) \\ &= - \sum_{p \in \text{prime}} \log \left( 1 - \frac{1}{p^s} \right) \\ &= - \sum_{p \in \text{prime}} \left( \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \left( - \frac{1}{p^s} \right)^n \right) \\ &= - \sum_{p \in \text{prime}} \left( - \sum_{n=1}^\infty \frac{1}{n p^{sn}} \right) \\ &= \sum_{p \in \text{prime}} \sum_{n=1}^\infty \frac{1}{n p^{sn}} \end{align*}

両辺を s について微分すると

    \[\frac{\zeta '(s)}{\zeta(s)} = \sum_{p \in \text{prime}} \sum_{n=1}^\infty n \cdot \frac{- \log p}{n p^{sn}} = - \sum_{p \in \text{prime}} \sum_{n=1}^\infty \frac{\log p}{p^{sn}}\]

ここで, s = 0 のとき

    \[\frac{\zeta '(0)}{\zeta(0)} = - \sum_{p \in \text{prime}} \sum_{n=1}^\infty \log p = - \sum_{p \in \text{prime}} \zeta(0) \log p\]

よって

    \begin{align*} \sum_{p \in \text{prime}} \log p &= - \frac{\zeta '(0)}{\zeta^2(0)} \\ &= - \frac{- \frac{1}{2} \log \frac{\pi}{2}}{(- \frac{1}{2})^2} \\ &= 2 \log (2 \pi) \\ &= \log (4 \pi^2) \end{align*}

また,

    \[\sum_{p \in \text{prime}} \log p = \log \prod_{p \in \text{prime}} p\]

より,

    \[\boxed{\prod_{p \in \text{prime}} p = 4 \pi^2}\]