二年級之夢

在座標上，兩公式的關係。

第一條公式，首先利用對數轉換和積分與級數順序變化[1]：

• 對數轉換 ${\displaystyle x^{x}=e^{x(\ln x)}}$
• 指數函數的自然展開 ${\displaystyle e^{x}=\sum _{n=0}^{\infty }{x^{n} \over n!}}$

得到 ${\displaystyle \int _{0}^{1}-x^{x}\,dx=\int _{0}^{1}\sum _{n=0}^{\infty }{\frac {(-x\ln x)^{n}}{n!}}\,dx=\sum _{n=0}^{\infty }\int _{0}^{1}{\frac {(-\ln x)^{n}*x^{n}}{n!}}\,dx.}$

設 ${\displaystyle u=(-\ln x)}$

所以 ${\displaystyle \sum _{n=0}^{\infty }\int _{0}^{1}{\frac {(-\ln x)^{n}*x^{n}}{n!}}\,dx=\sum _{n=0}^{\infty }\int _{\infty }^{0}{\frac {u^{n}e^{-nu}}{n!}}(-e^{-u})\,du=\sum _{n=0}^{\infty }\int _{0}^{\infty }{\frac {u^{n}e^{-nu}}{n!}}(e^{-u})\,du}$

再設 ${\displaystyle v=(n+1)u}$

${\displaystyle \sum _{n=0}^{\infty }\int _{0}^{\infty }{\frac {u^{n}e^{-nu}}{n!}}(e^{-u})\,du=\sum _{n=0}^{\infty }{\frac {1}{n!(n+1)^{n+1}}}\int _{0}^{\infty }v^{n}e^{-v}\,dv}$

根據Gamma函數 ${\displaystyle \int _{0}^{\infty }v^{n}e^{-v}\,dv=\Gamma (n+1)=n!}$

最終推得 ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n!(n+1)^{n+1}}}\int _{0}^{\infty }v^{n}e^{-v}\,dv=\sum _{n=0}^{\infty }{\frac {1}{(n+1)^{n+1}}}=\sum _{n=1}^{\infty }{\frac {1}{n^{n}}}}$

• 菲涅耳積分