二年级之梦

第一条公式，首先利用对数转换和积分与级数顺序变化^[1]：

• 对数转换 ${\displaystyle x^{x}=e^{x(\ln x)}}$ 
• 指数函数的自然展开 ${\displaystyle e^{x}=\sum _{n=0}^{\infty }{x^{n} \over n!}}$ 

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

设 ${\displaystyle u=(-\ln x)}$ 

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

再设 ${\displaystyle v=(n+1)u}$ 

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

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

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

• 菲涅耳积分