# 二年級之夢

{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {1}{x^{x}}}\,\mathrm {d} x&=\quad \sum _{n=1}^{\infty }{\frac {1}{n^{n}}}&&\scriptstyle {(=1.29128599706266354040728259059560054149861936827\dots })\\\int _{0}^{1}x^{x}\,\mathrm {d} x&=-\sum _{n=1}^{\infty }(-n)^{-n}&&\scriptstyle {(=0.78343051071213440705926438652697546940768199014\dots })\end{aligned}}}

## 證明

• 對數轉換 ${\displaystyle x^{-x}=e^{-x(\ln x)}}$
• 指數函數泰勒展開式 ${\displaystyle e^{x}=\sum _{n=0}^{\infty }{x^{n} \over n!}}$

${\displaystyle u=-(n+1)\ln x}$

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

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

## 參考

1. ^ BriTheMathGuy. The Integral of your Dreams (or Nightmares). YouTube. [2022-07-21]. （原始内容存档于2022-07-22）.页面存档备份，存于互联网档案馆