[@integralsbot](https://twitter.com/integralsbot)さんがツイートした[こちらの定理](https://twitter.com/integralsbot/status/1372853055062822915)の解説です．

&&&thm
以下の等式が成り立ちます．
$\d\i{z}{0}{1}{\i{y}{0}{1}{\i{x}{0}{1}{\f{x+y+z}}}}=1$
&&&
&&&prf 
<details><summary>解説</summary>
\begin{align*}
&\i{z}{0}{1}{\i{y}{0}{1}{\i{x}{0}{1}{\f{x+y+z}}}}\\
=&\i{z}{0}{1}{\i{y}{0}{1}{\i{t}{y+z}{1+y+z}{\f{t}}}}\\
=&\i{z}{0}{1}{\i{y}{0}{1}{\r{\i{t}{y+z}{1+\f{y+z}}{\f{t}}+\i{t}{1+\f{y+z}}{1+y+z}{\f{t}}}}}\\
=&\i{z}{0}{1}{\i{y}{0}{1}{\r{\i{t}{y+z}{1+\f{y+z}}{\f{y+z}}+\i{t}{1+\f{y+z}}{1+y+z}{\r{1+\f{y+z}}}}}}\\
=&\i{z}{0}{1}{\i{y}{0}{1}{\r{\f{y+z}\r{\r{1+\f{y+z}}-\r{y+z}}+\r{1+\f{y+z}}\r{\r{1+y+z}-\r{1+\f{y+z}}}}}}\\
=&\i{z}{0}{1}{\i{y}{0}{1}{\r{y+z}}}\\
=&\i{z}{0}{1}{\s{\frac{y^2}{2}+zy}_{y=0}^1}\\
=&\i{z}{0}{1}{\r{\frac{1}{2}+z}}\\
=&\s{\frac{z}{2}+\frac{z^2}{2}}_{z=0}^1\\
=&1
\end{align*}
なので，$\d\i{z}{0}{1}{\i{y}{0}{1}{\i{x}{0}{1}{\f{x+y+z}}}}=1$です．$\blacksquare$
</details>
&&&

解説14

この記事を高評価した人

この記事に送られたバッジ

投稿者

コメント

他の人のコメント