オイラーの和公式
}}f^{(k-1)}(x)\right]_{0}^{1}+(-1)^{m}\int _{0}^{n}{\frac {B_{m}(x-\lfloor {x}\rfloor )}{m!}}f^{(m)}(x)dx\end{aligned}}}
である。 B 1 ( 0 ) = − 1 2 , {\displaystyle B_{1}(0)=-\textstyle {\frac {1}{2}},} B 1 ( 1 ) = 1 2 , B 2 k ( 0 ) = B 2 k ( 1 ) = B 2 k , {\displaystyle B_{1}(1)=\textstyle {\frac {1}{2}},B_{2k}(0)=B_{2k}(1)=B_{2k},} B 2 k + 1 ( 0 ) = B 2 k + 1 ( 1 ) = B 2 k + 1 = 0 {\displaystyle B_{2k+1}(0)=B_{2k+1}(1)=B_{2k+1}=0} を代入すれば
∫ 0 n f ( x ) d x = ∑ j = 0 n − 1 f ( j ) − 1 2 f ( 0 ) + 1 2 f ( n ) − ∑ k = 2 m ( − 1 ) k B k k !
f ( m ) ( x ) d x {\displaystyle \int _{0}^{1}f(x)dx=\int _{0}^{1}B_{0}(x)f(x)dx=\sum _{k=1}^{m}\left[(-1)^{k-1}{\frac {B_{k}(x)}{k!}}f^{(k-1)}(x)\right]_{0}^{1}+(-1)^{m}\int _{0}^{1}{\frac {B_{m}(x)}{m!}}f^{(m)}(x)dx}
となるが、これは f ( x ) {\displaystyle f(x)} を f ( j + x ) {\displaystyle f(j+x)} に置き換えても成り立つから
∫ 0 n f ( x ) d x = ∑ j = 0 n − 1 ∫ 0 1 f ( j + x ) d x = ∑ j = 1 n − 1 ∑ k = 1 m [ ( − 1 ) k − 1 B k ( x ) k ! f ( k − 1 ) ( x ) ] 0 1 + ( − 1 ) m ∫ 0 n B m ( x − ⌊ x ⌋ ) m ! f ( m ) ( x ) d x {\displaystyle {\begin{aligned}\int _{0}^{n}f(x)dx&=\sum _{j=0}^{n-1}\int _{0}^{1}f(j+x)dx\\&=\sum _{j=1}^{n-1}\sum _{k=1}^{m}\left[(-1)^{k-1}{\frac {B_{k}(x)}{k!