令 $dps{B(m,n)=sum_{k=0}^n C_n^k cfrac{(-1)^k}{m+k+1}}$, $m,ninbN^+$. (1) 证明 $B(m,n)=B(n,m)$; (2) 计算 $B(m,n)$.
证明: (1) $$eex ea B(m,n)&=sum_{k=0}^n C_n^k (-1)^kint_0^1 x^{m+k} d x\ &= int_0^1 x^msum_{k=0}^n C_k^k 1^{n-k}(-x)^k d x\ &=int_0^1 x^m(1-x)^n d x\ &=int_0^1 (1-x)^mx^n d xquadsex{xleftrightsquigarrow 1-x}\ &=B(n,m). eea eeex$$ (2) $$eex ea B(m,n)&=cfrac{1}{n+1}int_0^1 (1-x)^m d x^{n+1}\ &=-cfrac{m}{n+1}int_0^1 (1-x)^{m-1}(-1)cdot x^{n+1} d x\ &=cfrac{m}{n+1}B(m-1,n+1)\ &=cfrac{m}{n+1}cdot cfrac{m-1}{n+2}cdot cdotscdot cfrac{1}{m+n}B(0,m+n)\ &=cfrac{m!n!}{(m+n+1)!}. eea eeex$$