[再寄小读者之数学篇](2014-06-20 Beta 函数)

令 $\dps{B(m,n)=\sum_{k=0}^n C_n^k \cfrac{(-1)^k}{m+k+1}}$, $m,n\in\bbN^+$. (1) 证明 $B(m,n)=B(n,m)$; (2) 计算 $B(m,n)$.

证明: (1) $$\beex \bea B(m,n)&=\sum_{k=0}^n C_n^k (-1)^k\int_0^1 x^{m+k}\rd x\\ &= \int_0^1 x^m\sum_{k=0}^n C_k^k 1^{n-k}(-x)^k\rd x\\ &=\int_0^1 x^m(1-x)^n\rd x\\ &=\int_0^1 (1-x)^mx^n\rd x\quad\sex{x\leftrightsquigarrow 1-x}\\ &=B(n,m). \eea \eeex$$ (2) $$\beex \bea B(m,n)&=\cfrac{1}{n+1}\int_0^1 (1-x)^m\rd x^{n+1}\\ &=-\cfrac{m}{n+1}\int_0^1 (1-x)^{m-1}(-1)\cdot x^{n+1}\rd x\\ &=\cfrac{m}{n+1}B(m-1,n+1)\\ &=\cfrac{m}{n+1}\cdot \cfrac{m-1}{n+2}\cdot \cdots\cdot \cfrac{1}{m+n}B(0,m+n)\\ &=\cfrac{m!n!}{(m+n+1)!}. \eea \eeex$$