不要悲观,可证.∑的范围是n从0到a-1, a是个整数.
[Σf(n) - ∫(0,a)f(x)dx]=∑∫(n,n+1)f(n+1)dx-∑∫(n,n+1)f(x)dx+f(0)
=∑∫(n,n+1)[f(n+1)-f(x)]dx+f(0)