本文对于积分from n=0 to 1 f(x,{Nx})dx带准确余项的渐近展开式from n=0 to 1 f(x,{Nx})dx=from n=0 to 1 from n=0 to 1f(x,y)dxdy+sum from k=1 to r 1/(k!) (1/N)~k from n=0 to 1[f^((k-1,0))(1,y)(?)_k(y-N)-f^((k-1,0))(O,y)B_k(y)]dy-1/(r|)(1/N)~r from n=0 to 1 from n=0 to 1 f^((r,O))(x,y)(?)_r(y-Nx)dxdy给出了一种简捷的推导,这种推导只需普通的分析知识,无需用到Euler-Maclaurin求和公式及Bernoulli多项式的Raabe乘积定理。
给出了 integral from n=0 to 1 f(x,Nx)dx 带准确余项的渐近展开式integral from n=0 to 1 f(x,Nx)dx=integral from n=0 to 1 integral from n=0 to 1 f(x,y)dxdy+sum from k=1 to r(1/k■(1/N)~k integral from n=0 to 1[f^(k-1),0(1,y)-f^(k-1,0)(0,y)]B_k(y)dy-(1/r■(1/N)~r integral from n=0 to 1 integral from n=0 to 1 f^(r,0)(x,y)■_r(y-Nx)dxdy的一种简单推导。
