令x=x,便有 rn(x0)=r(x0)=r(x)=…=r"(x)=0 逐次应用分部积分法,可得 r,(x)=r,(x)r()=rn(o)dt = r(t)d(t-x)=ri"(@(x-t)dt 21J7(d(-x) ∫7Xx-)d (1)(x-0)d f(t(x-t)dt令 x = 0 x ，便有 )( 0 xrn = )( 0 xrn′ = )( 0 xrn′′ = " = )( 0 )( xr n n = 0。 逐次应用分部积分法，可得 xr )( n = xr )( n - )( 0 xrn = ∫ ′ xx n ttr0 d)( = ∫ ′ − xx n xttr0 )d()( = ∫ ′′ − xx n ttxtr0 d))(( = - !2 1 ∫ ′′ − xx n xttr0 2 )d()( = !21 0 2 ( )( ) d x n x rt x t t ′′′ − ∫ …… = ! 1 n ∫ − + xx n n n ttxtr0 d))(()1( = !1n ∫ − + xx n n ttxtf 0 d))(()1(