设:
f(t)=∫abxtdx∫abxt+1dx
当 t=−1 且 t=−2 时,可化简为:
f(t)=(t+2)(bt+1−at+1)(t+1)(bt+2−at+2)
函数 f(t) 是单调不减的。
H(a,b)≤G(a,b)≤N(a,b)≤A(a,b)≤T(a,b)
f(−3)=−1(b−2−a−2)−2(b−1−a−1)=a+b2ab=H(a,b)
f(−1.5)=0.5(b−0.5−a−0.5)−0.5(b0.5−a0.5)=ab=G(a,b)
f(−0.5)=1.5(b0.5−a0.5)0.5(b1.5−a1.5)=3a+ab+b=N(a,b)
f(0)=2(b1−a1)1(b2−a2)=2a+b=A(a,b)
f(1)=3(b2−a2)2(b3−a3)=3(a+b)2(a2+ab+b2)=T(a,b)