应用两次施笃兹定理
lim an/n^2变为
(0,+∞)∫xsin[(3n-3)x]sin[(n-1)x]/(sinx)^2dx+(0,+∞)∫xcos[(4n-4)x]dx
=(0,+∞)∫xsin[(3n-3)x]sin[(n-1)x]/(sinx)^2dx
=(0,+∞)∫x{cos[(2n-2)x]-cos[4(n-1)x]}/(sinx)^2dx
(sinx)^2=-(cotx)'
洛朗级数展开得
(sinx)^2=1/x^2+1/3+x^/15+2x^4/189+o(x^4),高阶项的积分为0
同时(0,+∞)∫{cos[(2n-2)x]-cos[4(n-1)x]}x^2m=0
所以
=(0,+∞)∫x{cos[(2n-2)x]-cos[4(n-1)x]}/(sinx)^2dx
=(0,+∞)∫{cos[(2n-2)x]-cos[4(n-1)x]}/xdx
用收敛因子法解出
收敛因子e^(-tx)
(0,+∞)∫{cos[(2n-2)x]-cos[4(n-1)x]}/xdx=ln2