應用兩次施篤茲定理
lim an/n^2變?yōu)?br/>(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)'
洛朗級數(shù)展開得
(sinx)^2=1/x^2+1/3+x^/15+2x^4/189+o(x^4),高階項的積分為0
同時(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
用收斂因子法解出
收斂因子e^(-tx)
(0,+∞)∫{cos[(2n-2)x]-cos[4(n-1)x]}/xdx=ln2