Question

高等数学难题

Answer 1

利用积化和差、和伏锋差化积公式，还有一个重要反常积分：∫(0,+∞)sinx/xdx=π/2

sinxsin(x/3)sin(x/5)=(1/2)*[cos(2x/3)-cos(4x/3)]sin(x/5)
=(1/2)*cos(2x/3)sin(x/5)-(1/2)*cos(4x/3)sin(x/5)
=(1/4)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]
原式=(15/4)*∫(0,+∞)[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^3dx
=(-15/8)*∫(0,+∞)[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]d(1/x^2)
=(-15/斗顷8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2|(0,+∞)+(1/8)*∫(0,+∞)[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x^2dx
=lim(x->0) (15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2-(1/8)*∫(0,+∞)[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]d(1/x)
=lim(x->0) (15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2-(1/8)*[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x|(0,+∞)-(1/120)*∫(0,+∞)[169sin(13x/15)-49sin(7x/5)-529sin(23x/5)+289sin(17x/15)]/xdx
=lim(x->0) {(15/8)*[sin(13x/15)-sin(7x/15)-sin(23x/15)+sin(17x/15)]/x^2+(1/8)*[13cos(13x/15)-7cos(7x/5)-23cos(23x/5)+17cos(17x/15)]/x}-(π/240)*(169-49-529+289)
=π/2+(1/8)*lim(x->0) [15sin(13x/15)-15sin(7x/15)-15sin(23x/15)+15sin(17x/15)+13xcos(13x/15)-7xcos(7x/15)-23xcos(23x/15)+17xcos(17x/15)]/x^2
=π/2+(1/8)*lim(x->0) [13cos(13x/15)-7cos(7x/15)-23cos(23x/15)+17cos(17x/15)-(169/30)xsin(13x/15)+(49/30)xsin(7x/15)+(529/30)xsin(23x/15)-(289/空厅陆30)xsin(17x/15)]/x
=π/2+(1/8)*lim(x->0) [-(169/10)sin(13x/15)+(49/10)sin(7x/5)+(529/10)sin(13x/15)-(289/10)sin(17x/15)]
=π/2

Answer 2

积分结果是π/2.