求解反常积分已知积分(0->无穷)sinxdx/x=pi/2,求积分(0->正无穷)(sinx)^2dx/x^2

Question

求解反常积分
已知积分(0->无穷)sinxdx/x=pi/2,求积分(0->正无穷)(sinx)^2dx/x^2

Answer 1

∫[0-->+∞] (sinx)^2/x^2 dx
=(1/2)*∫[0-->+∞] (1-cos2x)/x^2dx
=(1/2)*∫[0-->+∞] 1/x^2dx-(1/2)*∫[0-->+∞] cos2x/x^2dx
=(1/2)*∫[0-->+∞] 1/x^2dx-(1/2)*∫[0-->+∞] cos2x/d(1/x)
=(1/2)*[0-->+∞] (-1/x)-(1/2)*[0-->+∞] cos2x/x+(1/2)*∫[0-->+∞] sin2x/xdx
后面那个积分作变量代换2x=t
=(1/2)*[0-->+∞] (cos2x-1)/x+[0-->+∞] ∫sint/tdt
前面这个式子,对于上限+∞和下限0,结果都为0
=∫[0-->+∞] sinx/xdx
=π/2