求解反常积分，谢谢！
3个回答
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∫[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
=(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
追问
不好意思，这不是这个反常积分吧？
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