若∫+∞0e∧-t∧2dt=√π/2,则∫+∞-∞x∧2e∧-x∧2dx
展开全部
∫(0->+∞)e^(-t^2)dt =√π/2
f(x)=e^(-x^2)
f(-x) =f(x)
∫(-∞->+∞)x^2.e^(-x^2)dx
=-(1/2)∫(-∞->+∞)xde^(-x^2)
=-(1/2)x.e^(-x^2)|(-∞->+∞) + (1/2) ∫(-∞->+∞)e^(-x^2) dx
=(1/2) ∫(-∞->+∞)e^(-x^2) dx
= ∫(0->+∞)e^(-x^2) dx
=√π/2
f(x)=e^(-x^2)
f(-x) =f(x)
∫(-∞->+∞)x^2.e^(-x^2)dx
=-(1/2)∫(-∞->+∞)xde^(-x^2)
=-(1/2)x.e^(-x^2)|(-∞->+∞) + (1/2) ∫(-∞->+∞)e^(-x^2) dx
=(1/2) ∫(-∞->+∞)e^(-x^2) dx
= ∫(0->+∞)e^(-x^2) dx
=√π/2
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
