高数 积分一题。急求高手啊!!!!
反常积分。下限是0,上限是+无穷大。被积函数是e^(-x^2)对x做积分。求解啊。大神!!!!!...
反常积分。下限是0,上限是+无穷大。被积函数是 e^(-x^2) 对x做积分。求解啊。大神!!!!!
展开
展开全部
这题要用重积分来求
令L=∫(0,∞ )e^(-x^2)dx
那么
∫(0,∞ )e^(-x^2)dx*∫(0,∞ )e^(-y^2)dy
=∫(0,∞ )∫(0,∞ )e^(-x^2-y^2)dxdy 做变换x=rcost y=rsint
=∫(0,π/2 )dt ∫(0,∞ )e^(-r^2)rdr
=π/2*1/2*∫(0,∞ )e^(-r^2) d(r^2)
=π/4*e^(-r^2) (0,∞)
=π/4
而∫(0,∞ )e^(-x^2)dx=∫(0,∞ )e^(-y^2)dy
所以∫(0,∞ )e^(-x^2)dx=√(π/4)=√π/2
令L=∫(0,∞ )e^(-x^2)dx
那么
∫(0,∞ )e^(-x^2)dx*∫(0,∞ )e^(-y^2)dy
=∫(0,∞ )∫(0,∞ )e^(-x^2-y^2)dxdy 做变换x=rcost y=rsint
=∫(0,π/2 )dt ∫(0,∞ )e^(-r^2)rdr
=π/2*1/2*∫(0,∞ )e^(-r^2) d(r^2)
=π/4*e^(-r^2) (0,∞)
=π/4
而∫(0,∞ )e^(-x^2)dx=∫(0,∞ )e^(-y^2)dy
所以∫(0,∞ )e^(-x^2)dx=√(π/4)=√π/2
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