Question

这道积分题应该怎么算？求过程

Answer 1

令A=∫(0,+∞) e^(-k^2*x^2)dx
则A^2=∫(0,+∞) e^(-k^2*x^2)dx*∫(0,+∞) e^(-k^2*y^2)dy
=∫∫(D) e^[-k^2*(x^2+y^2)]dxdy，其中D是第一象限
令x=rcost，y=rsint，则
A^2=∫(0,π/2)dt*∫(0,+∞) e^(-k^2*r^2)*rdr
=(π/2)*(-1/2k^2)*∫(0,+∞) e^(-k^2*r^2)d(-k^2*r^2)
=(-π/4k^2)*e^(-k^2*r^2)|(0,+∞)
=π/4k^2
因为A恒大于0，所以A=(√π)/(2k)
（1）2k^2*∫(0,+∞) x^2*e^(-k^2*x^2)dx
=-∫(0,+∞) x*d[e^(-k^2*x^2)]
=-x*e^(-k^2*x^2)|(0,+∞)+∫(0,+∞) e^(-k^2*x^2)dx
=0+A
=(√π)/(2k)
（2）2k^2*∫(0,+∞) x^3*e^(-k^2*x^2)dx
=-∫(0,+∞) x^2*d[e^(-k^2*x^2)]
=-x^2*e^(-k^2*x^2)|(0,+∞)+∫(0,+∞) e^(-k^2*x^2)d(x^2)
=0+(-1/k^2)*e^(-k^2*x^2)|(0,+∞)
=1/k^2