反常积分∫0到无穷e^(-x^2)dx=

Question

如题所述

Answer 1

k1 = ∫0到无穷e^(-x^2)dx

k2 = ∫0到无穷e^(-y^2)dy
k1*k2 =
∫0到无穷
∫0到无穷e^(-x^2)dx e^(-y^2)dy = ∫0到无穷 ∫0到无穷 e^[(-x^2)+(-y^2)dx dy
转到极坐标:
x^2 + y^2 = r^2 ; dxdy = r dr d(theta)
积分是在第一象限:

k1*k2 =
∫ 0到pi/2 [ ∫0到无穷 e^(-r^2)rdr ] d(theta)
=
∫ 0到pi/2 [(1/2) ∫0到无穷 e^(-r^2)d(r^2) ] d(theta)

let z=r^2,
k1*k2 =

∫ 0到pi/2 [(1/2) ∫0到无穷 e^(-z)dz ] d(theta) =

∫ 0到pi/2 (1/2) d(theta) = (1/2)*(pi/2)
= pi/4
so k1 = (pi/4)^(0.5)

Answer 2

楼主你好
二重积分的极坐标变换
解：∫<0,+∞>e^(-x²)dx=∫<0,+∞>e^(-y²)dy
故(∫<0,+∞>e^(-x²)dx)²
=∫<0,+∞>e^(-x²)dx∫<0,+∞>e^(-y²)dy
=∫<0,+∞>∫<0,+∞>e^[-(x²+y²)]dxdy
=∫<0,2π>dθ∫<0,+∞>e^(-r²)rdr
=2π∫<0,+∞>e^(-r²)rdr
=-π∫<0,+∞>e^(-r²)d(-r²)
=-πe^(-r²)|<0,+∞>
=π
即∫<0,+∞>e^(-x²)dx=√π

望采纳，谢谢本回答被提问者和网友采纳