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积分区域D1：边长√2R的1/4圆∫∫e^(-x^2-y^2)dxdy=(0,π/2)∫dθ(0,√2R)∫e(-r^2)rdr=π/4*[1-e^(-2R^2)]积分区域D2:边长R的1/4圆∫∫e^(-x^2-y^2)dxdy=(0,π/2)∫dθ(0,R)∫e(-r^2)rdr=π/4*[1-e^(-R^2)]积分区域D：边长为R的正方形（外圈是半径为√2R的圆，然后包着正方形，正方形内包着半径为R的圆）∫∫e^(-x^2-y^2)dxdy=(0,R)∫e(-x^2)dx(0,R)∫e(-y^2)dy=[(0,R)∫e(-x^2)dx]^2所以显然有(D2)∫∫e^(-x^2-y^2)dxdy<(D)∫∫e^(-x^2-y^2)dxdy<(D1)∫∫e^(-x^2-y^2)dxdy即π/4*[1-e^(-R^2)<[(0,R)∫e(-x^2)dx]^2<π/4*[1-e^(-2R^2)]夹逼准则，R→+∞lim [(0,R)∫e(-x^2)dx]^2=[(0,+∞)∫e(-x^2)dx]^2]^2=π/4所以(0,+∞)∫e(-x^2)dx=√π/2 或者不太严谨的做法积分区域为整个平面∫∫e^(-x^2-y^2)dxdy=[(0,+∞)∫e(-x^2)dx]^2=(0,π/2)∫dθ(0,+∞)∫e(-r^2)rdr=π/4(0,+∞)∫e(-x^2)dx=√π/2 |
