第1个回答 2019-12-19
不过现在已经没有任何时候任何事情发生
第2个回答 2019-12-19
微积分如果解不出来的话，可以请教一下你的辅导老师，他会给你详细解答。
第3个回答 2019-12-19
令f(x,y)=e^(-x^2)*cos(xy)，则∂f/∂y=(-x)*e^(-x^2)*sin(xy)
因为f(x,y)和∂f/∂y都在[0,+∞)×(-∞,+∞)上连续，所以微分和积分次序可交换
即dI(y)/dy=(d/dy)*∫(0,+∞) f(x,y)dx
=∫(0,+∞) (∂f/∂y)dx
=∫(0,+∞) (-x)*e^(-x^2)*sin(xy)dx
=(1/2)*∫(0,+∞) sin(xy)d[e^(-x^2)]
=(1/2)*e^(-x^2)*sin(xy)|(0,+∞)-(y/2)*∫(0,+∞) e^(-x^2)*cos(xy)dx
=-(y/2)*∫(0,+∞) f(x,y)dx
=-(y/2)*I(y)
整理得：dI(y)/I(y)=-(y/2)dy
∫dI(y)/I(y)=∫-(y/2)dy
ln|I(y)|=(-1/4)*y^2+C
I(y)=C*e^[(-1/4)*y^2]，其中C是任意常数
已知，当y=0时，I(0)=∫(0,+∞) e^(-x^2)dx=(√π)/2
所以I(0)=C=(√π)/2
I(y)=[(√π)/2]*e^[(-1/4)*y^2]本回答被网友采纳
因为f(x,y)和∂f/∂y都在[0,+∞)×(-∞,+∞)上连续，所以微分和积分次序可交换
即dI(y)/dy=(d/dy)*∫(0,+∞) f(x,y)dx
=∫(0,+∞) (∂f/∂y)dx
=∫(0,+∞) (-x)*e^(-x^2)*sin(xy)dx
=(1/2)*∫(0,+∞) sin(xy)d[e^(-x^2)]
=(1/2)*e^(-x^2)*sin(xy)|(0,+∞)-(y/2)*∫(0,+∞) e^(-x^2)*cos(xy)dx
=-(y/2)*∫(0,+∞) f(x,y)dx
=-(y/2)*I(y)
整理得：dI(y)/I(y)=-(y/2)dy
∫dI(y)/I(y)=∫-(y/2)dy
ln|I(y)|=(-1/4)*y^2+C
I(y)=C*e^[(-1/4)*y^2]，其中C是任意常数
已知，当y=0时，I(0)=∫(0,+∞) e^(-x^2)dx=(√π)/2
所以I(0)=C=(√π)/2
I(y)=[(√π)/2]*e^[(-1/4)*y^2]本回答被网友采纳