Question

这个积分方程怎么解？

麻烦大家了

Answer 1

=∫∫x*l^(-x²-y²) dD D={x>y} + ∫∫y*l^(-x²-y²) dD D={x<=y}
等号放哪都行，因为在交接点的左右极限相等，是连续函数
=∫(0~无穷)∫(y~无穷)x*l^(-x²-y²) dxdy + ∫(0~无穷)∫(x~无穷)y*l^(-x²-y²) dydx
=∫(0~无穷){(-0.5/lnl)*l^(-x²-y²) ] x: y~无穷 }dy +∫(0~无穷){(-0.5/lnl)l^(-x²-y²)] y: x~无穷 }dx
= ∫(0~无穷)(0.5/lnl)*l^(-2y²)dy+ ∫(0~无穷)(0.5/lnl)*l^(-2x²)dx

根据正态分布性质
∫(-无穷~无穷)(1/[(根号(2π))o])*e^(-t²/2o²)dt=1
∫(-无穷~无穷)e^(-t²/2o²)dt=(根号(2π))o
根据此类函数的对称性,取右面一半
∫(0~无穷) e^(-t²/2o²)dt=根号(π/2)o
∫(0~无穷) l^(-lnl*t²/2o²)dt=根号(π/2)o
lnl/(2o²)=2
o²=lnl/4

所以
∫(0~无穷) l^(-2t²)dt=根号(lnlπ/2)/2

此性质带回x,y的积分，得
= ∫(0~无穷)(0.5/lnl)*l^(-2y²)dy+ ∫(0~无穷)(0.5/lnl)*l^(-2x²)dx
=0.5根号(π/2lnl)/2+0.5根号(π/2lnl)/2
=根号(π/2lnl)/2