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3Â¥: Originally posted by peterflyer at 2014-05-22 20:24:45
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shy1992331: ½ð±Ò+5 2014-05-23 23:03:12
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4Â¥: Originally posted by shy1992331 at 2014-05-22 23:02:57
Ô­À´Èç´Ë£¬ÆäʵԭÌâÊÇg(x,y)=\int_{-\infty}^{0}e^{t(x+y)-t^{2}}dt£¬µÃµ½º¯Êýg(x,y)ºó£¬½«x,y»»³ÉËæ»ú±äÁ¿X£¬Y£¬ÆäÖÐX,Y·þ´Ó±ê×¼Õý̬·Ö²¼ÇÒÏ໥¶ÀÁ¢£¬Çóg(X,Y)µÄ·Ö²¼?²»ÖªµÀÄãÓÐûÓÐ˼·£¿...

g(x,y)=Integral{e^[t*(x+y)-t^2]*dt , -¡Þ, 0}=e^[(x+y)^2/4]*Integral{e^{-[t-(x+y)/2]^2}*dt , -¡Þ, 0}
=e^[(x+y)^2/4]*Integral{e^{-u^2}*du , -¡Þ, -(x+y)/2}=e^[(x+y)^2/4]*Integral{e^{-u^2}*du , (x+y)/2, ¡Þ}
=e^[(x+y)^2/4]*sqrt(¦Ð)/2*erfc[(x+y)/2]
5Â¥2014-05-23 00:27:46
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