純數-積分代入法

朋友﹐動下腦筋就做到架啦
x+y+xy=1=>x=(1-y)/(1+y),y=(1-x)/(1+x)
when x=1,y=0 when x=0,y=1
Also dy/dx=-(1+y)^2/2
The original integration becomes
∫[from 1 to 0] [(1-y)/(1+y)]^p[(2y)/(1+y)]^q[2/(1+y)]^r[-2/(1+y)^2]dy
=2^(q+r+1)∫[from 0 to 1] y^q(1-y)^p(1+y)^(-p-q-r-2) dy
=2^(q+r+1)J(q,p,-p-q-r-2)

2009-03-28 20:34:57 補充：
I go to dinner with my friend la

感謝2位的ans...
我一看到這個代入就有d不知所措添...=.=

Oh, 好彩check check，慢了一步

2009-03-28 20:42:37 補充：
慢了一步，不過都upload我嘅答案。

2009-03-28 20:45:00 補充：
http://i601.photobucket.com/albums/tt95/physicsworld9999/physicsworld05Mar282040.jpg