Discussion Section 3 Math 31B 4C Practice Integration by Parts The problem is to get a closed equation for I a b a b non negative integers defined by 1 Z xa 1 x b dx I a b 0 a We will first show I a b I b a Notice that 1 Z xa 1 x b dx let u 1 x du dx and x 1 u I a b 0 0 Z 1 u a ub du 1 Z 1 1 u a ub du 0 Z 1 ub 1 u a du 0 I b a Where the final equation follows because it is the same as the original integral just with relabelling the variable x as u b Now let s find a formula for I a 0 1 Z xa dx I a 0 0 1 a 1 And by part a I a 0 I 0 a and therefore 1 a 1 a c Now let s show assuming a 0 I a b I a 1 b 1 b 1 I 0 a Z I a b 1 xa 1 x b dx 0 Now let u 1 1 x b 1 u0 1 x b v xa v 0 axa 1 b 1 Page 1 of 2 Discussion Section 3 Math 31B 4C Z 1 xa 1 x b dx I a b 0 Z 0 1 1 vu0 uv 0 Z 1 uv 0 0 Z 1 1 1 1 b 1 a 1 x x 1 x b 1 axa 1 dx b 1 0 b 1 0 Z 1 a a 1 x b 1 xa 1 dx I a 1 b 1 0 0 b 1 0 b 1 d Finishing up Now take any I a b by c a I a 1 b 1 b 1 a a 1 I a 2 b 2 b 1 b 2 I a b and repeating this for I a 3 b 3 I a 4 b 4 until the first argument of the I function is 0 and using b a a 1 2 1 I 0 b a b 1 b 2 b a 1 b a a b 1 a b b a b a 1 b a 1 I a b So finally we get I a b a b b a 1 Page 2 of 2