How can you prove this equality? I am trying to figure out the following equality, but cannot seem

Hugo Brady

Hugo Brady

Answered question

2022-05-21

How can you prove this equality?
I am trying to figure out the following equality, but cannot seem to get anywhere. I tried integrating by parts, but that blew up when I set u = (log x)^n and tried to take log (0). I also tried differentiating the right side but got stuck when I did not know the derivative of n !
How can we prove that the following is true:
0 1 x a ( log x ) n d x = ( 1 ) n ( n ! ) ( a + 1 ) n + 1
Any suggestions would be greatly appreciated.

Answer & Explanation

Asa Austin

Asa Austin

Beginner2022-05-22Added 4 answers

Applying integration by parts:
0 1 x a ( l o g x ) n d x = lim b 0 + [ ( log ( x ) ) n 1 a + 1 x a + 1 | b 1 ] n a + 1 0 1 x a ( log x ) n 1 d x
Now if a > 1, you can show with L'Hopital's rule that
lim b 0 + [ ( log ( b ) ) n b a + 1 ] = 0
Hence
0 1 x a ( l o g x ) n d x = n a + 1 0 1 x a ( log x ) n 1 d x
See the inductive pattern now? Why you will have n factors of ( 1 ) when you're finished?
vaganzahi

vaganzahi

Beginner2022-05-23Added 1 answers

0 1 x a ln n ( x ) d x = ( 1 ) n n ! ( a + 1 ) n + 1 :   ?
With a > 1 :
0 1 x a ln n ( x ) d x = lim μ 0 n μ n 0 1 x a + μ d x = lim μ 0 n μ n ( 1 a + μ + 1 ) = n a n ( 1 a + 1 ) = n 1 a n 1 ( 1 ( a + 1 ) 2 ) = n 2 a n 2 ( 2 ( a + 1 ) 3 ) = n 3 a n 3 ( 3 × 2 ( a + 1 ) 4 ) = = n k a n k ( ( 1 ) k k ! ( a + 1 ) k + 1 ) = = ( 1 ) n n ! ( a + 1 ) n + 1

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