I would like some help to find a closed form for the following integral: int_0^1 ((log (1+x))^3)/(x)dx I was told it could be calculated in a closed form. I've already proved that int_0^1 (log (1+x))/(x)dx=(pi^2)/(12) using power series expansion. Thank you.

raapjeqp 2022-10-14 Answered
A closed form for 0 1 ( log ( 1 + x ) ) 3 x d x?
I would like some help to find a closed form for the following integral:
0 1 ( log ( 1 + x ) ) 3 x d x
I was told it could be calculated in a closed form. I've already proved that
0 1 log ( 1 + x ) x d x = π 2 12
using power series expansion.
Thank you.
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Answers (1)

RamPatWeese2w
Answered 2022-10-15 Author has 15 answers
Letting u = log ( 1 + x )
0 1 log 3 ( 1 + x ) x   d x = 0 log 2 u 3 e u 1 e u   d u = 0 log 2 u 3 1 e u   d u = 0 log 2 u 3 n = 0 e n u   d u = n = 0 0 log 2 u 3 e n u   d u = 0 log 2 u 3   d u + n = 1 0 log 2 u 3 e n u   d u = log 4 ( 2 ) 4 + n = 1 0 log 2 u 3 e n u   d u .
Then integrating by parts 3 times,
0 1 log 3 ( 1 + x ) x   d x = log 4 ( 2 ) 4 n = 1 e n u ( 6 n 4 + 6 u n 3 + 3 u 2 n 2 + u 3 n ) | 0 log 2 = log 4 ( 2 ) 4 n = 1 [ 1 2 n ( 6 n 4 + 6 log 2 n 3 + 3 log 2 ( 2 ) n 2 + log 3 ( 2 ) n ) 6 ζ ( 4 ) ] = 3 log 4 ( 2 ) 4 6 Li 4 ( 1 2 ) 6 log ( 2 )   Li 3 ( 1 2 ) 3 log 2 ( 2 ) Li 2 ( 1 2 ) + 6 ζ ( 4 ) 0.1425141979 .
The answer could of course be simplified using the known values of Li 2 ( 1 2 ) , Li 3 ( 1 2 ) , and ζ ( 4 )
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