Evaluate int_0^1 log^2(1-x) log^2(x) dx

Maia Pace 2022-08-13 Answered
Evaluate 0 1 log 2 ( 1 x ) log 2 ( x ) d x
I have no idea where to even start. WolframAlpha cant compute it either.
0 1 log 2 ( 1 x ) log 2 ( x ) d x
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
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Answers (1)

Alaina Mcintosh
Answered 2022-08-14 Author has 16 answers
Using the series of ln 2 ( 1 x )
0 1 ln 2 x ln 2 ( 1 x )   d x = n = 1 2 H n n + 1 0 1 x n + 1 ln 2 x   d x = n = 1 4 H n ( n + 1 ) ( n + 2 ) 3
Then integrate f ( z ) = ( γ + ψ 0 ( z ) ) 2 ( z + 1 ) ( z + 2 ) 3 along an infinitely large square. The integral vanishes which implies the sum of its residues is zero. At the positive integers,
n = 1 R e s ( f , n ) = n = 1 Res z = n [ 1 ( z + 1 ) ( z + 2 ) 3 ( z n ) 2 + 2 H n ( z + 1 ) ( z + 2 ) 3 ( z n ) ] = n = 1 2 H n ( n + 1 ) ( n + 2 ) 3 n = 1 4 n + 5 ( n + 1 ) 2 ( n + 2 ) 4 = n = 1 2 H n ( n + 1 ) ( n + 2 ) 3 + π 4 30 + 2 ζ ( 3 ) 91 16
At z=0,
R e s ( f , 0 ) = Res z = 0 1 z 2 ( z + 1 ) ( z + 2 ) 3 = 5 16
At z=−2,
R e s ( f , 2 ) = 1 2 lim z 2 d 2 d z 2 ( γ + ψ 0 ( z ) ) 2 z + 1 = π 4 36 + 2 ζ ( 3 ) + 2 π 2 3 6
Therefore,
0 1 ln 2 x ln 2 ( 1 x )   d x = 2 [ π 4 30 2 ζ ( 3 ) + 91 16 + 5 16 + π 4 36 2 ζ ( 3 ) 2 π 2 3 + 6 ] = π 4 90 8 ζ ( 3 ) 4 π 2 3 + 24
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