We already know that int_(0)^(pi/2) x ln(2 cos x)dx=(7)/(16)zeta(3), int_(0)^(pi/2)x^2 ln^2(2cos x)dx=(11pi)/(16)zeta(4). Does the following integral admit a closed form? int_(0)^(pi/2)x^3 ln^3(2cos x)dx

Teagan Huffman 2022-09-24 Answered
A closed form for 0 π / 2 x 3 ln 3 ( 2 cos x ) d x
We already know that
0 π / 2 x ln ( 2 cos x ) d x = 7 16 ζ ( 3 ) , 0 π / 2 x 2 ln 2 ( 2 cos x ) d x = 11 π 16 ζ ( 4 ) .
Does the following integral admit a closed form?
0 π / 2 x 3 ln 3 ( 2 cos x ) d x
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Answers (2)

Ashly Sanford
Answered 2022-09-25 Author has 9 answers
Proposition.
0 π / 2 x 3 ln 3 ( 2 cos x ) d x = 45 512 ζ ( 7 ) 3 π 2 16 ζ ( 5 ) 5 π 4 64 ζ ( 3 ) 9 4 ζ ( 5 ¯ , 1 , 1 )
where ζ ( p ¯ , 1 , 1 ) is the colored MZV (Multi Zeta Values) function of depth 3 and weight p + 2 given by
ζ ( p ¯ , 1 , 1 ) := n = 1 ( 1 ) n n p k = 1 n 1 H k 1 k
belonging to a family of functions introduced by L. Euler and also called Euler(-Zagier) sums.
We have a general result.
Theorem. Let be any positive integer. Then
0 π / 2 x 2 + 1 ln 2 + 1 ( 2 cos x ) d x Q ( ζ ( 4 + 3 ) , ζ ( 2 ) ζ ( 4 + 1 ) , . . . , ζ ( 4 ) ζ ( 3 ) , ζ ( 2 + 3 ¯ , { 1 } 2 ) )
It is remarkable that there is only one constant
ζ ( 2 + 3 ¯ , { 1 } 2 ) = n 1 > . . . > n 2 + 1 > 0 ( 1 ) n 1 n 1 2 + 3 n 2 n 2 + 1
for each integral of the considered form. The question of whether one can reduce this constant to colored MZVs/MZVs of lower depths is still subject to a conjecture (Zagier).
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easternerjx
Answered 2022-09-26 Author has 3 answers
We can try the harmonic analysis path. Since:
log ( 2 cos x ) = n = 1 + ( 1 ) n + 1 n cos ( 2 n x ) ,
(1) log ( 2 sin x ) = n = 1 + cos ( 2 n x ) n ,
we have, as an example:
0 π / 2 log 3 ( 2 sin x ) d x = 3 π 4 ζ ( 3 )
since
(2) 0 π / 2 cos ( 2 n 1 x ) cos ( 2 n 2 x ) cos ( 2 n 3 x ) d x = π 8 δ 2 max n i = ( n 1 + n 2 + n 3 )
Now:
(3) π / 2 x = π 4 + 2 π m = 0 + cos ( ( 4 m + 2 ) x ) ( 2 m + 1 ) 2
hence by multiplying ( 1 ) and ( 3 ) we can write the Fouries cosine series of ( π / 2 x ) log ( 2 sin x ) over ( 0 , π / 2 ) and grab from ( 2 ) and grab from ( 2 ) a combinatorial equivalent for
0 π / 2 ( ( π / 2 x ) log ( 2 sin x ) ) 3 d x .
With the aid of Mathematica I got:
( π / 2 x ) log ( 2 sin x ) = n = 0 + cos ( ( 2 n + 1 ) x ) π ( 2 n + 1 ) 2 ( ( 2 n + 1 ) ( ψ ( 2 n + 1 2 ) ψ ( 2 n + 1 2 ) ) + 2 ( 2 γ + ψ ( 2 n + 1 2 ) + ψ ( 2 n + 1 2 ) ) ) ,
( π / 2 x ) log ( 2 sin x ) = n = 0 + cos ( ( 2 n + 1 ) x ) π ( 2 n + 1 ) 2 ( 2 H 2 n + 1 2 ( 2 n + 1 ) ( ψ ( 2 n + 1 2 ) ψ ( 2 n + 1 2 ) ) ) .
(1) ( π / 2 x ) log ( 2 sin x ) = n = 0 + cos ( ( 2 n + 1 ) x ) π ( 2 n + 1 ) 2 ( 4 log 2 + j = 0 n 4 2 j + 1 + 4 2 n + 1 + ( 2 n + 1 ) j = 0 n 8 ( 2 j + 1 ) 2 ) .
So we have the Fourier cosine series of ( π / 2 x ) log ( 2 sin x ) but the path does not look promising from here. However, if we replace π / 2 x with a periodic continuation we get the way nicer identity:
(2) ( π / 2 x ) log ( 2 sin x ) = ( n = 1 + sin ( 2 n x ) n ) ( n = 1 + cos ( 2 n x ) n )
that directly leads to:
(3) f ( x ) = ( π / 2 x ) log ( 2 sin x ) = n = 1 + H n 1 n sin ( 2 n x ) .
Now since 0 π / 2 sin ( 2 m x ) d x = 1 m 1 ( mod 2 ) m and 0 π / 2 sin ( 2 a x ) sin ( 2 b x ) d x = π 4 δ a , b , the first two identites are easily proven. Now the three-terms integral
0 π / 2 sin ( 2 a x ) sin ( 2 b x ) sin ( 2 c x ) d x
is a linear combination of 1 a + b + c , 1 a + b + c , 1 a b + c , 1 a + b c depending on the parity of a , b , c, so it is quite difficult to find, explicitly, the Fourier cosine series of , f ( x ) 2 or the integral 0 π / 2 f ( x ) 3 d x, but still not impossible. In particular, we know that the Taylor coefficients of the powers of log ( 1 x ) depends on the generalized harmonic numbers. In our case,
log ( 1 x ) = n = 1 + 1 n x n ,
log ( 1 x ) 2 = n = 2 + 2 H n 1 n x n ,
log ( 1 x ) 3 = n = 3 + 3 H n 1 2 3 H n 1 ( 2 ) n x n ,
(4) log ( 1 x ) 4 = n = 4 + 4 H n 1 3 + 8 H n 1 ( 3 ) 12 H n 1 H n 1 ( 2 ) n x n
hence we can just find a closed form for
0 π / 2 x 3 ( 1 2 cos x ) n d x
and sum everything through the third previous identity. Ugh.
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