A closed form for ∫ 0 π / 2 x 3 ln 3 ⁡ ( 2 cos ⁡ x ) d xWe 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

Question

A closed form for       ∫          0              π              /            2            x    3        ln    3    ⁡  (  2  cos  ⁡  x  )        d    xWe 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

Ashly Sanford · Accepted Answer

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   ℓ be any positive integer. Then      ∫          0              π              /            2                x          2      ℓ      +      1              ln          2      ℓ      +      1        ⁡  (  2  cos  ⁡  x  )            d        x  ∈      Q          (    ζ    (    4    ℓ    +    3    )    ,    ζ    (    2    )    ζ    (    4    ℓ    +    1    )    ,    .    .    .    ,    ζ    (    4    ℓ    )    ζ    (    3    )    ,    ζ    (                  2        ℓ        +        3            ¯        ,    {    1          }              2        ℓ              )    )  It is remarkable that there is only one constant  ζ  (            2      ℓ      +      3        ¯    ,  {  1      }          2      ℓ        )  =      ∑                  n                  1                    &amp;gt;      .      .      .      &amp;gt;              n                  2          ℓ          +          1                    &amp;gt;      0                  (      −      1              )                              n            1                                              n        1                  2          ℓ          +          3                            n        2            ⋯              n                  2          ℓ          +          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).

easternerjx · Accepted Answer

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  xis 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  xand sum everything through the third previous identity. Ugh.

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

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