Closed form for ∫ 0 1 2 − x ( 1 − x ) x log ⁡ ( ( 2 − x ) x 1 − x ) d xIs it possible to find a closed form for this integral? Q = ∫ 0 1 2 − x ( 1 − x ) x log ⁡ ( ( 2 − x ) x 1 − x ) d x

Question

Closed form for       ∫    0    1                      2        −        x                    (        1        −        x        )                x                log  ⁡      (                  (        2        −        x        )                x                    1        −        x              )    d  xIs it possible to find a closed form for this integral?  Q  =      ∫    0    1                      2        −        x                    (        1        −        x        )                x                log  ⁡      (                  (        2        −        x        )                x                    1        −        x              )    d  x

Martha Mcclain · Accepted Answer

First, we transform the integral into a more computable form by using some substitutions.                              Q                                    =                  ∫          0          1                                                    2              −              x                                      (              1              −              x              )                            x                                              log        ⁡                  (                                    (              2              −              x              )                            x                                      1              −              x                                )                d        x                                          =                  ∫          0          1                                                    1              +              u                                      u              (              1              −              u              )                                      log        ⁡                  (                                    (              1              +              u              )              (              1              −              u              )                        u                    )                d        u                          where           u          =          1          −          x                                                  =                  ∫          0          1                                      1            +            u                                u            (            1            −                          u              2                        )                          log        ⁡                  (                                    1              −                              u                2                                      u                    )                d        u                                          =                  1          2                          ∫          0          1                                      1            +                          t                                                          t                                                3                  4                                                                    1              −              t                                      log        ⁡                  (                                    1              −              t                                      t                                )                d        t                          where           t          =                      u            2                                                            =                  1          2                          ∫          0          1                                      log            ⁡            (            1            −            t            )                                              t                                                3                  4                                                                    1              −              t                                      d        t        −                  1          4                          ∫          0          1                                      log            ⁡            (            t            )                                              t                                                3                  4                                                                    1              −              t                                      d        t                            (1)                                          +                  1          2                          ∫          0          1                                      log            ⁡            (            1            −            t            )                                              t                                                1                  4                                                                    1              −              t                                      d        t        −                  1          4                          ∫          0          1                                      log            ⁡            (            t            )                                              t                                                1                  4                                                                    1              −              t                                      d        t            These four integrals can be evaluated by calculating derivatives of beta function in terms of digamma function. For e.g.                              ∫          0          1                                      log            ⁡            (            1            −            t            )                                              t                                                3                  4                                                                    1              −              t                                      d        t                            =                  d                      d            z                                                {                          ∫              0              1                                      t                              −                                  3                  4                                                      (            1            −            t                          )                              z                −                1                                                  d            t            }                                z            =                          1              2                                                                        =                  d                      d            z                                                {                                          Γ                                  (                                      1                    4                                    )                                Γ                (                z                )                                            Γ                                  (                                      1                    4                                    +                  z                  )                                                      }                                z            =                          1              2                                                                        =                              Γ                          (                              1                4                            )                                      π                                            Γ                          (                              3                4                            )                                                {                      ψ            0                                (                          1              2                        )                    −                      ψ            0                                (                          3              4                        )                    }                                    (2)                                  =                  π                      3                          /                        2                                                2                                Γ                                          (                                  3                  4                                )                            2                                                {          log          ⁡          2          −                      π            2                    }                    To get the last expression, I used the special values                              ψ          0                          (                      3            4                    )                                    =        −        γ        +                  π          2                −        3        log        ⁡        2                                      ψ          0                          (                      1            2                    )                                    =        −        γ        −        2        log        ⁡        2            Using the same technique, the other three integrals can be evaluated:                    (3)                              ∫          0          1                                      log            ⁡            (            t            )                                              t                                                3                  4                                                                    1              −              t                                      d        t                            =        −                  π                      5                          /                        2                                                2                                Γ                                          (                                  3                  4                                )                            2                                                          (4)                              ∫          0          1                                      log            ⁡            (            t            )                                              t                                                1                  4                                                                    1              −              t                                      d        t                            =                              (            4            π            −            16            )            Γ                                          (                                  3                  4                                )                            2                                            2            π                                              (5)                              ∫          0          1                                      log            ⁡            (            1            −            t            )                                              t                                                1                  4                                                                    1              −              t                                      d        t                            =                              2            (            −            8            +            π            +            2            log            ⁡            2            )            Γ                                          (                                  3                  4                                )                            2                                            2            π                              Substituting the results of equations   (  2  )  ,  (  3  )  ,  (  4  ) and   (  5  ) in   (  1  ) gives   Q  =            Γ                        (                      3            4                    )                          −          2                            π        2            log      ⁡      2      −      Γ                        (                      3            4                    )                2            (      4      −      2      log      ⁡      2      )              2            π

Jude Hunt · Accepted Answer

Q  =            Γ                        (                      3            4                    )                          −          2                            π        2            ln      ⁡      2      −      Γ                        (                      3            4                    )                2            (      4      −      ln      ⁡      4      )              2            π

Closed form for <msubsup> ∫ 0 1 </msubsup> <msqrt>

Answered question

Answer & Explanation

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