Integral ∫ 0 1 x 9 ( x 4 + x 2 − x − 1 − 5 ln ⁡ x ) ( x 10 − 1 ) ln ⁡ x d x

Question

Integral       ∫    0    1                      x        9                    (                  x          4                +                  x          2                −        x        −        1        −        5        ln        ⁡        x        )                            (                  x                      10                          −        1        )            ln      ⁡      x            d    x

Louis Lawrence · Accepted Answer

Let&#039;s denote the integral in question as                    (1)                    I        =                  ∫          0          1                                                    x              9                                      (                              x                4                            +                              x                2                            −              x              −              1              −              5              ln              ⁡              x              )                                                          (                              x                                  10                                            −              1              )                        ln            ⁡            x                          d        x        .            Changing the variable   x  =      y          1              /            10       and renaming y back to x we get                    (2)                    I        =                  ∫          0          1                                                    x                              2                                  /                                5                                      +                          x                              1                                  /                                5                                      −                          x                              1                                  /                                10                                      −            1            −            ln            ⁡                          x                                            (            x            −            1            )            ln            ⁡            x                          d        x        .            Some elementary transformations show that                    (3)                    I        =                  J                (        2                  /                5        )        +                  J                (        1                  /                5        )        −                  J                (        1                  /                10        )        ,            where we introduced notation                    (4)                              J                (        q        )        =                  ∫          0          1                                                    x              q                        −            1            −            q                        ln            ⁡            x                                (            x            −            1            )            ln            ⁡            x                          d        x        .            The integral       J    (  q  ) can be evaluated as follows:                    (5)                                                                        J                            (              q              )              =                              ∫                0                1                                            ∫                0                q                                                                                  x                    p                                    −                  1                                                  x                  −                  1                                            d              p                            d              x              =                              ∫                0                q                                                                                                                                ∫                        0                        1                                                                                                                          x                            p                                                    −                          1                                                                          x                          −                          1                                                                    d                      x                                        ⏟                                                                    DLMF 5.9.16                                                          d              p                                                          =                              ∫                0                q                                            H                p                                          d              p              =              q              ⋅              γ              +              ln              ⁡              Γ              (              q              +              1              )              ,                                          where       H    p   are harmonic numbers:     +    and        ψ    0   is the digamma function:       ψ    0    (  x  )  =      d          d      x        ln   . Let me mention that the formula DLMF 5.9.16 becomes particularly obvious for positive integer p, when        H    p    =      ∑          n      =      1        p        n          −      1      Pluging (5) back into (3), we get                    (6)                    I        =                  1          2                γ        +        ln        ⁡                  4          5                +        ln        ⁡                              Γ                          (                              1                5                            )                        Γ                          (                              2                5                            )                                            Γ                          (                              1                10                            )                                      .             From the formula (74) on this MathWorld page we know that                    (7)                                          Γ                          (                              1                5                            )                        Γ                          (                              2                5                            )                                            Γ                          (                              1                10                            )                                      =                                            2              5                                                  π                                                          5              4                                                  ϕ                                      .            (see the paper Raimundas Vidūnas, Expressions for values of the gamma function for a proof).Making use of this formula, we get the final result                    (8)                    I        =                  1          2                γ        +                  11          5                ln        ⁡        2        −                  5          4                ln        ⁡        5        +                  1          2                ln        ⁡        π        −                  1          2                ln        ⁡        ϕ        .

Integral <msubsup> ∫ 0 1 </msubsup>

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