Integral ∫ 0 1 ln ⁡ x ln ⁡ ( 1 − z x ) 1 − x d xHow can I evaluate following logarithmic integral: ∫ 0 1 ln ⁡ x ln ⁡ ( 1 − z x ) 1 − x d x

Question

Integral       ∫    0    1              ln      ⁡      x      ln      ⁡      (      1      −      z      x      )              1      −      x        d  xHow can I evaluate following logarithmic integral:      ∫    0    1              ln      ⁡      x      ln      ⁡      (      1      −      z      x      )              1      −      x        d  x

Kaiden Stevens · Accepted Answer

Let  I  (  z  )  =      ∫    0    1              ln      ⁡      x      ln      ⁡      (      1      −      z      x      )              1      −      x           d  x    ;    for      z  &amp;lt;  1then                              I          ′                (        z        )                            =        −                  ∫          0          1                                      x            ln            ⁡            x                                (            1            −            x            )            (            1            −            z            x            )                                   d        x                                          =        −                  1                      1            −            z                                    ∫          0          1                          [                                    x              ln              ⁡              x                                      1              −              x                                −                                    z              x              ln              ⁡              x                                      1              −              z              x                                ]                         d        x                                          =        −                  1                      1            −            z                                    ∫          0          1                          [                      ∑                          n              =              0                        ∞                                x                          n              +              1                                ln          ⁡          x          −                      ∑                          n              =              0                        ∞                    (          z          x                      )                          n              +              1                                ln          ⁡          x          ]                         d        x                                          =                  1                      1            −            z                                    [                      ∑                          n              =              0                        ∞                                1                          (              n              +              2                              )                2                                              −                      ∑                          n              =              0                        ∞                                              z                              n                +                1                                                    (              n              +              2                              )                2                                              ]                                                  =                  1                      1            −            z                                    [                                    π              2                        6                    −          1          −                                                    Li                2                            ⁡              (              z              )                        z                    +          1          ]                                    I        (        z        )                            =        −                              π            2                    6                ln        ⁡        (        1        −        z        )        −        ∫                                            Li              2                        ⁡            (            z            )                                z            (            1            −            z            )                                   d        z                                          =        −                              π            2                    6                ln        ⁡        (        1        −        z        )        −        ∫                                            Li              2                        ⁡            (            z            )                    z                         d        z        −        ∫                                            Li              2                        ⁡            (            z            )                                1            −            z                                   d        z                                          =                  −                                    π              2                        6                    ln          ⁡          (          1          −          z          )          −                      Li            3                    ⁡          (          z          )          −          2                      Li            3                    ⁡          (          1          −          z          )          +          2                      Li            2                    ⁡          (          1          −          z          )          ln          ⁡          (          1          −          z          )                                                                     +                      Li            2                    ⁡          (          z          )          ln          ⁡          (          1          −          z          )          +          ln          ⁡          z                      ln            2                    ⁡          (          1          −          z          )          +          2          ζ          (          3          )                ,            where   I  (  0  )  =  0 implying   C  =  2      Li    3    ⁡  (  1  )  =  2  ζ  (  3  )Notes :  ∫                    Li        k            ⁡      (      x      )        x       d  x  =      Li          k      +      1        ⁡  (  x  )  +  Cand  ∫                    Li        2            ⁡      (      x      )              1      −      x           d  x  =  2      Li    3    ⁡  (  1  −  z  )  −  2      Li    2    ⁡  (  1  −  z  )  ln  ⁡  (  1  −  z  )  −      Li    2    ⁡  (  z  )  ln  ⁡  (  1  −  z  )  −  ln  ⁡  z      ln    2    ⁡  (  1  −  z  )  +  C

Karsyn Stafford · Accepted Answer

It appears that if   z is a negative integer number,   z  =  −  m                    I        =                            −                    ζ        (        2        )        log        ⁡        (        m        +        1        )        +                  1          2                log        ⁡        m                  log          2                ⁡        (        m        +        1        )        −                  1          2                          log          3                ⁡        (        m        +        1        )                                          −                    log        ⁡        (        m        +        1        )                  Li          2                ⁡                  (                      1                          m              +              1                                )                −                  Li          3                ⁡                  (                      1                          m              +              1                                )                +                  Li          3                ⁡                  (                      m                          m              +              1                                )                +        ζ        (        3        )        .            Proof is straightforward through Euler-Landen&#039;s identities.

How can I evaluate following logarithmic integral: int_0^1 (ln x ln ( 1 - zx ))/(1 - x) dx

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