Evaluating improper integral ∫ 0 1 log ⁡ ( x ) x + α d x for small positive α

Question

Evaluating improper integral       ∫    0    1              log      ⁡      (      x      )              x      +      α          d  x for small positive   α

alomjabpdl0 · Accepted Answer

∫    0    1              ln      ⁡      (      x      )              a      +      x        d  x  =      ∫    0          1              /            a                  ln      ⁡      (      a      y      )              1      +      y        d  y  =                              ln          ⁡          (          1          +          y          )          ln          ⁡          (          a          y          )                                    |                        0                          1                              /                            a                                      ⏟                    0        −      ∫    0          1              /            a                  ln      ⁡      (      1      +      y      )        y    d  x  =      Li    2    (  −  y  )            |        0          1              /            a        =      Li    2    (  −  1      /    a  )and the result follows on using the dilogarithm inversion formula:      Li    2    (  −  1      /    z  )  =  −            π      2        6    −      1    2        ln    2    ⁡  (  z  )  −      Li    2    (  −  z  )  .To prove the last identity, differentiate       Li    2    (  −  1      /    z  ) then integrate back.In geeneral we have      ∫    0    1                      ln        p            ⁡      (      x      )              a      +      x        d  x  =  (  −  1      )          p      +      1          p  !        Li          p      +      1        (  −  1      /    a  )which follows from the integral represenation of the polylogarithm function      Li          p      +      1        (  z  )  =            (      −      1              )                  p                            p      !            ∫    0    1              z              ln                  p                    ⁡      (      t      )              1      −      z      t        d  tupon replacing z by   −  1      /    a

dikcijom2k · Accepted Answer

Changing the integration variable to   t  :=  x      /    α, we get  I  =      ∫    0          1              /            α                  log      ⁡      α      +      log      ⁡      t              1      +      t          d  t  .The integration of the   log  ⁡  α term is easy:      ∫    0          1              /            α                  log      ⁡      α              1      +      t          d  t  =  log  ⁡  α  log  ⁡            1      +      α        α    .For the   log  ⁡  t term, changing the integration variable to   u  :=  1  +  t                              ∫          0                      1                          /                        α                                                log            ⁡            t                                1            +            t                                  d        t                            =                  ∫          1                      1            +            1                          /                        α                                                log            ⁡            (            u            −            1            )                    u                        d        u                                          =        −                  [          Φ                      (            1            +                          1              α                        )                    −          Φ          (          1          )          ]                    where   Φ is defined as in this Wikipedia article:  Φ  (  x  )  :=  −      ∫    0    x                      |            1      −      u              |              u      d  u  =      {                                        Li            2                    ⁡          (          x          )                          (          x          ≤          1          )                                                                π              2                        3                    −                      Li            2                    ⁡                      (                          1              x                        )                    −                      1            2                    (          log          ⁡          (          x          )                      )            2                                    (          x          &amp;gt;          1          )                        Combining the two and using       Li    2    ⁡  (  1  )  =      π    2        /    6 yield  I  =  −      1    2    (  log  ⁡  α      )    2    −            π      2        6    +      Li    2    ⁡      (          α              1        +        α              )    +      1    2    (  log  ⁡  (  1  +  α  )      )    2    .Finally, this identity:      Li    2    ⁡  (  z  )  +      Li    2    ⁡      (          z              z        −        1              )    =  −      1    2    (  log  ⁡  (  1  −  z  )      )    2  simplifies the last two terms into   −      Li    2    ⁡  (  −  α  ), giving the desired result.

Evaluating improper integral <msubsup> ∫ 0 1 </msubsup>

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