Integral ∫ 0 ∞ 1 x 3 ( 1 + log ⁡ 1 + e x − 1 1 + e x ) d xIs it possible to evaluate this integral in a closed form?

Question

Integral       ∫    0    ∞        1          x      3            (    1    +    log    ⁡                  1        +                  e                      x            −            1                                      1        +                  e          x                      )    d  xIs it possible to evaluate this integral in a closed form?

Govorei9b · Accepted Answer

We first remark the following identity:                              ∫                      0                                ∞                                                x                          s              −              1                                            z                          e                              x                                      −            1                                  d        x                            =                  ∫                      0                                ∞                                                              z                              −                1                                                    x                              s                −                1                                                    e                              −                x                                                          1            −                          z                              −                1                                                    e                              −                x                                                            d        x                                          =                  ∑                      n            =            1                                ∞                                    z                      −            n                                    ∫                      0                                ∞                                    x                      s            −            1                                    e                      −            n            x                                  d        x                                          =        Γ        (        s        )                  ∑                      n            =            1                                ∞                                                z                          −              n                                            n                          s                                      =        Γ        (        s        )                              L            i                                s                          (                  z                      −            1                          )            which initially holds for       |    z      |    &amp;gt;  1, and then extends holomorphically for       z          −      1        ∉  (  1  ,  ∞  ] since both sides define holomorphic functions on this region. Now, by integrating by parts,                              ∫                      0                                ∞                                    1                      x            3                                    (          1          +          log          ⁡                                    1              +                              e                                  x                  −                  1                                                                    1              +                              e                                  x                                                              )                        d        x                            =                  3          2                          ∫                      0                                ∞                                    x                      2                          /                        3                                    (                      1                          (              −              1              )                              e                                  x                                            −              1                                −                      1                          (              −                              e                                  −                  1                                            )                              e                                  x                                            −              1                                )                        d        x                                          =                  3          2                Γ                  (                      5            3                    )                          {                                    L              i                                      5                              /                            3                                (          −          1          )          −                                    L              i                                      5                              /                            3                                (          −          e          )          }                                                  =        −                  3          2                Γ                  (                      5            3                    )                          {          (          1          −                      2                          −              2                              /                            3                                )          ζ          (          5                      /                    3          )          +                                    L              i                                      5                              /                            3                                (          −          e          )          }                .            I can hardly believe that             L      i              5              /            3        (  −  e  ) can be simplified further.

Integral ∫ 0 ∞ 1 x 3 ( 1 + log ⁡ 1 + e...

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