A closed form for ∫ 0 ∞ ln ⁡ x ⋅ ln ⁡ ( 1 + 1 2 cosh ⁡ x ) d xIs it possible to evaluate this integral in a closed form? ∫ 0 ∞ ln ⁡ x ⋅ ln ⁡ ( 1 + 1 2 cosh ⁡ x ) d x = ∫ 0 ∞ ln ⁡ x ⋅ ln ⁡ ( 1 + 1 e − x + e x ) d xI tried to evaluate it with a CAS, and looked up in integral tables, but was not successful.

Question

A closed form for       ∫    0    ∞    ln  ⁡  x  ⋅  ln  ⁡      (    1    +          1              2        cosh        ⁡        x              )    d  xIs it possible to evaluate this integral in a closed form?      ∫    0    ∞    ln  ⁡  x  ⋅  ln  ⁡      (    1    +          1              2        cosh        ⁡        x              )    d  x  =      ∫    0    ∞    ln  ⁡  x  ⋅  ln  ⁡      (    1    +          1                        e                      −            x                          +                  e          x                      )    d  xI tried to evaluate it with a CAS, and looked up in integral tables, but was not successful.

kejohananws · Accepted Answer

Define  I  (  α  )  =      ∫          0              ∞        log  ⁡  x  log  ⁡  (  1  −      e          −      α      x        )    d  x  .Integrating by parts, followed by the substitution   α  x  ↦  x, we have                    I        (        α        )                            =        α                  ∫                      0                                ∞                                                x            −            x            log            ⁡            x                                              e                              α                x                                      −            1                                  d        x                                          =                  1          α                          ∫                      0                                ∞                                                (            1            +            log            ⁡            α            )            x            −            x            log            ⁡            x                                              e                              x                                      −            1                                  d        x                                          =                  1          α                          {          (          1          +          log          ⁡          α          )          ζ          (          2          )          −                                                                                    d                  ζ                  (                  s                  )                  Γ                  (                  s                  )                                s                            |                                      s              =              2                                }                                                  =                  1          α                          {          (          γ          +          log          ⁡          α          )          ζ          (          2          )          −                      ζ            ′                    (          2          )          }                .            Then it follows that                              ∫                      0                                ∞                          log        ⁡        x        log        ⁡                  (          1          +                      1                          2              cosh              ⁡              x                                )                        d        x                            =                  ∫                      0                                ∞                          log        ⁡        x        log        ⁡                  (                                    1              −                              e                                  −                  3                  x                                                                    1              −                              e                                  −                  x                                                              ⋅                                    1              −                              e                                  −                  2                  x                                                                    1              −                              e                                  −                  4                  x                                                              )                        d        x                                          =        I        (        2        )        +        I        (        3        )        −        I        (        1        )        −        I        (        4        )                                          =                  5          12                          ζ          ′                (        2        )        −                  5          72                γ                  π                      2                          +                  1          18                          π                      2                          log        ⁡        (        3        )        .            Addendum - Something you might want to know:The following identity played the key role in this proof.      ∫          0              ∞                  x              s        −        1                            e                  x                    −      1          d  x  =  Γ  (  s  )  ζ  (  s  )  ,which holds for   ℜ  s  &amp;gt;  1

A closed form for <msubsup> ∫ 0 <mi mathvariant="normal">&#x221E

Answered question

Answer & Explanation

New Questions in Algebra II