How do you evaluate: ∫ 0 ∞ log ⁡ x e x + e − x + 1 d x

Question

How do you evaluate:       ∫          0              ∞                  log      ⁡      x                      e        x            +              e                  −          x                    +      1               d    x

Eleanor Luna · Accepted Answer

I  =            ∫              0                    ∞                            log        ⁡        x                              e          x                +                  e                      −            x                          +        1                         d        x    =          1      2              ∫              0                    ∞                            ln        ⁡        (        x        )                    cosh        ⁡        x        +                  1          2                      d    x  Consider the &quot;discrete&quot; Laplace transform      ∑          k      =      1              ∞            e          −      k      t        sin  ⁡  (  k  z  )  =      1    2              sin      ⁡      z              cosh      ⁡      t      −      cos      ⁡      z          t  &amp;gt;  0If we put   z  =            2      π        3        ∑          k      =      1              ∞            e          −      k      t        sin  ⁡      (                  2        π        k            3        )    =      1    2                      3            2              cosh      ⁡      t      +              1        2            Therefore                    I        =                              2                      3                                    ∫                      0                                ∞                                    1          2                                                    3                        2                                cosh            ⁡            x            +                          1              2                                      ln        ⁡        (        x        )        d        x                            =                              2                      3                                    ∫                      0                                ∞                                    ∑                      k            =            1                                ∞                                    e                      −            k            x                          sin        ⁡                  (                                    2              π              k                        3                    )                ln        ⁡        (        x        )        d        x                            =                              2                      3                                    ∑                      k            =            1                                ∞                          sin        ⁡                  (                                    2              π              k                        3                    )                          ∫                      0                                ∞                                    e                      −            k            x                          ln        ⁡        (        x        )        d        x            This last integral is the Laplace transform of the   ln  ⁡  (  x  ) function      L        {    ln    ⁡    (    x    )    }    =      ∫          0              ∞            e          −      s      t        ln  ⁡  (  t  )  d  t  =  −            ln      ⁡      (      s      )      +      γ        s      ℜ  (  s  )  &amp;gt;  0where   γ is the Euler constant.Hence                    I        =                              2                      3                                    ∑                      k            =            1                                ∞                          sin        ⁡                  (                                    2              π              k                        3                    )                          ∫                      0                                ∞                                    e                      −            k            x                          ln        ⁡        (        x        )        d        x                            =                            −                  2                      3                                    ∑                      k            =            1                                ∞                                                sin            ⁡                          (                                                2                  π                  k                                3                            )                                      (              ln              ⁡              (              k              )              +              γ              )                                k                                    =                            −                  2                      3                                                                                                ∑                                      k                    =                    1                                                        ∞                                                                                        sin                    ⁡                                          (                                                                        2                          π                          k                                                3                                            )                                        ln                    ⁡                    (                    k                    )                                    k                                            ⏟                                                          S                              1                                                    −                              2            γ                                3                                                                                                ∑                                      k                    =                    1                                                        ∞                                                                                        sin                    ⁡                                          (                                                                        2                          π                          k                                                3                                            )                                                        k                                            ⏟                                                          S                              2                                                        Recall the Kummer&#039;s Fourier series for the log-gamma function:  ln  ⁡  Γ  (  t  )  =      1    2    ln  ⁡      (          π              sin        ⁡        (        π        t        )              )    +      (    γ    +    ln    ⁡    (    2    π    )    )        (          1      2        −    t    )    +      1    π        ∑          k      =      1              ∞                  sin      ⁡      (      2      π      k      t      )      ln      ⁡      (      k      )        k  If we put       t    =          1      3            S          1        =      ∑          k      =      1              ∞                  sin      ⁡      (                        2          π          k                3            )      ln      ⁡      (      k      )        k    =  π  ln  ⁡  Γ      (          1      3        )    −      π    2    ln  ⁡      (                  2        π                    3              )    −            π      ln      ⁡      (      2      π      )        6    −            γ      π        6  For       S          2       take the principal branch of the   ln  ⁡  (  z  ) function      S          2        =      ∑          k      =      1              ∞                  sin      ⁡              (                              2            π            k                    3                )              k    =  ℑ      (          ∑              k        =        1                    ∞                            e                                            2              π              i              k                        3                              k        )    =  ℑ      (    −    ln    ⁡    (    1    −          e                                    2            π            i                    3                      )    )    =  ℑ      (    −                  ln        ⁡        (        3        )            2        +                  i        π            6        )    =      π    6  Therefore  I  =  −      2          3            S          1        −            2      γ              3            S          2        =  −            2      π              3        ln  ⁡  Γ      (          1      3        )    +      π          3        ln  ⁡      (                  2        π                    3              )    +      π          3              3              ln  ⁡  (  2  π  )Therefore, we can conclude                                          ∫                          0                                      ∞                                                          log              ⁡              x                                                      e                x                            +                              e                                  −                  x                                            +              1                                                       d                    x          =                      π                          3                                ln          ⁡                      (                                          2                π                                            3                                      )                    +                      π                          3                              3                                              ln          ⁡          (          2          π          )          −                                    2              π                                      3                                ln          ⁡          Γ                      (                          1              3                        )

How do you evaluate: <msubsup> ∫ <mrow class="MJX-TeXAtom-ORD"> 0

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