How to calculate integral ∫ 0 ∞ x ( e 2 π x − 1 ) ( x 2 + 1 ) 2 d x ?

Question

How to calculate integral       ∫    0    ∞        x          (              e                  2          π          x                    −      1      )      (              x        2            +      1              )        2              d  x ?

knolsaadme · Accepted Answer

Recall Binet&#039;s second   ln  ⁡  Γ formula:      ∫          0              ∞                  arctan      ⁡      (      t      z      )                      e                  2          π          t                    −      1          d  t  =      1    2    ln  ⁡  Γ      (          1      z        )    +      1    2        (          1      z        −          1      2        )    ln  ⁡  (  z  )  +      1          2      z        −      1    4    ln  ⁡  (  2  π  )Consider the following integral:  ∫      t          (              e                  2          π          t                    −      1      )      (      1      +              t        2                    z        2                    )        2                d  z  =            t      z              2      (      1      +              t        2                    z        2            )      (              e                  2          π          t                    −      1      )        +            arctan      ⁡      (      t      z      )              2      (              e                  2          π          t                    −      1      )      Since            t      z              2      (      1      +              t        2                    z        2            )      (              e                  2          π          t                    −      1      )        =      z    2        ∂          ∂      z            (                  arctan        ⁡        (        t        z        )                              e                      2            π            t                          −        1              )      ⟹        ∫          0              ∞        ∫      t          (              e                  2          π          t                    −      1      )      (      1      +              t        2                    z        2                    )        2                d  z    d  t  =      ∫          0              ∞            z    2        ∂          ∂      z            (                  arctan        ⁡        (        t        z        )                              e                      2            π            t                          −        1              )      d  t  +      ∫          0              ∞                  arctan      ⁡      (      t      z      )              2      (              e                  2          π          t                    −      1      )          d  tUsing Binet&#039;s formula we determine then:      ∫          0              ∞        ∫      t          (              e                  2          π          t                    −      1      )      (      1      +              t        2                    z        2                    )        2                d  z    d  t  =      1          4      z        −      1    8    ln  ⁡  (  2  π  z  )  −            ψ              (                  1          z                )                    4      z        +      1    4    ln  ⁡      (    Γ          (              1        z            )        )    −      1    8  where   ψ is the digamma function.Taking the derivative with respect to z then the limit as   z  →  1 we determine:                                          ∫                          0                                      ∞                                            x                          (                              e                                  2                  π                  x                                            −              1              )              (                              x                2                            +              1                              )                2                                                        d          x          =                                    π              2                        24                    −                      3            8

Feinsn · Accepted Answer

∫                              0                                            ∞                                                                    x                                                      (                                                                                                                                      e                                                                          2                          π                          x                                                                                          −                      1                                                            )                                                                              (                                                                                              x                                                      2                                                                          +                        1                                                                  )                                                              2                                                                                                          d                        x                                              =                                                   1            4                                    [                                −            2                          ∫                              0                                            ∞                                                                                      ℑ                                      (                                                                                                                [                                                                                1                            +                                                          i                                                        x                                                                              ]                                                                          −                          2                                                                                                          )                                                                                                                                e                                                              2                      π                      x                                                                          −                  1                                                                    d                        x                              ]                    The brackets-[] enclosed expression can be evaluated with the Abel-Plana Formula. Namely,                                                              ∫                              0                                            ∞                                                                    x                                                      (                                                                                                                                      e                                                                          2                          π                          x                                                                                          −                      1                                                            )                                                                              (                                                                                              x                                                      2                                                                          +                        1                                                                  )                                                              2                                                                                                          d                        x                                              =                                                   1            4                                    [                                              ∑                              n                =                0                                            ∞                                                                    1                                                      (                                                              1                      +                      n                                                            )                                                        2                                                                        −                          ∫                              0                                            ∞                                                                                                          d                                    n                                                                      (                                                              1                      +                      n                                                            )                                                        2                                                                        −                                                                                                1                    2                                                                                        1                                                                  (                                                                          1                          +                          n                                                                        )                                                                    2                                                                                            |                                            n                                 =                                 0                                                        ]                                    =                                          1            4                                    (                                                                                π                                      2                                                  6                                      −            1            −                                          1                2                                                        )                =                                                                              π                                      2                                                  24                                      −                                          3                8                                                    ≈        0.0362

How to calculate integral <msubsup> ∫ 0 <mi mathvariant="normal">&#x

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