I need help computing this integral: (1) ∫ 0 ∞ x cos ⁡ ( a x ) x 2 + 1 coth ⁡ ( s x ) d x

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I need help computing this integral:                    (1)                              ∫          0          ∞                                      x            cos            ⁡            (            a            x            )                                              x              2                        +            1                          coth        ⁡        (        s        x        )                  d                x

percolarse2rzd · Accepted Answer

I  =      ∫          0              ∞                  x      cos      ⁡      (      a      x      )                      x        2            +      1        coth  ⁡  (  s  x  )    d  xRecall the Mittag-Leffler pole expansion of   cot  ⁡  (  z  )  cot  ⁡  (  z  )  =      1    z    +  2  z      ∑          k      =      1              ∞            1                  z        2            −      (      k      π              )        2            Letting   z  →  i  s  x and multiplying by i gives:  coth  ⁡  (  s  x  )  =      1          s      x        +  2  s  x      ∑          k      =      1              ∞            1          (      s      x              )        2            +      (      k      π              )        2            Substituting this expression in, we have then:  I  =      1    s        ∫          0              ∞                  cos      ⁡      (      a      x      )                      x        2            +      1          d  x  +      ∑          k      =      1              ∞            ∫          0              ∞                  2      s              x        2            cos      ⁡      (      a      x      )              (              x        2            +      1      )      (      (      s      x              )        2            +      (      k      π              )        2            )          d  xWe can separate the integrals because each converges individually. The first integral has a well-known result:      1    s        ∫          0              ∞                  cos      ⁡      (      a      x      )                      x        2            +      1          d  x  =            π              e                  −          a                            2      s      The second integral is a little less trivial, but it can be done.Mathematica gives:      ∫          0              ∞                  2      s              x        2            cos      ⁡      (      a      x      )              (              x        2            +      1      )      (      (      s      x              )        2            +      (      k      π              )        2            )          d  x  =                    π        2            k              e                  −                                    π              a              k                        s                              −      π              e                  −          a                    s                      π        2                    k        2            −              s        2            Furthermore, Mathematica gives:      ∑          k      =      1              ∞                          π        2            k              e                  −                                    π              a              k                        s                              −      π              e                  −          a                    s                      π        2                    k        2            −              s        2              =      1    2        e          −                        π          a                s                  (    Φ          (              e                  −                                    a              π                        s                              ,      1      ,      1      −              s        π            )        +    Φ          (              e                  −                                    a              π                        s                              ,      1      ,      1      +              s        π            )        )    −            π              e                  −          a                            2      s        +      π    2        e          −      a        cot  ⁡  (  s  )Where   Φ is the Lerch transcendent.Thus summing the two results, we have:  I  =      1    2        e          −                        π          a                s                  (    Φ          (              e                  −                                    a              π                        s                              ,      1      ,      1      −              s        π            )        +    Φ          (              e                  −                                    a              π                        s                              ,      1      ,      1      +              s        π            )        )    +      π    2        e          −      a        cot  ⁡  (  s  )However, this expression does not hold for s being an integer multiple of   π due to   cot  ⁡  (  s  ) causing it to be an indeterminate when one takes the limit, so there needs to be some finesse for evaluating these cases. I&#039;m not sure if there is simplification possible for the result I obtained for I involving   Φ.

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