I would like to evaluate the integral below ∫ 0 ∞ sin 3 ⁡ x cosh ⁡ x + cos ⁡ x d x x

Question

I would like to evaluate the integral below      ∫    0          ∞                          sin        3            ⁡      x              cosh      ⁡      x      +      cos      ⁡      x                  d      x        x

scipionhi · Accepted Answer

Since            sin      ⁡      (      x      )              cosh      ⁡      (      x      )      +      cos      ⁡      (      x      )        =  2      ∑          n      ≥      1        (  −  1      )          n      −      1            e          −      n      x        sin  ⁡  (  n  x  )we can combine the above with      sin    2    ⁡  (  x  )  sin  ⁡  (  n  x  )  =      1    4        [    2    sin    ⁡    (    n    x    )    +    sin    ⁡    (    (    2    −    n    )    x    )    −    sin    ⁡    (    (    2    +    n    )    x    )    ]   and       ∫    0    ∞        e          −      a      x                  sin      ⁡              (        b        x        )              x        d    x  =  arctan  ⁡      (          b      a        )   to get                    I                            =                  1          2                          ∑                      n            ≥            1                          (        −        1                  )                      n            −            1                                    ∫                      0                                ∞                                                e                          −              n              x                                x                          [          2          sin          ⁡          (          n          x          )          +          sin          ⁡          (          (          2          −          n          )          x          )          −          sin          ⁡          (          (          2          +          n          )          x          )          ]                                  d                x                                          =                  1          2                          ∑                      n            ≥            1                          (        −        1                  )                      n            −            1                                    [                      π            2                    +          arctan          ⁡                      (                                          2                −                n                            n                        )                    −          arctan          ⁡                      (                                          2                +                n                            n                        )                    ]                                                  =                  1          2                          ∑                      n            ≥            1                          (        −        1                  )                      n            −            1                                    [          arctan          ⁡          (          n          +          1          )          −          arctan          ⁡          (          n          −          1          )          ]                                                  =                  1          2                          [          −          arctan          ⁡          (          0          )          +          arctan          ⁡          (          1          )          ]                                                  =                  π          8                    as desired.

pokoljitef2 · Accepted Answer

It was shown in this answer that if n is a nonnegative integer, then      ∫          0              ∞                  sin      ⁡              (        (        2        n        +        1        )        x        )                    cosh      ⁡      (      x      )      +      cos      ⁡      (      x      )                              d            x        x    =      π    4    .Therefore,                              ∫                      0                                ∞                                                              sin                              3                                      ⁡            (            x            )                                cosh            ⁡            (            x            )            +            cos            ⁡            (            x            )                                                                      d                        x                    x                                    =                  1          4                          ∫                      0                                ∞                                                3            sin            ⁡            (            x            )            −            sin            ⁡            (            3            x            )                                cosh            ⁡            (            x            )            +            cos            ⁡            (            x            )                                                                      d                        x                    x                                                  =                  1          4                          (                                    3              π                        4                    −                      π            4                    )                                                  =                  π          8                .            In general, if m is a nonnegative integer, then                              ∫                      0                                ∞                                                              sin                              2                m                +                1                                      ⁡            (            x            )                                cosh            ⁡            (            x            )            +            cos            ⁡            (            x            )                                                                      d                        x                    x                                              =                      (            1            )                                    1                      2                          2              m                                                ∑                      k            =            0                                m                          (        −        1                  )                      m            −            k                                                              (                                                          2              m              +              1                        k                                              )                                                ∫                      0                                ∞                                                sin            ⁡                          (              (              2              m              +              1              −              2              k              )              x              )                                            cosh            ⁡            (            x            )            +            cos            ⁡            (            x            )                                                              d                        x                    x                                                  =                  π                      2                          2              m              +              2                                                ∑                      k            =            0                                m                          (        −        1                  )                      m            −            k                                                              (                                                          2              m              +              1                        k                                              )                                      .

I would like to evaluate the integral below <msubsup> ∫ 0 <mrow cla

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