I = ∫ 0 ∞ sin ⁡ ( k x ) x ( x 2 + 1 ) d xWhere k ∈ R + What methods can be employed to solve this integral?

Name: I = ∫ 0 ∞ sin ⁡ ( k x ) x ( x 2 + 1 ) d xWhere k ∈ R + What methods can be employed to solve this integral?
Uploaded: 2022-02-23T17:22:57+00:00
Description: I = ∫ 0 ∞ sin ⁡ ( k x ) x ( x 2 + 1 ) d xWhere k ∈ R + What methods can be employed to solve this integral?

Question

I  =      ∫          0              ∞                  sin      ⁡      (      k      x      )              x              (                  x          2                +        1        )                d  xWhere   k  ∈            R              +      What methods can be employed to solve this integral?

Nodussimj · Accepted Answer

The method I took was:Let  I  (  t  )  =      ∫          0              ∞                  sin      ⁡      (      k      x      t      )              x              (                  x          2                +        1        )                d  xTake the Laplace Transform                              L                          [          I          (          t          )          ]                                    =                  ∫                      0                                ∞                                                              L                                      [              sin              ⁡              (              k              x              t              )              ]                                            x                          (                              x                2                            +              1              )                                              d        x                                          =                  ∫                      0                                ∞                                                k            x                                              (                              k                2                                            x                2                            +                              s                2                            )                        x                          (                              x                2                            +              1              )                                              d        x                                          =                  ∫                      0                                ∞                                    k                                    (                              k                2                                            x                2                            +                              s                2                            )                                      (                              x                2                            +              1              )                                              d        x                                          =                  ∫                      0                                ∞                                    [                                    k              3                                                      (                                  k                  2                                −                                  s                  2                                )                                            (                                  k                  2                                                  x                  2                                +                                  s                  2                                )                                              −                      k                                          (                                  k                  2                                −                                  s                  2                                )                                            (                                  x                  2                                +                1                )                                              ]                        d        x                                          =                              [                                          k                3                                            (                                  k                  2                                −                                  s                  2                                )                                                                    arctan                ⁡                                  (                  k                  x                  )                                                            k                s                                      −                          k                              (                                  k                  2                                −                                  s                  2                                )                                      arctan            ⁡            (            x            )            ]                                0                                ∞                                                            =                              k            2                                s                          (                              k                2                            −                              s                2                            )                                                π          2                −                  k                      (                          k              2                        −                          s              2                        )                                    π          2                                                  =                  k                                    k              2                        −                          s              2                                                [                      k            s                    −          1          ]                          π          2                                                  =                  k                      s                          (              k              +              s              )                                                π          2                    And thus,  I  (  t  )  =            L              −      1            [          k              s                  (          k          +          s          )                            π      2        ]    =      [    1    −          e              −        k        t              ]        π    2  Lastly,  I  =  I  (  1  )  =      [    1    −          e              −        k              ]        π    2

curukksm · Accepted Answer

You can also use differentiation under the integral sign. Let  f  :      R    →      R      ,    f  (  k  )  =      ∫    0    ∞              sin      ⁡      (      k      x      )              x      (      1      +              x        2            )              d    x    .Clearly,   f  (  0  )  =  0 . The dominated convergence theorem yields   f  ∈      C    1    (      R    ) and      f    ′    (  k  )  =      ∫    0    ∞              cos      ⁡      (      k      x      )              1      +              x        2                    d    x    ,so       f    ′    (  0  )  =      π    2   . This integral has already been discussed on this site many times, but in order to keep the answer self-contained we can just differentiate one more time. As suggested here, for k&amp;gt;0 we let kx=t to find      f    ′    (  k  )  =      ∫    0    ∞              k      cos      ⁡      (      t      )                      k        2            +              t        2                    d    t    .Thus we obtain   f  ∈      C    2    (            R        +    ) and                              f          ″                (        k        )                            =        −                  ∫          0          ∞                                                    k              2                        −                          t              2                                            (                          k              2                        +                          t              2                                      )              2                                      cos        ⁡        (        t        )                          d                t        =        −                  ∫          0          ∞                                      t            sin            ⁡            (            t            )                                              k              2                        +                          t              2                                                                d                t        =        −                  ∫          0          ∞                                      x            sin            ⁡            (            k            x            )                                1            +                          x              2                                                        d                x                                          =                  ∫          0          ∞                                      sin            ⁡            (            k            x            )                                x            (            1            +                          x              2                        )                                            d                x        −                  ∫          0          ∞                                      sin            ⁡            (            k            x            )                    x                                  d                x        =        f        (        k        )        −                  π          2                        ,                k        &amp;gt;        0                ,            using the dominated convergence theorem, integration by parts and the known value of the Dirichlet integral. We conclude that   f      |          [      0      ,      ∞      )       is the (unique) solution of the initial value problem      {                            −                      g            ″                    +          g          =                      π            2                                                      i            n                               (          0          ,          ∞          )                                      g          (          0          )          =          0                                                  g            ′                    (          0          )          =                      π            2                                      ,so   f  (  k  )  =      π    2    (  1  −            e              −      k        )    ,    k  ≥  0 . Since f is obviously antisymmetric, the correct continuation to negative values is given by  f  (  k  )  =      π    2    sgn  ⁡  (  k  )  (  1  −            e              −              |            k              |              )    ,    k  ∈      R      .

I=int_0^(oo) (sin(kx))/(x(x^2+1))dx Where k in RR^+ What methods can be employed to solve this integral?

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