What is the Laplace transform of ∫ 0 t sin ⁡ x x d xI'm thinking about approaching it as a convolution but I am not sure how. Could I define it as the convolution of 1 and sin ⁡ x x ?

Question

What is the Laplace transform of      ∫    0    t              sin      ⁡      x        x    d  xI&#039;m thinking about approaching it as a convolution but I am not sure how. Could I define it as the convolution of 1 and             sin      ⁡      x        x  ?

driogairea1 · Accepted Answer

We need  F  (  s  )  =      ∫    0          ∞            e          −      s      t            ∫    0    t                      sin        ⁡        (        x        )            x        d  x  d  t  =      ∫    0          ∞            ∫    x          ∞            e          −      s      t                          sin        ⁡        (        x        )            x        d  t  d  x  =      ∫    0          ∞                          sin        ⁡        (        x        )            x        ⋅                    e                  −          s          x                    s        d  xWe have      F    ′    (  s  )  =      ∫    0          ∞                          sin        ⁡        (        x        )            x        ⋅                    −        x                  e                      −            s            x                              s        d  x  −      ∫    0          ∞                          sin        ⁡        (        x        )            x        ⋅                    e                  −          s          x                            s        2              d  xThis gives us                    s                  F          ′                +        F                            =        −                  ∫          0                      ∞                                    e                      −            s            x                          sin        ⁡        (        x        )        d        x        =        −                              1                                          s                2                            +              1                                              ⟹                                      d                          d              s                                                (          s          F          (          s          )          )                =        −                              1                                          s                2                            +              1                                                              s        F        (        s        )        −                  lim                      y            →            0                          y        F        (        y        )                            =        −                  ∫          0          s                                                    d              t                                                      t                2                            +              1                                      =        −        arctan        ⁡        (        s        )            We have      lim          y      →      0        y  F  (  y  )  =      lim          y      →      0            ∫    0          ∞                          sin        ⁡        (        x        )            x            e          −      y      x        d  x  =      ∫    0          ∞            lim          y      →      0            e          −      y      x                          sin        ⁡        (        x        )            x        d  x  =      ∫    0          ∞                          sin        ⁡        (        x        )            x        d  x  =            π      2      Hence,                                          F            (            s            )            =                                                                                                      π                      2                                                        −                  arctan                  ⁡                  (                  s                  )                                s                                      =                                          1                s                                      arctan            ⁡                          (                                                1                  s                                            )

Paloma Sanford · Accepted Answer

Integrate by parts:      ∫    0          ∞            e          −      s      x            (          ∫      0      x                      sin        ⁡                  t                    t            d    t    )      d  x  =            [      −                        e                      −            s            x                          s                    (                  ∫          0          x                                      sin            ⁡                          t                                t                        d        t        )            ]              0              ∞        +      1    s        ∫    0          ∞            e          −      s      x                  sin      ⁡              x              x      d  xThe first term is obviously zero since the integral converges. Now consider  A  (  s  )  =      ∫    0          ∞            e          −      s      x                  sin      ⁡              x              x      d  x  .We clearly have      A    ′    (  s  )  =  −      ∫    0          ∞            e          −      s      x        sin  ⁡      x      d  x  .It is well-known how to do this (parts, complex exponentials...), and we find      A    ′    (  s  )  =  −      1          1      +              s        2            Then  A  (  s  )  =  A  (  ∞  )  +      ∫    s          ∞                  d      t              1      +              t        2              =  arctan  ⁡            (              1        s            )        .Hence we have the Laplace Transform as      1    s    arctan  ⁡            (              1        s            )        .

What is the Laplace transform of int_0^t (sin x)/(x) dx

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