Find the Laplace transform of the function H ( t − a ) t n . That's what I have done so far: L { H ( t − a ) t n } = ∫ 0 ∞ e − s t H ( t − a ) t n d t = ∫ a ∞ e − s t t n d t = ∫ 0 ∞ e − s ( t + a ) ( t + a ) n d tCould you tell me how I can continue?

Question

Find the Laplace transform of the function   H  (  t  −  a  )      t          n      . That&#039;s what I have done so far:  L  {  H  (  t  −  a  )      t          n        }  =      ∫          0              ∞            e          −      s      t        H  (  t  −  a  )      t          n        d  t  =      ∫          a              ∞            e          −      s      t            t          n        d  t  =      ∫          0              ∞            e          −      s      (      t      +      a      )        (  t  +  a      )          n        d  tCould you tell me how I can continue?

t5an1izqp · Accepted Answer

You can obtain directly from the integral      L    {  H  (  t  −  a  )      t          n        }  =      ∫          a              ∞            e          −      s      t            t          n        d  t  =      1          s              n        +        1                  ∫          a      s              ∞            e          −      z            z          n        d  z  =      1          s              n        +        1              Γ  (  n  +  1  ,  a  s  )recalling the integral expression of incomplete Gamma function for integer n  Γ  (  n  ,  x  )  =      ∫          x              ∞            e          −      z            z          n      −      1        d  z  .NOTE Another useful methodology may be the following.Observing that  H  (  t  −  a  )      t    n    =  H  (  t  −  a  )  (  t  −  a  +  a      )    n    =  H  (  t  −  a  )      ∑          k      =      0        n                      (                    n      k                      )              (  t  −  a      )    k        a          n      −      k      the Laplace transform becomes      L    {  H  (  t  −  a  )      t    n    }  =      ∑          k      =      0        n                      (                    n      k                      )                  a          n      −      k            L    {  H  (  t  −  a  )  (  t  −  a      )    k    }recalling that       L    {  H  (  t  −  a  )  f  (  t  −  a  )  }  =      e          −      a      s        F  (  s  ). Using       L    {      t    n    }  =            n      !              s              n        +        1            , we obtain      L    {  H  (  t  −  a  )      t    n    }  =      e          −      a      s            ∑          k      =      0        n                      (                    n      k                      )                  a          n      −      k                  k      !              s              k        +        1              =            n      !              e                  −          a          s                            s              n        +        1                  ∑          k      =      0        n              (      a      s              )                  n          −          k                            (      n      −      k      )      !        =      1          s              n        +        1              n  !      e          −      a      s            ∑          ν      =      0        n              (      a      s              )                  ν                            ν      !      and recalling that  Γ  (  n  ,  z  )  =  (  n  −  1  )  !      e          −      z            ∑          ν      =      0              n      −      1                  z              ν                    ν      !      is the incomplete Gamma function for integer n, we obtain finally      L    {  H  (  t  −  a  )      t    n    }  =      1          s              n        +        1              Γ  (  n  +  1  ,  a  s  )  .

Pellagra3d · Accepted Answer

You know that       L    {      U    (  t  −  a  )  }  =            e              −        a        s              s    ,  a  &amp;gt;  0. And      L    {      t    n    f  (  t  )  }  =  (  −  1      )    n              (            L    (  f  (  t  )  )                    )                    (      n      )      Now let think about             (                  e              −        a        s              s                      )                    (      n      )      . Using Maple, I found:  n  =  1  →            (                        e                      −            a            s                          s            )        ′    =  −  exp  ⁡  (  −  a  s  )            a      s      +      1              s      2          n  =  2  →            (                        e                      −            a            s                          s            )        ″    =  exp  ⁡  (  −  a  s  )                    a        2                    s        2            +      2      a      s      +      2              s      3          n  =  3  →            (                        e                      −            a            s                          s            )        ‴    =  −  exp  ⁡  (  −  a  s  )                    a        3                    s        3            +      3              a        2                    s        2            +      6      a      s      +      6              s      4      And so I think the possible pattern could be  (  −  1      )    n                      ∑                  k          =          0                          n                                      n          !                          (          n          −          k          )          !                            s                  n          −          k                            a                  n          −          k                            s              n        +        1

Find the Laplace transform of the function H(t−a)t^n

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