Antiderivative of x a e ( b x ) I need to know the primitive function (Antiderivative) of this function: f ( x ) = x a e ( b x ) where a and b is a positive constants please how could I find the primitive of function ? is there any technique concerning this types?

Question

Antiderivative of       x    a        e          (      b      x      )      I need to know the primitive function (Antiderivative) of this function:  f  (  x  )  =      x    a        e          (      b      x      )       where a and b is a positive constants please how could I find the primitive of function ? is there any technique concerning this types?

grindweg1v · Accepted Answer

Explanation:Note that if   a  ∈      N  , then   f  (  x  )  =      x    a        e          (      b      x      )      .can be written as   f  (  x  )  =            ∂      a              ∂              b        a                  e          b      x        ..So you can first find an an antiderivative of       e          b      x       and then differentiate it a times w.r.t b. Otherwise you need the incomplete gamma function as said by others.

dglennuo · Accepted Answer

Step 1The fundamental theorem of calculus states                    (1)                              d                      d            x                                    (                      ∫                          g              (              x              )                                      ∞                                h          (          t          )                    d          t          )                =        −        h        (        g        (        x        )        )                  g          ′                (        x        )        .            The definition of the Incomplete gamma function is                     (2)                    Γ        (        a        ,        x        )        =                  ∫          x                      ∞                                    t                      a            −            1                                            e                      −            t                                  d        t        .            Combining these two gives                    (3)                              d                      d            x                          Γ        (        a        +        1        ,        −        b        x        )                                            =                                      (              2              )                                                d                      d            x                                    (                      ∫                          −              b              x                                      ∞                                            t                          a                                                      e                          −              t                                          d          t          )                                                    =                                      (              1              )                                      −        (        −        b        x                  )          a                          e                      b            x                          (        −        b        )        =        b        (        −        b                  )                      a                                    x          a                          e                      b            x                          .            Step 2Therefore                    ∫                  x          a                          e                      b            x                                  d        x                            =        ∫        b        (        −        b                  )                      a                                    x          a                          e                      b            x                                    b                      −            1                          (        −        b                  )                      −            a                                  d        x                                                                              =                                      (              3              )                                                b                      −            1                          (        −        b                  )                      −            a                          ∫                  (                      d                          d              x                                Γ          (          a          +          1          ,          −          b          x          )          )                        d        x                                          =                  b                      −            1                          (        −        b                  )                      −            a                          Γ        (        a        +        1        ,        −        b        x        )        +        C        .

Antiderivative of x a </msup> e ( b x )

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