How to evaluate the integral ∫ a ∞ x ( 1 − exp ⁡ ( − x − a ) ) d x

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How to evaluate the integral       ∫    a    ∞    x            (        1  −  exp  ⁡  (  −      x          −      a        )            )        d  x

concludirgt · Accepted Answer

Denoting the integral as   I  =      ∫    a    ∞    x            (        1  −      e          −              x                  −          a                                )              d    x, we can take the substitution   t  =      x          −      a      . We get       d    t  =  −      a          x              a        +        1                  d    x which gives  I  =      1    a        ∫          0                      a                  −          a                                1      −              e                  −          t                            t              1        +                  2          a                          d    t  =      1    a        ∫          0                      a                  −          a                          (    1    −          e              −        t              )              (      −              a                  2                      t                                          2                a                                                        )        ′        d    tThe above integral is in perfect form to apply integration by parts. We thus get   I  =      1          a            [                  a            2              (                                    e                          −              t                                −          1                          t                                    2              a                                          )                                |                            0                              a                      −            a                                +                  a            2              ∫              0                                          a                          −              a                                                  t                        (                      1            −                          2              a                                )                −        1                    e              −        t                        d        t    ]  Recalling that the lower incomplete Gamma function is given by   γ  (      α    ,      x    )  :=      ∫          0                      x                  t                  α            −      1            e          −      t              d    t we get  I  =            a      2        2        (          e              −                  a                      −            a                                −    1    )    +      1    2                                            lim                          t              →                              0                                  +                                                                                        1              −                              e                                  −                  t                                                                    t                                                2                  a                                                                    ⏟                            L              +      1    2    γ      (    1    −          2      a        ,          a              −        a              )  We just need to evaluate the limit L to conclude the problem, and this is the part that will give us the restriction on the possible values of a. By applying L&#039;Hôpital&#039;s rule we get  L  =      a    2        lim          t      →              0                  +                                e              −        t                    t                                    2            a                    −          1                    And here the value of the limit entirely depends on if                     2        a            −      1       is positive, negative or 0. If       2    a    −  1  &amp;gt;  0, then we have a       1    0   situation and the limit is infinity, so I diverges. If we have       2    a    −  1  =  0 then   γ      (    1    −          2      a        ,          a              −        a              )    =  γ  (  0  ,      2          −      2        ) diverges, so here I again diverges. If             2      a        −    1    &amp;lt;    0   then       1          t                        2          a                −        1              →  0 as   t  →      0    +  and thus L=0. We can finally conclude that  I  =                                                        a              2                        2                                (                          e                              −                                  a                                      −                    a                                                                        −            1            )                    +                      1            2                    γ                      (            1            −                          2              a                        ,                          a                              −                a                                      )                    ,                                a            &amp;gt;            2

How to evaluate the integral <msubsup> ∫ a <mi mathvariant="normal">

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