Is there a closed-form solution for the integral ∫ 0 ∞ log 2 ⁡ ( 1 + a x ) ⋅ e − b x d xwith a , b ≥ 0?

Question

Is there a closed-form solution for the integral      ∫          0              ∞            log          2        ⁡  (  1  +  a  x  )  ⋅      e          −      b      x              d    xwith   a  ,  b  ≥  0?

Haylie Park · Accepted Answer

You may express your integral in terms of a known special function, here in terms of the incomplete gamma function defined by  Γ  (  α  ,  z  )  :=      ∫    α    ∞        u          z      −      1            e          −      u        d  uIndeed, you have                              ∫                      0                                ∞                                    log                      2                          ⁡        (        1        +        a        x        )        ⋅                  e                      −            b            x                                            d                x                            =                              e                          b                              /                            a                                            a            ln            ⁡            2                                    ∫          1                      ∞                          ln        ⁡        v        ⋅                  e                      −            b            u                          /                        a                                            d                v                (        v        =        1        +        a        x        )                                                        =                              [                                          e                                  b                                      /                                    a                                                            a                ln                ⁡                2                                      ln            ⁡            v                          (              −                              a                b                                            e                                  −                  b                  v                                      /                                    a                                            )                        ]                    1                      ∞                          +                              e                          b                              /                            a                                            b            ln            ⁡            2                                    ∫          1                      ∞                                    v                      −            1                          ⋅                  e                      −            b            v                          /                        a                                            d                v                                                        =                              e                          b                              /                            a                                            b            ln            ⁡            2                                    ∫                      b                          /                        a                                ∞                                    u                      0            −            1                          ⋅                  e                      −            u                                            d                u                (        u        =        b        v                  /                a        )                                                        =                              e                          b                              /                            a                                            b            ln            ⁡            2                                  Γ                  (          0          ,                      b            a                    )                .

Is there a closed-form solution for the integral int_0^(oo) log_2 (1+ax) * e^(-bx)dx with a,b >=0?

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Answer & Explanation

New Questions in Calculus 1