Recursion Formulas of ∫ x α ln ⁡ x dx and ∫ ln β ⁡ x x dx

Question

Recursion Formulas of    ∫      x          α        ln  ⁡  x     dx and   ∫                    ln                  β                    ⁡      x        x       dx

Zoe Andersen · Accepted Answer

For   α  =  −  1, we have                    ∫                  x                      −            1                          ln        ⁡        (        x        )        d        x                            =                                                            ∫                                                                            ln                      ⁡                      (                      x                      )                                        x                                                  d                x                =                ∫                t                d                t                            ⏞                                            t            =            ln            ⁡            (            x            )                          =                                            t              2                        2                          +        constant                                          =                                                            ln                2                            ⁡              (              x              )                        2                          +        constant        =                                                            ln                                  1                  +                  1                                            ⁡              (              x              )                                      1              +              1                                      +        constant            The expression you have is                    ∫                  x                      α                          ln        ⁡        (        x        )        d        x                            =                                                            x                                  α                  +                  1                                            ln              ⁡              (              x              )                                      1              +              α                                      −                                            x                              α                +                1                                                    (              1              +              α                              )                2                                                    +        constant            You cannot directly take the limit as   α  →  −  1. But note that you can draw some constants out from the constant term to help you.                    ∫                  x                      α                          ln        ⁡        (        x        )        d        x                            =                                            (                              x                                  α                  +                  1                                            −              1              )              ln              ⁡              (              x              )                                      1              +              α                                      +                                            ln              ⁡              (              x              )                                      1              +              α                                      +                                            1              −                              x                                  α                  +                  1                                                                    (              1              +              α                              )                2                                                                                                        −                                                      1                                          (                      1                      +                      α                                              )                        2                                                                                            +                constant                            ⏟                                            new constant                                                            =                                            (                              x                                  α                  +                  1                                            −              1              )              ln              ⁡              (              x              )                                      1              +              α                                      +                                            ln              ⁡              (              x              )                                      1              +              α                                      +                                            1              −                              x                                  α                  +                  1                                                                    (              1              +              α                              )                2                                                    +        constant                                        (        ♠        )            Now note that      x          1      +      α        =  exp  ⁡  (  (  1  +  α  )  ln  ⁡  (  x  )  )  =  1  +  (  1  +  α  )  ln  ⁡  (  x  )  +                    (        1        +        α                  )          2                    2            ln    2    ⁡  (  x  )  +      O    (  (  1  +  α      )    3    )Hence,                                                        1              −                              x                                  1                  +                  α                                                                    1              +              α                                                          =        −        ln        ⁡        (        x        )        −                                            1              +              α                        2                                    ln          2                ⁡        (        x        )        +                  O                (        (        1        +        α                  )          2                )                            ln        ⁡        (        x        )        +                                            1              −                              x                                  1                  +                  α                                                                    1              +              α                                                          =        −                                            1              +              α                        2                                    ln          2                ⁡        (        x        )        +                  O                (        (        1        +        α                  )          2                )                                                                ln              ⁡              (              x              )                                      1              +              α                                      +                                            1              −                              x                                  1                  +                  α                                                                    (              1              +              α                              )                2                                                                        =        −                                                            ln                2                            ⁡              (              x              )                        2                          +                  O                (        (        1        +        α        )        )            Now letting   α  →  −  1, we get that      lim          α      →      −      1        ∫      x          α        ln  ⁡  (  x  )  =      lim          α      →      −      1                                    ln          2                ⁡        (        x        )            2        +      O    (  (  1  +  α  )  )  =                              ln          2                ⁡        (        x        )            2      which matches with our original integral.

Recursion Formulas of int x^a ln x dx and int ln^b x/x dx

Answered question

Answer & Explanation

New Questions in Calculus 1