If: λ = ∫ 0 1 d x 1 + x 3 ;Then evaluate: p = lim n → ∞ ( ∏ r = 1 n ( n 3 + r 3 ) n 3 n ) 1 / n ;

Question

If:  λ  =      ∫          0              1                  d      x              1      +              x        3              ;Then evaluate:  p  =      lim          n      →      ∞                  (                                    ∏                          r              =              1                                      n                                            (                          n              3                        +                          r              3                        )                                    n                      3            n                              )              1              /            n        ;

Pradellalo · Accepted Answer

First, to evaluate   λ, use partial fractions:      1          1      +              x        3              =      1    3        (          1              1        +        x              −                  x        −        2                              x          2                −        x        +        1              )  The second piece in the parentheses is more amenable to integration when expressed as follows:            x      −      2                      x        2            −      x      +      1        =            x      −      1              /            2              (      x      −      1              /            2              )        2            +      3              /            4        −            3              /            2              (      x      −      1              /            2              )        2            +      3              /            4      Now, let  A  =      ∫    0    1              d      x              1      +      x        =  log  ⁡      2    B  =      ∫    0    1    d  x              x      −      1              /            2              (      x      −      1              /            2              )        2            +      3              /            4        =            [      log      ⁡                        [                                    (              x              −                              1                2                            )                        2                    +                      3            4                    ]                    ]        0    1    =  0  C  =      ∫    0    1    d  x              3              /            2              (      x      −      1              /            2              )        2            +      3              /            4        =      3              [      arctan      ⁡                        [                      2                          3                                            (            x            −                          1              2                        )                    ]                    ]        0    1    =      π          3      Therefore  λ  =      1    3    (  A  −  B  +  C  )  =      1    3    log  ⁡      2    +      π          3              3            Now for the limit, which is evaluated by taking logs of both sides:  log  ⁡      p    =      lim          n      →      ∞            1    n        ∑          r      =      0        n    log  ⁡            (      1      +                        r          3                          n          3                    )      This is a Riemann sum, in the sense that      lim          n      →      ∞            1    n        ∑          r      =      0        n    f      (          r      n        )    =      ∫    0    1    d  x    f  (  x  )Therefore                    log        ⁡                  p                                    =                  ∫          0          1                d        x                log        ⁡                  (          1          +                      x            3                    )                                                  =        [        x        log        ⁡                  (          1          +                      x            3                    )                          ]          0          1                −                  ∫          0          1                d        x                                      3                          x              3                                            1            +                          x              3                                                                        =        log        ⁡                  2                −        3                  ∫          0          1                d        x                          (          1          −                      1                          1              +                              x                3                                              )                                                  =        log        ⁡                  2                −        3        +        3        λ                                          =        2        log        ⁡                  2                −        3        +                  π                      3                                              ∴        p                            =        4                  e                      −            (            3            −                          π                              3                                      )

If: lambda=int_0^1 dx/(1+x^3); then evaluate p=lim_(n->infty)(prod_(r=1)^n(n^3+r^3))^(1/n)/(n^(3n))

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