According to my calculations ∫ 0 ∞ d x ( 1 + x 3 ) n = ( 3 n − 4 ) × ( 3 n − 7 ) × ⋯ × 5 × 2 3 n + 1 / 2 ( n − 1 ) ! 2 π

Question

According to my calculations      ∫    0    ∞                      d            x              (      1      +              x        3                    )        n              =            (      3      n      −      4      )      ×      (      3      n      −      7      )      ×      ⋯      ×      5      ×      2                      3                  n          +          1                      /                    2                    (      n      −      1      )      !        2  π

Allison Pena · Accepted Answer

∏          k      =      1              n      −      1        3  k  −  1  =                    3                  n          −          1                    Γ      (      n      −      1              /            3      )              Γ      (      2              /            3      )      so now you can use Stirling&#039;s series to find an asymptotic for your integral. I get that      ∫    0          ∞            1          (      1      +              x        3                    )        n              d  x  =            Γ      (      4              /            3      )              n      3            (    1    +          2              9        n              +          O        (          n              −        2              )    )    .So you can combine this result with             ∑              k        =        1            n              k      p        =                  n                  p          +          1                            p        +        1              +                  n        p            2        +          O        (          n              p        −        1              )   to find an asymptotic for your sum.

Audrina Jackson · Accepted Answer

The integral is the beta function in disguise. Let   x  =            (              z                  1          −          z                    )              1              /            3      . Then                              I          n                                    =                              ∫          0          ∞                                      d            x                                (            1            +                          x              3                                      )              n                                                                        =                              1          3                          ∫          0          1                d        z                          z                      −            2                          /                        3                          (        1        −        z                  )                      n            −            4                          /                        3                                                            =                              1          3                B        (        n        −        1                  /                3        ,        1                  /                3        )                                          =                              1          3                Γ        (        1                  /                3        )                              Γ            (            n            −            1                          /                        3            )                                Γ            (            n            )                                                            =                    Γ        (        4                  /                3        )                              Γ            (            n            −            1                          /                        3            )                                Γ            (            n            )                          .            If the upper limit is finite, a closed expression can be found for your sum,      ∑          n      =      1        N        I    n    =      3    2    Γ  (  4      /    3  )            Γ      (      N      +      2              /            3      )              Γ      (      N      )        .

According to my calculations <msubsup> ∫ 0 <mi mathvariant="normal"

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