How to separate ∏ k = 1 ∞ ( 16 k − 15 ) 1 16 k − 15 ( 16 k − 1 ) 1 16 k − 1 from Log-gamma function.

Question

How to separate       ∏          k      =      1              ∞                  (      16      k      −      15              )                              1                          16              k              −              15                                                  (      16      k      −      1              )                              1                          16              k              −              1                                           from Log-gamma function.

Karla Hull · Accepted Answer

Let  f  (  z  )  =  π      (    ln    ⁡    Γ    (    z    )    +          (      z      −              1        2            )        (    γ    +    ln    ⁡    2    )    +    (    z    −    1    )    ln    ⁡    π    +          1      2        ln    ⁡    sin    ⁡    (    π    z    )    )    =      ∑          k      =      1              ∞                  sin      ⁡      (      2      π      k      z      )        k    ln  ⁡  kThen assuming   α  ∈      N   and   α  &amp;gt;  2  ln  ⁡      ∏          k      =      1              ∞                  (      α      k      −      α      +      1              )                              1                          α              k              −              (              α              −              1              )                                                  (      α      k      −      1              )                              1                          α              k              −              1                                            =      ∑          k      =      1        ∞        (                  ln        ⁡        (        α        k        −        α        +        1        )                    α        k        −        α        +        1              −                  ln        ⁡        (        α        k        −        1        )                    α        k        −        1              )    =      ∑          k      =      1        ∞    s  (  k  )            ln      ⁡      k        k    s  (  k  )  =      {                            1                          k          ≡          1                    (          mod                    α          )                                      −          1                          k          ≡          −          1                    (          mod                    α          )                                      0                          otherwise                        We can express s(k) as a sum of sin functions using the discrete fourier transform:  s  (  k  )  =      1    α        ∑          n      =      1              α      −      1            s    n    sin  ⁡      (                  2        π        n        k            α        )  where      s    n    =      ∑          k      =      1              α      −      1        s  (  k  )  sin  ⁡      (                  2        π        n        k            α        )    =  sin  ⁡      (                  2        π        n            α        )    −  sin  ⁡      (                  2        π        n        (        α        −        1        )            α        )    =  2  sin  ⁡      (                  2        π        n            α        )  Hence,      ∑          k      =      1        ∞    s  (  k  )            ln      ⁡      k        k    =      2    α        ∑          n      =      1              α      −      1        sin  ⁡      (                  2        π        n            α        )        ∑          k      =      1        ∞    sin  ⁡      (                  2        π        n        k            α        )              ln      ⁡      k        k    =      2    α        ∑          n      =      1              α      −      1        sin  ⁡      (                  2        π        n            α        )    f      (          n      α        )  Constant terms in f are annihilated in this formula, and  sin  ⁡  (  2  x  )  ln  ⁡  (  sin  ⁡  (  x  )  )  =  −  sin  ⁡  (  2  (  π  −  x  )  )  ln  ⁡  sin  ⁡  (  π  −  x  )  ,so the lnsin terms are annihilated as well to give            2      π        α        ∑          n      =      1              α      −      1        sin  ⁡      (                  2        π        n            α        )        (    ln    ⁡    Γ          (              n        α            )        +          (              n        α            )        (    γ    +    ln    ⁡    2    π    )    )  Using      ∑          n      =      1              α      −      1        n  sin  ⁡      (                  2        π        n            α        )    =  −      α    2    cot  ⁡      (          π      α        )  which comes from taking the imaginary part of       ∑          n      =      0              α      −      1        n      z    n    =            2      z              (      1      −      z              )        2              +            (      α      −      1      )      z              1      −      z       with   z  =  exp  ⁡      (                  2        π        i            α        )  , we get            2      π        α        ∑          n      =      1              α      −      1        sin  ⁡      (                  2        π        n            α        )        (          n      α        )    (  γ  +  ln  ⁡  2  π  )  =  −      π    α    (  γ  +  ln  ⁡  2  π  )  cot  ⁡      (          π      α        )  so we can finally write      ∏          k      =      1              ∞                  (      α      k      −      α      +      1              )                              1                          α              k              −              α              +              1                                                  (      α      k      −      1              )                              1                          α              k              −              1                                            =            (      2      π              e                  γ                    )              −              π        α            cot      ⁡              (                  π          α                )                  ∏          n      =      1              α      −      1        Γ            (              n        α            )                                2          π                α            sin      ⁡              (                              2            π            n                    α                )

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