How to separate <munderover> &#x220F;<!-- ∏ --> <mrow class="MJX-TeXAtom-ORD"> k

babyagelesszj 2022-07-03 Answered
How to separate k = 1 ( 16 k 15 ) 1 16 k 15 ( 16 k 1 ) 1 16 k 1 from Log-gamma function.
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Answers (1)

Karla Hull
Answered 2022-07-04 Author has 20 answers
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 k
Then assuming α N and α > 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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