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Boilanubjaini8f 2022-06-16 Answered
Just as the topic ask how to evaluate
k = 1 ( 18 ) [ ( k 1 ) ! ] 2 ( 2 k ) ! .
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Answers (2)

gaiageoucm5p
Answered 2022-06-17 Author has 20 answers
Notice that
( k 1 ) ! 2 ( 2 k ) ! = Γ ( k ) Γ ( k ) Γ ( 2 k + 1 ) = 1 2 k B ( k , k ) = 1 2 k 0 1 x k 1 ( 1 x ) k 1 d x
Thus
k = 1 ( k 1 ) ! 2 ( 2 k ) ! = 0 1 ( k = 1 1 2 k x k 1 ( 1 x ) k 1 ) d x = 0 1 ln ( 1 x + x 2 ) 2 x ( 1 x ) d x = 2 1 / 2 1 / 2 ln ( 3 / 4 + u 2 ) 1 4 u 2 d u = 2 0 1 ln ( ( 3 + u 2 ) / 4 ) 1 u 2 d u = π 2 18
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mravinjakag
Answered 2022-06-18 Author has 4 answers
In order to fill in on the last equality, define
f ( t ) = 2 0 1 ln ( 1 t ( 1 u 2 ) ) 1 u 2 d u
Clearly f ( 0 ) = 0, and we are interested in computing f ( 1 4 )
f ( t ) = 2 0 1 d u 1 + t ( 1 u 2 ) = u = 1 t t tan ( ϕ ) 0 arcsin ( t ) 2 d ϕ t ( 1 t ) = 2 arcsin ( t ) t ( 1 t ) = 2 d d t arcsin 2 ( t )
Thus
f ( 1 4 ) = 0 1 4 2 d d t arcsin 2 ( t ) = 2 arcsin 2 ( 1 2 ) = π 2 18
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