Theorem 2. This Equation holds true for $p = 2, \beta = 1$ and for the case when each values of $a_i$ is greater than 1.
H ( { 1 a i } i = 1 n ; 2 r ( n ) ; 1 ) = ( − 1 ) n + 1 ∑ c y c tanh − 1 ( 1 a 1 ) a 1 ∏ 1 < j ≤ n ( a 1 − a j ) ( ∏ i = 1 n a i ) ; a i > 1 H ( { a i 1 } i = 1 n ; 2 r ( n ) ; 1 ) = ( − 1 ) n + 1 cyc ∑ a 1 ∏ 1 < j ≤ n ( a 1 − a j ) tanh − 1 ( a 1 1 ) ( i = 1 ∏ n a i ) ; a i > 1
Proof. As per the definition of Hyder series we can write the expression
H ( { 1 a i } i = 1 n ; 2 r ( n ) ; 1 ) = ∑ k 1 , k 2 , k 3 , … , k n ≥ 0 1 a 1 k 1 a 2 k 2 ⋯ a n k n ( 2 k 1 + 2 k 2 + ⋯ + 2 k n + 1 ) = ∫ 0 1 ∑ k 1 , k 2 , k 3 , … , k n ≥ 0 ( x 2 a 1 ) k 1 ( x 2 a 2 ) k 2 ⋯ ( x 2 a n ) k n d x = ( − 1 ) n ( ∏ i = 1 n a i ) ∫ 0 1 1 ( x 2 − a 1 ) ( x 2 − a 2 ) ⋯ ( x 2 − a n ) d x H ( { a i 1 } i = 1 n ; 2 r ( n ) ; 1 ) = k 1 , k 2 , k 3 , … , k n ≥ 0 ∑ a 1 k 1 a 2 k 2 ⋯ a n k n ( 2 k 1 + 2 k 2 + ⋯ + 2 k n + 1 ) 1 = ∫ 0 1 k 1 , k 2 , k 3 , … , k n ≥ 0 ∑ ( a 1 x 2 ) k 1 ( a 2 x 2 ) k 2 ⋯ ( a n x 2 ) k n d x = ( − 1 ) n ( i = 1 ∏ n a i ) ∫ 0 1 ( x 2 − a 1 ) ( x 2 − a 2 ) ⋯ ( x 2 − a n ) 1 d x
The following integral can be solved by using partial fraction. It is quite
intresting to note that while using partial fraction for this product,we can
see a cyclic pattern in it. i.e
1 ( x 2 − a 1 ) ( x 2 − a 2 ) … ( x 2 − a n ) = ∑ c y c 1 ( x 2 − a 1 ) ∏ 1 < j ≤ n ( a 1 − a j ) ( x 2 − a 1 ) ( x 2 − a 2 ) … ( x 2 − a n ) 1 = cyc ∑ ( x 2 − a 1 ) ∏ 1 < j ≤ n ( a 1 − a j ) 1
therefore
∫ 0 1 1 ( x 2 − a 1 ) ( x 2 − a 2 ) … ( x 2 − a n ) d x = ∫ 0 1 ∑ c y c 1 ( x 2 − a 1 ) ∏ 1 < j ≤ n ( a 1 − a j ) d x = ∑ c y c 1 ∏ 1 < j ≤ n ( a 1 − a j ) ∫ 0 1 1 ( x 2 − a 1 ) d x ∫ 0 1 ( x 2 − a 1 ) ( x 2 − a 2 ) … ( x 2 − a n ) 1 d x = ∫ 0 1 cyc ∑ ( x 2 − a 1 ) ∏ 1 < j ≤ n ( a 1 − a j ) 1 d x = cyc ∑ ∏ 1 < j ≤ n ( a 1 − a j ) 1 ∫ 0 1 ( x 2 − a 1 ) 1 d x
since $\int_0^1 \frac{1}{(x^2 - a_1)} dx = \frac{-\tanh^{-1}\left(\frac{1}{\sqrt{a_1}}\right)}{\sqrt{a_1}}$
