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1: 25.5 Integral Representations
25.5.1 ζ ( s ) = 1 Γ ( s ) 0 x s - 1 e x - 1 d x , s > 1 .
25.5.2 ζ ( s ) = 1 Γ ( s + 1 ) 0 e x x s ( e x - 1 ) 2 d x , s > 1 .
25.5.3 ζ ( s ) = 1 ( 1 - 2 1 - s ) Γ ( s ) 0 x s - 1 e x + 1 d x , s > 0 .
25.5.4 ζ ( s ) = 1 ( 1 - 2 1 - s ) Γ ( s + 1 ) 0 e x x s ( e x + 1 ) 2 d x , s > 0 .
25.5.5 ζ ( s ) = - s 0 x - x - 1 2 x s + 1 d x , - 1 < s < 0 .
2: 25.14 Lerch’s Transcendent
25.14.5 Φ ( z , s , a ) = 1 Γ ( s ) 0 x s - 1 e - a x 1 - z e - x d x , s > 0 , a > 0 , z [ 1 , ) .
25.14.6 Φ ( z , s , a ) = 1 2 a - s + 0 z x ( a + x ) s d x - 2 0 sin ( x ln z - s arctan ( x / a ) ) ( a 2 + x 2 ) s / 2 ( e 2 π x - 1 ) d x , s > 0 if | z | < 1 ; s > 1 if | z | = 1 , a > 0 .
3: 25.12 Polylogarithms
25.12.14 F s ( x ) = 1 Γ ( s + 1 ) 0 t s e t - x + 1 d t , s > - 1 ,
25.12.15 G s ( x ) = 1 Γ ( s + 1 ) 0 t s e t - x - 1 d t , s > - 1 , x < 0 ; or s > 0 , x 0 ,
4: 25.11 Hurwitz Zeta Function
25.11.5 ζ ( s , a ) = n = 0 N 1 ( n + a ) s + ( N + a ) 1 - s s - 1 - s N x - x ( x + a ) s + 1 d x , s 1 , s > 0 , a > 0 , N = 0 , 1 , 2 , 3 , .
25.11.25 ζ ( s , a ) = 1 Γ ( s ) 0 x s - 1 e - a x 1 - e - x d x , s > 1 , a > 0 .
25.11.26 ζ ( s , a ) = - s - a x - x - 1 2 ( x + a ) s + 1 d x , - 1 < s < 0 , 0 < a 1 .
25.11.29 ζ ( s , a ) = 1 2 a - s + a 1 - s s - 1 + 2 0 sin ( s arctan ( x / a ) ) ( a 2 + x 2 ) s / 2 ( e 2 π x - 1 ) d x , s 1 , a > 0 .
25.11.31 1 Γ ( s ) 0 x s - 1 e - a x 2 cosh x d x = 4 - s ( ζ ( s , 1 4 + 1 4 a ) - ζ ( s , 3 4 + 1 4 a ) ) , s > 0 , a > - 1 .
5: 25.16 Mathematical Applications
25.16.6 H ( s ) = - ζ ( s ) + γ ζ ( s ) + 1 2 ζ ( s + 1 ) + r = 1 k ζ ( 1 - 2 r ) ζ ( s + 2 r ) + n = 1 1 n s n B ~ 2 k + 1 ( x ) x 2 k + 2 d x ,
25.16.7 H ( s ) = 1 2 ζ ( s + 1 ) + ζ ( s ) s - 1 - r = 1 k ( s + 2 r - 2 2 r - 1 ) ζ ( 1 - 2 r ) ζ ( s + 2 r ) - ( s + 2 k 2 k + 1 ) n = 1 1 n n B ~ 2 k + 1 ( x ) x s + 2 k + 1 d x .
6: 25.2 Definition and Expansions
25.2.8 ζ ( s ) = k = 1 N 1 k s + N 1 - s s - 1 - s N x - x x s + 1 d x , s > 0 , N = 1 , 2 , 3 , .
25.2.9 ζ ( s ) = k = 1 N 1 k s + N 1 - s s - 1 - 1 2 N - s + k = 1 n ( s + 2 k - 2 2 k - 1 ) B 2 k 2 k N 1 - s - 2 k - ( s + 2 n 2 n + 1 ) N B ~ 2 n + 1 ( x ) x s + 2 n + 1 d x , s > - 2 n ; n , N = 1 , 2 , 3 , .
7: 28.32 Mathematical Applications
defines a solution of Mathieu’s equation, provided that (in the case of an improper curve) the integral converges with respect to z uniformly on compact subsets of . …