SUT Journal of Mathematics Vol 43, No 7, 51 61 Some expressions of double triple sine functions Hidekazu Tanaka Received March, 7 Abstract We show some expressions of double triple sine functions Then we apply the results to special values of Dirichlet L-functions ζ3 AMS Mathematics Subject Classification Primary 11M6 Key words phrases multiple sine function, multiple gamma function, multiple Hurwitz zeta function 1 Introduction The double sine function S x the triple sine function S 3 x see [] are defined as S x := Γ x 1 Γ x S 3 x := Γ 3 x 1 Γ 3 3 x 1 respectively, where Γ x := expζ, x = exp s ζ s, x s= Γ 3 x := expζ 3, x = exp s ζ 3s, x s= are the double gamma function the triple gamma function We notice that the double Hurwitz zeta function the triple Hurwitz zeta function are constructed by ζ s, x := x + n 1 + n s n 1,n 51
5 H TANAKA ζ 3 s, x := x + n 1 + n + n 3 s n 1,n,n 3 We recall the classical objects: S 1 x := Γ 1 x 1 Γ 1 1 x 1, Γ 1 x := expζ 1, x = exp s ζ 1s, x s= Then we have where we used ζ 1 s, x = ζs, x = S 1 x = n + x s n= π ΓxΓ1 x Γ 1 x = Γx π which was obtained by Lerch [1] First we describe the double gamma function Γ x the double sine function S x using the logarithm of the usual gamma function Γx as follows: Theorem 11 We have Γ x = Γx1 x π S x = 1 xπ e sinπx x x exp + ζ 1 + x 1 x exp log Γt dt x log Γt dt, We notice that this result is considered to be an analogue of Raabe s formula +1 x log Γt dt = x log x x + 1 logπ proved in [1] 1844 We refer to [6, 7, 9] for another kind of generalization of Raabe s formula using the generalized gamma function Γ r x := exp ζs, x s s=1 r suggested by Milnor [11] the generalized sine function S r x := Γ r x 1 Γ r 1 x 1r
SOME EXPRESSIONS OF DOUBLE AND TRIPLE SINE FUNCTIONS 53 introduced in [7] Here we remark that this result is considered as a kind of Raabe s formula from the view point of the generalized gamma function the generalized sine function see [6, 7, 9] for details From Theorem 11 we get expressions for special values of some Dirichlet L-functions: Theorem 1 We obtain L, χ 3 = 8 3π 9 4π log 3e L, χ 4 = 3π 3π log π e 4 3π 3 7 4 1 4 5 3 1 3 log Γt dt, log Γt dt, where χ 3 χ 4 are non-trivial Dirichlet characters of modulo 3 4 respectively Remark 11 We can rewrite Theorem 1 as L, χ 3 = 4 3π ζ 1, 1 3 3 ζ 1, 5 3 + 8 3π 9 L, χ 4 = π ζ 1, 1 4 ζ 1, 7 4 + 3π 3 log 4 log, 3 Next we consider the triple gamma function Γ 3 x the triple sine function S 3 x Theorem 13 We have x 1x Γx Γ 3 x = exp x3 π 4 + 7 8 x 17 4 x + 3 ζ 1 + ζ t + log Γu dudt + 3 x log Γt dt, x 1x e sinπx S 3 x = π πx 1x 3 x t exp + log Γu dudt + 3 x 3 x log Γt dt x 3ζ 1 ζ From this we get the following result: Theorem 14 We obtain ζ3 = 8π 7 7 4 log 9 4 log π + 1 4 + 6ζ 1 + 4 3 t log Γu dudt
54 H TANAKA Remark 1 We show also that 3 t log Γu dudt = 3 So we can rewrite Theorem 14 as ζ3 = 3π 7 ζ 1, t dt + 9 16 logπ 3 ζ 1 ζ 1, t dt Lemma 1 ζ 1, x = Proofs of results log Γt dt + x 1 + logπ x + ζ 1 Proof of Lemma 1 Let γ = lim 1 + 1 + + 1n log n = 577156649 n be the Euler constant, then by the infinite product expression for the gamma function log Γx = log x + γx + log 1 + x x n n Hence we have Also d 3 dx 3 log Γt dt = n + x n= 3 x 3 ζs, x = ss + 1s + converges absolutely for Rs > Therefore 3 s x s= 1 3 ζs, x = n= n + x s 3 n + x n= So we can write ζ 1, x log Γt dt = ax + bx + c,
SOME EXPRESSIONS OF DOUBLE AND TRIPLE SINE FUNCTIONS 55 where a, b, c are some constant numbers variable, it is easy to verify that log Γt + 1 dt = = +1 1 +1 Using log Γt + 1 log Γt = log t, we get log Γt dt +1 log Γt dt log Γt dt Here under the change of the log Γt dt = x log x + x Moreover, from ζs, x + 1 ζs, x = x s we have Thus we obtain ζ 1, x + 1 ζ 1, x ζ 1, x + 1 ζ 1, x = x log x +1 log Γt dt + x To decide b we recall Euler s integral see [1, 5, 8]: π Note that log Γt dt = logsin ϕ dϕ = π log log Γ1 t dt log Γt dt Denote the integral of the left-h side by I Then we know Also we can calculate I = = logγtγ1 t dt log = log π π sinπt logsinπt dt = 1 π = π dt logsinπt dt π π log Γt dt log Γt dt = ax + a + b, log Γt dt = ax + a + b logsin ϕ dϕ logsin ϕ dϕ
56 H TANAKA Therefore from Euler s integral we find that log Γt dt = 1 logπ a = 1, b = 1 1 logπ Putting x = 1 we note c = ζ 1, where we used ζ 1, 1 = ζ 1 Then we obtain the result Now we show Theorems Proof of Theorem 11 We put fx := Γx1 x x x exp + ζ 1 + π Hence we have log Γt dt log fx = 1 x log Γx 1 logπ + x x + ζ 1 + f x 1 x = + 1 x Γ f Γ x On the other h, we know x ζ s, x = sζ s + 1, x = s ζs, x + 1 xζs + 1, x = sζs, x + 1 x ζs, x, x where we used ζs, x = sζs + 1, x x Since ζs, x is analytic in a region containing s =, we can write By x ζ s, x = s ζ, x + ζ, xs + + 1 x x ζ, x = 1 x ζ, x + ζ, xs + log Γt dt,
SOME EXPRESSIONS OF DOUBLE AND TRIPLE SINE FUNCTIONS 57 the Lerch s formula see [1] Γx ζ, x = log, π we obtain Also, we know So we have Naturally we see Γ x = x 1 + 1 x Γ Γ Γ x f1 = 1 exp ζ 1 + π f1 = e ζ 1 Γ 1 = e ζ 1,1 = e ζ 1 log Γt dt Moreover by the definition of S x, the proof of Theorem 11 is completed Proof of Theorem 1 The following examples were known by Kurokawa- Koyama [, 4]: S 1 3 = 3 1 3 3 exp 4π L, χ 3, S 1 4 = 3 8 exp 1 π L, χ 4 Then, applying Lemma 1 to Theorem 1 we obtain the result in Remark 11 Proof of Theorem 13 We define gx := x 1x Γx π + ζ exp x3 + t 4 + 7 8 x 17 4 + log Γu dudt + 3 x ζ 1, t dt x + 3 ζ 1 log Γt dt Then we show Theorem 13 similarly as in the proof of Theorem 11 Immediately we obtain d dx x 3 log gx = Γ Γ x + x 3x + Γ xγx Γ x Γ 3 x x + 7 4
58 H TANAKA On the other h, we note log Γ 3 x = ζ 3, x where we used ζ 3 s, x = Here we have Then we get = 1 = 1 So we can write = 1 ζ, x + 3 x ζ 1, x + n 1,n,n 3 n 1 + n + n 3 + x s n + 1n + n + x s n= 3 x ζs, x + ζs 1, x + ζs, x = ss + 1ζs +, x x x ζ 3s, x = ss + 1ζ 3 s +, x x ζ 3s, x = Hence we obtain x 1x ζ, x, x 1x ζs, x = ss + 1 1 3 x ζs, x + ζs + 1, x x 1x + ζs +, x ss + 1 x 3s + 1 = ζs, x + ζs, x x x 1x + ζs, x x ss + 1 ζ, x + ζ, xs + x 3s + 1 + x x 1x + x ζ, x + ζ, xs + ζ, x + ζ, xs + d dx log Γ 3x = x 3 Γ Γ x + x 3x + Γ xγx Γ x Γ 3 x x + 7 4
SOME EXPRESSIONS OF DOUBLE AND TRIPLE SINE FUNCTIONS 59 Therefore for some constants a, b Moreover by Lemma 1 we have t t log Γu dudt = Hence we can calculate gx Γ 3 x = eax+b ζ 1, t dt + 1 1 + logπ ζ 1, 4 log Γt dt = logπ 1 log Γu dudt = 7 1 + logπ ζ 1 + On the other h, treating ζ g1 = exp ζ g = exp + ζ 1 ζ 1 ζ 3 s, 1 = 1 ζs + ζs 1 ζ 1, t dt we have Finally we show 1 ζ 3 s, = 1 ζs ζs 1, ζ Γ 3 1 = exp ζ Γ 3 = exp ζ k, t dt = + ζ 1 ζ 1 where k 1 be an integer To prove 1 we use the following formula see [3, 7]
6 H TANAKA Lemma Generalized Kummer s Formula Let k 1 be an integer < x < 1 1 When k is odd, k+1 { ζ 1 k! log n cosπnx k, x = π k+1 n k+1 + logπ + γ π When k is even, ζ k, x = 1 k k! π k+1 } sinπnx n k+1 { + logπ + γ + π } cosπnx n k+1 1 + 1 + + 1 k log n sinπnx n k+1 1 + 1 + + 1 k cosπnx n k+1 sinπnx n k+1 Here we notice that sinπnt dt = cosπnt dt = Thus we obtain 1 by Lemma From the definition of S 3 x, Theorem 13 is proved Proof of Theorem 14 Koyama [, 5]: The following example was shown by Kurokawa- S 3 3 = 1 8 exp 3 16π ζ3 So by Theorem 13 we have the result Theorem 14, we obtain Since ζ3 = 8π 7 3 we have Remark 1 1 ζ 1, tdt = Applying Lemma 1 1 to 1 log + 1 3 4 + 4 ζ 1, t dt 1 = ζ 1, t dt + t log t dt ζ 1, t dt 1 8 log 1 16,
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