Log-Tangent Integrals and the Riemann Zeta Function

. We show that integrals involving the log-tangent function, with respect to any square-integrable function on (cid:0) 0 , π 2 (cid:1) , can be evaluated by the harmonic series. Consequently, several formulas and algebraic properties of the Riemann zeta function at odd positive integers are discussed. Furthermore, we show among other things, that the log-tangent integral with respect to the Hurwitz zeta function deﬁnes a mero-morphic function and its values depend on the Dirichlet series ζ h ( s ) := (cid:80) n ≥ 1 h n n − s , where h n = (cid:80) nk =1 (2 k − 1) − 1 .


Introduction
The theory of the Riemann zeta function is a fascinating topic in number theory with many beautiful results and open questions. The Riemann zeta function is defined by ζ(s) := (n = 1, 2, . . .) must be irrational, and that at least one of the eight numbers ζ(2n + 1) (n = 2, . . . , 9) must be irrational. In the same direction and in order to study some algebraic properties of the set {ζ(2n + 1), n ∈ N}, the current authors [8] showed recently that these numbers appear, both explicitly and implicitly, in the value of integrals involving the log-tangent function. More precisely, we showed that, for any square-integrable function f on (0, π/2), the integral can be approximated by a finite sum involving the Riemann zeta function at odd positive integers. In particular, for any polynomial P , we have [8,Theorem 2] π 2 0 P 2 π x log(tan x)dx = deg P +1 2 k=1 (−1) k−1 π 2k−1 c k (P )ζ(2k + 1), (1.1) where c k (P ) = 1 − 1 2 2k+1 P (2k−1) (1) + P (2k−1) (0) and where P (k) (x) denotes the k-th derivative of P at point x.
Besides the theory of numbers, integrals involving the log-tangent function have various important applications in many different fields of mathematics. In physics, logarithmic-trigonometric integrals also have some applications in the evaluation of classical, semi-classical and quantum entropies of position and momentum, see for example [15].
The purpose of this paper is to show that the integrals involving the logtangent function can be evaluated by some series regarding the harmonic number. That is, for certain square-integrable functions f on (0, π/2), we have Here, H n := n k=1 1 k is the so-called harmonic number. Furthermore, by combining this result with those obtained in [8], we will be able to express many different sums involving the Riemann zeta function at odd integer in terms of variant Euler sums, where (α k ) ∞ k=1 is a real sequence. The mentioned results will appear in Section 2. It is worth noting that Chen [6] studied the Euler series (1.2), when (α k ) ∞ k=1 is a power sequence, by utilizing the following generating function In Section 3, we provide a general study of the Dirichlet series ζ h (s), or the h-zeta function, defined by Finally, we conclude the paper with a brief discussion of the corresponding results and remarks.

Recursive formulas involving zeta numbers
It is well known that the Fourier system 1 forms an orthonormal basis for the Hilbert space L 2 (0, π/2) with respect to the inner product It follows that any square-integrable function f on (0, π/2) can be expressed as the Fourier series where a 0 (f ) := f, 1 √ 2 , a n (f ) := f, cos(4n·) and b n (f ) := f, sin(4n·) . Let It is not hard to check that H even is the orthogonal complement subspace of the closed subspace H odd in L 2 (0, π/2) and that the system {sin(4nx)} ∞ n=1 forms an orthonormal basis for H odd .
Recall that, for any square-integrable function f on (0, π/2), The function x → log(tan x) belongs to H odd and the operator L is a linear functional on L 2 (0, π/2), so we have Therefore, it suffices to study the linear functional L on the Hilbert space H odd .
Clearly, if f ∈ H odd , then the Fourier expansion of f is reduced to It follows immediately that for any f ∈ H odd .
Now we are ready to state our main theorem.
holds in L 2 (0, π/2) and, for any f ∈ H odd , we have Specializing Theorem 1 for f (x) = log(tan x), we obtain We also know that Therefore, we must have Note that this last formula can also be obtained by applying the Parseval's identity. Now let It is clear that f belongs to H odd , so we have By Theorem 1, we obtain More generally, we have the following result.
Proof. Let Clearly, f r belongs to H odd . Since, for each n ∈ N, we have it follows from Theorem 1 and (2.2) that Finally, the fact that completes the proof of Corollary 1.
We should mention that Bradley [5] studied the transformation for all r ∈ [0, 1/2]. He showed that which can also be deduced from our Corollary 1. Moreover, by using the identities listed in [ Another interesting Euler sum is given in the following corollary.
Corollary 2. Let m be a positive integer. Then we have Proof. From [13, eq. (1. 3)], we know that, for any positive integer m, the (2m − 1)th Bernoulli polynomial can be expanded as In other words, we have, for all x ∈ [0, π/2), By applying Theorem 1, we obtain On the other hand, we find from (1.1) that

Thus, by combining this last equation with (2.3), we obtain
as required.
It is worth noting that Corollary 2 can be rewritten as It is not hard to see that our Corollary 2 is a generalization of Chen's formula (2.2). For some historical background of this kind of recursive formulas, it is worth mentioning that Euler [9] proved the following formula where H n := n k=1 1/k is the harmonic number. In addition, Georghiou and Philippou [10] showed that Another well-known recursive formula for zeta function at even positive integers is given in [7, p. 167] as (2n + 1)ζ(2n) = 2 For more relevant explicit Euler sums, the reader is advised to see [3,4].
Next we show another explicit formulation of (1.1).
Theorem 2. For any polynomial P such that P (1 − x) = −P (x), we have where P (k) (x) denotes the k-th derivative of P at point x.
Proof. It is clear, from the functional equation of P , that P is of odd degree, say 2m − 1 (m ∈ N). Thus, by using integration by parts, it is easy to see that Since the polynomial P 2 π · is in H odd , it follows that its expansion is Finally, we apply Theorem 1 to complete the proof.
We note that another way to prove Theorem 2 is by employing the following fact. Each polynomial P of degree 2m − 1 such that P (x) = −P (1 − x) can be written as Then we apply (2.3) to prove Theorem 2. Next if we take P (x) = E 2m−1 (x) to be the (2m − 1)-th Euler polynomial, we obtain the following inversion formula of Corollary 2. Proof. By applying Theorem 2 to the Euler polynomial E 2m−1 , we obtain It follows that

Now, by (2.3) and the fact that
for all |z| < 2. We compare this last equation with [8, p. 11] to obtain Note that we can also deduce Corollary 2 and Corollary 3 directly from (2.5). Clearly, the integral π/2 0 e 2xz log(tan x)dx exists for any complex number z. Moreover, we have (2.6) The case that z = 2ik (k ∈ Z) is included as z tends to 2ik; that is, The proof of (2.6) is based on the expansion of the function x → exp(2xz) as in (2.1); namely, n sin(4nx) n 2 + ( z 2 ) 2 .
Then Theorem 1 helps complete the proof of (2.6). On the other hand, the authors showed in [8, p. 11] that π 2 0 e 2xz log(tan x)dx = e πz + 1 4z where ψ is the digamma function. Therefore, we obtain the following identity

Analytic continuation of ζ h (s)
Let s=σ+it (σ, t ∈ R) be a complex number, we define the h-zeta function by where h n := n k=1 1/(2k − 1). Since h n = 1 2 log(4n) + 1 2 γ + O(1/n 2 ) as n tends to ∞, we know that ζ h is analytic in the half-plane σ > 1. On the other hand, it follows from the generating function (1.3) that, for any real x > 0, Thus, for all σ > 1, we have That is, for all σ > 1, we have for all |x| < π, where w n = (−1) n n 2 2n−1 − 1 r n , n ∈ N and where is absolutely convergent for all σ > 1. Moreover, we have for all σ > 1. Therefore, Since K is a regular function, ζ h (s) has an analytic continuation to the whole complex plane, except at the pole s = 1 of order 2 and at simple poles s = 1−2n (n ∈ N). Moreover, the residue of ζ h (s) at s = 1 equals It is interesting to mention that the h-zeta function is closely related to the Hurwitz zeta function. Their first connection is that, for all σ > 1, Recall that the Hurwitz zeta function is defined for all σ > 1 and for all x > 0 by and it can be extended by analytic continuation to the whole complex plane, except at the simple pole s = 1 with residue 1. Also, it is well known that [16, p. 269], for all σ < 0, It is worth remarking that ζ h (s) can be extended analytically by using (3.1). Indeed, let z be a complex number such that z = 1. For each 0 < x < 1, we write ζ(z, x) = x −z + ζ * (z, x).
Then it follows from (3.1) that the integral exists for all z < 1. From [11, 1.518, eq. 3, p. 53], we have so, by using integration by parts, we obtain, for all z < 1, However, the right-hand side of the equality above converges absolutely for any complex number z not equal to odd positive integers (i.e. z = 1, 3, 5, . . .), and this defines an analytic continuation of the function On the other hand, the function ζ * (z, x) is defined on the half-plane z > 1 by ∞ 0 e −y y z−1 1 − e −y e −xy dy, 0 < x < 1.
It follows that, for all z > 1,

Conclusions
The value distribution of the h-zeta function can be an interesting topic to pursue. Besides, it is important to study the algebraic aspects of the numbers ζ(2n + 1) as we have shown in Corollary 2 and Corollary 3 that the h-zeta function appears in several integrals involving the Riemann zeta function and the Hurwitz zeta function.
For the sake of completeness, we present the following interesting example. On the other hand, the analogue of the Parseval's formula for the Mellin transforms (see for example [12, p. 484 We have seen that the h-zeta function vanishes at s = −2n (n = 0, 1, 2, . . .) and that ζ h (2n) is closely related to the numbers ζ(2n + 1) (n ∈ N). Hence, it is natural to ask if ζ h (s) satisfies certain functional equation. Maybe, by answering this question, one can find the expression of h-zeta function, as a complex-valued function, in terms of the Riemann zeta function.