EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 7023 ISSN 1307-5543 – ejpam.com Published by New York Business Global Generalization of an Integral Related to Stieltjes Moment Problem Irshad Ayoob Department of Mathematics and Sciences, Prince Sultan University, P.O. Box 66833, Riyadh 11586, Saudi Arabia Abstract. In connection with the non-uniqueness of the Stieltjes moment problem on (0,∞), Stieltjes constructed the nontrivial function f(x) = e−x1/4 sin ( x1/4 ) satisfying ∫∞ 0 xnf(x) dx = 0 for all integers n ≥ 0. We extend this by considering I(k) = ∫ ∞ 0 e−x1/4 sin ( x1/4 ) xk dx for real k ≥ 0, evaluating I(k) explicitly and proving I(k) = 0 if and only if k ∈ Z≥0. More generally, for parameters m > 0, α > 0, β ∈ R, q, k ∈ R we analyze Im,q(k, α, β) = ∫ ∞ 0 e−αx1/m sin ( βx1/m ) xk(x1/m)q dx, derive a closed form, and give necessary and sufficient conditions for its vanishing. We also establish cosine analogues, both for the Stieltjes example and for the generalized integral mentioned above. As a consequence, we obtain integral representations of Γ(A) for suitable A > 0, as well as integral formulas for several classical constants arising from gamma function. To understand the importance of integrals that vanish for every value of a continuous parameter, we will also discuss Salem’s equivalence of the Riemann hypothesis, which is formulated in terms of such a parameter-dependent integral. 2020 Mathematics Subject Classifications: 44A60, 33B15, 26B15 Key Words and Phrases: Moment problem, gamma function, definite integral 1. Introduction Definite integrals are widely used in both pure and applied mathematics. While many can be solved with basic methods like substitution or integration by parts, numerous in- tegrals are non-elementary and cannot be expressed using standard functions. Evaluating these requires advanced techniques, such as transformations (Laplace, Mellin), complex DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.7023 Email address: iayoub@psu.edu.sa (I. Ayoob) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 2 of 16 analysis (contour integration, residues), and special functions (Gamma, Beta, hypergeo- metric). Exact solutions are valuable for precise predictions, deeper theoretical insight, and verifying numerical methods. Consequently, research on advanced methods for definite integrals remains active, with comprehensive compilations and recent studies highlighting ongoing developments and continued interest in the field. For the extensive list of definite integrals and some recent recsults, one may look for the references [1–10]. This paper deals with the evaluation of an integral related to the Steiltjes moment problem. We recall few definitions from the moment problem. The classical moment problem asks [11]: given a sequence of real numbers {mn}∞n=0, does there exist a positive Borel measure µ on some subset of R such that mn = ∫ xn dµ(x), n = 0, 1, 2, . . .? Depending on the support allowed for µ, one obtains three classical versions as stated below. Definition 1 (Hamburger moment problem). This problem seeks the measure µ on the entire real line (−∞,∞). That is, given {mn}, does there exist a positive measure µ with mn = ∫ ∞ −∞ xn dµ(x), n = 0, 1, 2, . . .? This is the most general form of the problem. Definition 2 (Stieltjes moment problem). This problem seeks the measure µ on the half– line [0,∞). That is, one asks whether there exists µ with mn = ∫ ∞ 0 xn dµ(x), n = 0, 1, 2, . . .? Such sequences {mn} are called Stieltjes moment sequences. This version is closely con- nected to continued fractions and orthogonal polynomials on [0,∞). Definition 3 (Hausdorff moment problem). Here the measure µ is confined to the compact interval [0, 1]. That is, does there exist µ with mn = ∫ 1 0 xn dµ(x), n = 0, 1, 2, . . .? In every version of the moment problem, the issue of uniqueness is crucial. For the Stieltjes and Hamburger cases, whenever a solution exists it need not be unique, there may be infinitely many measures producing the same moment sequence (the indeterminate case). By contrast, the Hausdorff moment problem, when solvable, always yields a single I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 3 of 16 measure, making it determinate. Thus, in the indeterminate situations, one encounters infinitely many distinct measures that share the same prescribed moments. The non- uniqueness of moments is classically studied through Carleman’s condition and Krein’s condition. Related to Hausdorff moment problem, we have the classical result proved through Weierstrass approximation theorem, Proposition 1. [12] Let f ∈ C([0, 1]). If∫ 1 0 xnf(x) dx = 0 for all n = 0, 1, 2, . . . , then f ≡ 0 on [0, 1]. About the infinite case, we may pose a similar question: Question 1. Suppose f ∈ C([0,∞)) and∫ ∞ 0 xnf(x) dx = 0 for all n = 0, 1, 2, . . . . Does it follow that f ≡ 0 on [0,∞)? In relation to the non-uniqueness of the moment problem, Stieltjes showed [13](see p.506) that function f in Question 1 may be non-zero by giving the following example: Example 1. The integral I(n) = ∫ ∞ 0 xnf(x) dx = 0 (n = 0, 1, 2, . . . ), (1) for the non-trivial function, f(x) = e−x1/4 sin ( x1/4 ) , x ≥ 0. (2) This paper studies the following questions related to Stieltjes Example 1: Question 2. The integral I(n) = ∫∞ 0 xnf(x) dx = 0 for f(x) = e−x1/4 sin ( x1/4 ) , x ≥ 0 for all n = 0, 1, 2, . . . . What if non-negative integer n is replaced by non-negative real number k, that is, for what real numbers k ≥ 0, we have that I(k) = ∫∞ 0 xke−x1/4 sin ( x1/4 ) dx = 0. The second question deals with the following generalization of the Steiltjes integral. Question 3. Evaluate the following generalized integral explicitly (for appropriate param- eters) and determine the necessary and sufficient conditions on the parameters so that the integral vanishes, Im,q(k) = ∫ ∞ 0 e−αx1/m sin ( βx1/m ) xk (x1/m)q dx, (3) where k ≥ 0 is a real number. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 4 of 16 We will solve the integrals I(k) and Im,q(k) given in Questions 2 and 3 explicitly, and establish the necessary and sufficient conditions for the vanishing of these of integrals. This question is interesting as there exist integrals which vanish for all values of the continuous parameter (not only on discrete subsets of reals as in the case of Example 1 with parameter n.) For instance, Example 2. I(a) = ∫ ∞ 0 sin(ax)− sin((a+ 1)x) x dx = 0, a > 0. This follows from the Dirichlet integral ∫ ∞ 0 sin(αx) x dx = π 2 sgn(α), and since a > 0 and a+ 1 > 0, I(a) = π 2 − π 2 = 0. Thus I(a) = 0 for a > 0. Another motivating example for studying Question 1 and Question 2, and more broadly, integrals that vanish for every value of a continuous parameter, is Salem’s equivalence of the Riemann Hypothesis. This equivalence is expressed through an integral involving a continuous parameter, as stated below: Theorem 1. [14] The Riemann hypothesis is true if and only if the integral equation∫ ∞ 0 xr−1f(x) ext + 1 dx = 0, for all t > 0, admits only the trivial bounded measurable solution f(x) ≡ 0, for a fixed parameter r satisfying 1 2 < r < 1. Equivalently, we we may write Theorem 1 as following: Theorem 2. The Riemann hypothesis is false if and only if there exists a nontrivial bounded measurable function f(x) ̸≡ 0 satisfying the integral equation∫ ∞ 0 xr−1f(x) ext + 1 dx = 0, for all t > 0, for fixed parameter r such that 1 2 < r < 1. Salem’s criterion has been the subject of recent investigations by several authors ([15– 18]) using a variety of integral transform techniques. In particular, the following result was obtained in [17] (see Corollary 3.3): Theorem 3. Let f(x) be a bounded measurable function on R+ such that f(x) = O(x1/2) as x → 0+. If f(x) satisfies the integral equation given in Theorem 1 for given k with 1 2 < k < 1 and for all t > 0, then f(x) vanishes almost everywhere on R+. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 5 of 16 It is important to note that if there exists a nontrivial bounded measurable function f(x) satisfying the integral equation presented in Theorem 2 for all continuous parameters t > 0, then such an f(x) would constitute a counterexample to Salem’s equivalence of the Riemann Hypothesis. Emphasis should be placed on the fact that this nontrivial bounded function must cause the integral on the left-hand side of Theorem 2 to vanish for all t > 0. Consequently, the study of such parameter-dependent improper integrals that vanish for every value of a continuous parameter becomes particularly interesting depending upon the context. This paper examines a parameter-dependent improper integral, inspired by the Stieltjes moment problem, which vanishes for all continuous parameter values. 2. Main Results We begin with the explicit evaluation of Steiltjes integral. Theorem 4. For a real number k > −1 define I(k) = ∫ ∞ 0 e−x1/4 sin ( x1/4 ) xk dx. (4) Then I(k) is absolutely convergent for k > −1 and, I(k) = 2−2k Γ(4k + 4) sin ( (k + 1)π ) , (5) and, in particular, I(k) = 0 if and only if k ∈ {0, 1, 2, . . . }. Proof. Absolute convergence of (4) holds for k > −1 because on [0, 1],∣∣e−x1/4 sin(x1/4)xk ∣∣ ≤ x1/4xk = xk+1/4 ≤ xk, ∫ 1 0 xk dx = 1 k + 1 < ∞, and on [1,∞) we have | sin(x1/4)| ≤ 1 and the factor e−x1/4 decays super–polynomially, so∫∞ 1 e−x1/4 xk dx < ∞. We make the substitution, u = x1/4, x = u4, dx = 4u3 du, (6) to obtain I(k) = ∫ ∞ 0 e−u sin(u) (u4)k (4u3) du = 4 ∫ ∞ 0 e−u sin(u)u4k+3 du. (7) We denote real and imaginary parts by ℜ and ℑ, and write sinu = ℑ(eiu), we have e−u sinu = ℑ ( e−ueiu ) = ℑ ( e−(1−i)u ) , ⇒ ∫ ∞ 0 u4k+3e−u sinu du = ℑ ∫ ∞ 0 u4k+3e−(1−i)u du. (8) Combining (7) and (8) yields the identity, I(k) = 4ℑ ∫ ∞ 0 u4k+3 e−(1−i)u du. (9) I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 6 of 16 Now we set, a = 4k + 4 > 0, λ = 1− i (ℜλ = 1 > 0). (10) By the Gamma function ( with the substitution t = λu), we obtain∫ ∞ 0 ua−1e−λu du = ∫ ∞ 0 ( t λ )a−1 e−t dt λ = λ−a ∫ ∞ 0 ta−1e−t dt = Γ(a) λa , (11) we have, with a = 4k + 4, ∫ ∞ 0 u4k+3e−(1−i)u du = Γ(4k + 4) (1− i) 4k+4 . (12) Next we compute the complex power explicitly using the polar form 1− i = √ 2 e−iπ/4: (1− i)−(4k+4) = (√ 2 e−iπ/4 )−(4k+4) = 2−(2k+2) e i(4k+4)π/4 = 2−(2k+2) e i(k+1)π. (13) Substituting (12) and (13) into (9) and taking imaginary parts gives, I(k) = 4ℑ ( Γ(4k + 4) (1− i)−(4k+4) ) = 4Γ(4k + 4) 2−(2k+2)ℑ ( e i(k+1)π ) = 2−2k Γ(4k + 4) sin ( (k + 1)π ) , (14) which proves (5). Since 2−2k > 0 and Γ(4k + 4) > 0 for k > −1, the equality I(k) = 0 holds if and only if sin((k + 1)π) = 0, i.e., (k + 1)π = mπ for some m ∈ Z ⇐⇒ k = m− 1. Under the constraint k > −1, this forces m ≥ 1, so precisely k ∈ {0, 1, 2, . . . } yield I(k) = 0, and conversely any such k makes sin((k + 1)π) = 0, establishing the stated equivalence. Now we have the following generalization of Theorem 4. Remark 1. The above theorem provides an explicit closed-form evaluation of the integral I(k) in terms of the Gamma function and the sine function. It demonstrates that the vanishing of I(k) occurs only for nonnegative integer values of k. Theorem 5. Let m > 0, α > 0, β ∈ R, q ∈ R, and k ∈ R. Define Im,q(k, α, β) = ∫ ∞ 0 e−αx1/m sin ( βx1/m ) xk (x1/m)q dx, (15) and set A = mk + q +m. (16) I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 7 of 16 Then the integral in (15) converges for every A > −1. Moreover, if A > 0 (which we assume for the evaluation below), one has the closed form Im,q(k, α, β) = mΓ(A) (α2 + β2)−A/2 sin ( A arctan β α ) . (17) Consequently, if β ̸= 0 then Im,q(k, α, β) = 0 if and only if A arctan β α ∈ πZ; (18) if β = 0 then Im,q(k, α, 0) = 0 for all admissible parameters. Proof. Write the integrand as e−αx1/m sin(βx1/m)xk+q/m. For convergence near x = 0, use | sin t| ≤ min{1, |t|}. As x ↓ 0, we have t = x1/m ↓ 0, thus | sin(βx1/m)| ≤ |β|x1/m, which gives ∣∣e−αx1/m sin(βx1/m)xk+q/m ∣∣ ≪ xk+q/m+1/m (x ↓ 0). (19) Hence ∫ 1 0 |e−αx1/m sin(βx1/m)xk+q/m| dx < ∞ provided k + q/m+ 1/m > −1, i.e. mk + q +m > −1 ⇐⇒ A > −1. (20) As x → ∞ we have | sin(βx1/m)| ≤ 1 and the factor e−αx1/m decays faster than any power, so ∫ ∞ 1 ∣∣e−αx1/m sin(βx1/m)xk+q/m ∣∣ dx < ∞ for every k, q when α > 0. (21) Therefore (15) converges for all A > −1. For evaluation assume A > 0. We make the substitution u = x1/m, x = um, dx = mum−1 du, (22) and note xk(x1/m)q = umk+q. Then Im,q(k, α, β) = ∫ ∞ 0 e−αu sin(βu)umk+q (mum−1) du = m ∫ ∞ 0 e−αu sin(βu)uA−1 du. (23) Using sin(βu) = ℑ(eiβu) and the absolute integrability guaranteed by A > −1 and α > 0, we may pass the imaginary part inside the integral: e−αu sin(βu) = ℑ ( e−(α−iβ)u ) =⇒ Im,q(k, α, β) = mℑ ∫ ∞ 0 uA−1 e−(α−iβ)u du. (24) Put λ = α− iβ (ℜλ = α > 0). (25) I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 8 of 16 For A > 0 the Gamma gamma function (by the substitution t = λu) gives,∫ ∞ 0 uA−1e−λu du = ∫ ∞ 0 ( t λ )A−1 e−t dt λ = λ−A ∫ ∞ 0 tA−1e−t dt = Γ(A) λA . (26) Applying (26) with (25) in (24) yields Im,q(k, α, β) = mℑ ( Γ(A) (α− iβ)A ) . (27) We write α− iβ in polar form, with r = √ α2 + β2, θ = arctan β α ∈ ( −π 2 , π 2 ) , (28) we have α− iβ = r e−iθ and hence (α− iβ)−A = r−A e iAθ. (29) Substituting (29) into (27) and using ℑ(eiAθ) = sin(Aθ) gives Im,q(k, α, β) = mΓ(A)ℑ ( r−Ae iAθ ) = mΓ(A) r−A sin(Aθ) = mΓ(A) (α2 + β2)−A/2 sin ( A arctan β α ) , (30) which is (17). Since m > 0, Γ(A) > 0 for A > 0, and (α2 + β2)−A/2 > 0, the vanishing Im,q(k, α, β) = 0 (when β ̸= 0) is equivalent to sin(Aθ) = 0, i.e. (18). If β = 0 then sin(βx1/m) ≡ 0 and (15) gives Im,q(k, α, 0) = 0. The following corollary recovers Theorem 4 from Theorem 5. Corollary 1. Let m = 4, α = 1, β = 1, q = 0. For k > −1, I4,0(k, 1, 1) = ∫ ∞ 0 e−x1/4 sin ( x1/4 ) xk dx = 2−2k Γ(4k + 4) sin ( (k + 1)π ) . Proof. Here A = mk+ q+m = 4k+4, r = √ α2 + β2 = √ 2, and θ = arctan(1) = π/4. Substituting these into (17) gives I4,0(k, 1, 1) = 4Γ(4k + 4) ( √ 2)−(4k+4) sin ( (4k + 4)π4 ) = 2−2k Γ(4k + 4) sin ( (k + 1)π ) . We also have the following special cases. Corollary 2 (Equal parameters α = β > 0). For any m > 0 and any k, q with A > 0, Im,q(k, α, α) = mΓ(A) 2−A/2 α−A sin ( A π 4 ) . In particular, Im,q(k, α, α) = 0 if and only if A ∈ 4Z. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 9 of 16 Proof. When α = β, we have r = √ 2α and θ = π/4, so (17) reduces to the displayed formula. The zero condition follows from sin(Aπ/4) = 0. Corollary 3 (q = 0). If q = 0 and k > −1, then A = m(k + 1) and Im,0(k, α, β) = mΓ ( m(k + 1) ) (α2 + β2)−m(k+1)/2 sin ( m(k + 1) arctan β α ) . Proof. Put q = 0 in (17). Next, we give the cosine variant of Theorem 4. The proof is similar as in Theorem 5 but for the completeness, we write it here. Theorem 6. Let m > 0, α > 0, β ∈ R, q ∈ R, and k ∈ R. Define Jm,q(k, α, β) = ∫ ∞ 0 e−αx1/m cos ( βx1/m ) xk (x1/m)q dx, (31) and set A = mk + q +m. (32) Then the integral in (31) converges for every A > −1. Moreover, if A > 0 (which we assume for the evaluation below), one has the closed form Jm,q(k, α, β) = mΓ(A) (α2 + β2)−A/2 cos ( A arctan β α ) . (33) Consequently, if β ̸= 0 then Jm,q(k, α, β) = 0 if and only if A arctan β α ∈ π 2 + πZ; (34) if β = 0 then Jm,q(k, α, 0) = mΓ(A)α−A. (35) Proof. We write the integrand as e−αx1/m cos(βx1/m)xk+q/m. For x ↓ 0 we use | cos t| ≤ 1 to obtain ∣∣e−αx1/m cos(βx1/m)xk+q/m ∣∣ ≪ xk+q/m (x ↓ 0), (36) so ∫ 1 0 |e−αx1/m cos(βx1/m)xk+q/m| dx < ∞ provided k + q/m > −1, i.e. mk + q > −1 ⇐⇒ A−m > −1 ⇐⇒ A > −1. (37) As x → ∞ we have | cos(βx1/m)| ≤ 1 and the factor e−αx1/m yields super–polynomial decay, hence∫ ∞ 1 ∣∣e−αx1/m cos(βx1/m)xk+q/m ∣∣ dx < ∞ for every k, q when α > 0. (38) Therefore (31) converges for all A > −1. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 10 of 16 Assume A > 0 for evaluation. We make the substitution u = x1/m, x = um, dx = mum−1 du, (39) and note xk(x1/m)q = umk+q, to obtain Jm,q(k, α, β) = ∫ ∞ 0 e−αu cos(βu)umk+q (mum−1) du = m ∫ ∞ 0 e−αu cos(βu)uA−1 du. (40) Using cos(βu) = ℜ ( eiβu ) and absolute integrability for A > −1, we may pass the real part inside the integral: e−αu cos(βu) = ℜ ( e−(α−iβ)u ) =⇒ Jm,q(k, α, β) = mℜ ∫ ∞ 0 uA−1 e−(α−iβ)u du. (41) Let λ = α− iβ (ℜλ = α > 0). (42) For A > 0, the Gamma function, obtained by the substitution t = λu, gives∫ ∞ 0 uA−1e−λu du = ∫ ∞ 0 ( t λ )A−1 e−t dt λ = 1 λA ∫ ∞ 0 tA−1e−t dt = Γ(A) λA . (43) Applying (43) in (41) yields Jm,q(k, α, β) = mℜ ( Γ(A) (α− iβ)A ) . (44) We write α− iβ in polar form. With r = √ α2 + β2, θ = arctan β α ∈ ( −π 2 , π 2 ) , (45) we have α− iβ = r e−iθ, hence (α− iβ)−A = r−A e iAθ. (46) Substituting (46) into (44) and using ℜ(eiAθ) = cos(Aθ) gives Jm,q(k, α, β) = mΓ(A)ℜ ( r−Ae iAθ ) = mΓ(A) r−A cos(Aθ) = mΓ(A) (α2 + β2)−A/2 cos ( A arctan β α ) , (47) which proves (33). The zero condition (34) follows since m > 0, Γ(A) > 0 for A > 0, and (α2 + β2)−A/2 > 0, so Jm,q(k, α, β) = 0 if and only if cos(Aθ) = 0, i.e. Aθ ∈ π 2 + πZ. When β = 0, we have θ = 0 and cos(Aθ) = cos 0 = 1, and (47) reduces to (35). We have the following cosine analog of the Steiltjes example. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 11 of 16 Corollary 4. Let m = 4, α = 1, β = 1, q = 0. For k > −1, G(k) = ∫ ∞ 0 e−x1/4 cos ( x1/4 ) xk dx = 2−2k Γ(4k + 4) cos ( (k + 1)π ) . Furthermore, G(k) = 0 if and only if k = n− 1 2 , n ∈ {0, 1, 2, 3....}. Proof. Here A = 4k + 4, r = √ 2, and θ = arctan(1) = π/4. Substituting into (33), J4,0(k, 1, 1) = 4Γ(4k + 4) ( √ 2)−(4k+4) cos ( (4k + 4)π4 ) = 2−2k Γ(4k + 4) cos ( (k + 1)π ) . Note that in Corollary 4, for the vanishing of G(k), the values of k are not non-negative fractions, whereas the Steiltjes type moment problem requires k to be non-negative in- tegers. To have k non-negative integers and a cosine variant of Steiltjes , we have the following result. Corollary 5. Let m > 2 and α > 0, and set θ = π m , β = α tan θ = α tan ( π m ) , q = −m 2 . (48) For k ∈ R define Jm,q(k, α, β) = ∫ ∞ 0 e−αx1/m cos ( βx1/m ) xk (x1/m)q dx. (49) Then Jm,q(k, α, β) converges for k > −1 2 and Jm,q(k, α, β) = mΓ ( m(k + 1 2) ) (α2 + β2)− m(k+1/2) 2 cos ( π ( k + 1 2 )) . (50) In particular, Jm,q(k, α, β) = 0 ⇐⇒ k ∈ Z≥0. (51) Proof. By Theorem 6, for A = mk + q +m and θ = arctan(β/α) ∈ (−π 2 , π 2 ) one has Jm,q(k, α, β) = mΓ(A) (α2 + β2)−A/2 cos ( Aθ ) , (52) whenever A > 0 (and convergence holds for A > −1). With the parameter choice (48) we have θ = π/m ∈ (0, π2 ), A = mk + q +m = m ( k + 1 2 ) , and hence Aθ = m ( k + 1 2 ) · π m = π ( k + 1 2 ) . (53) Substituting (53) and A = m(k + 1 2) into (52) yields (50). Since A > 0 is equivalent to k > −1 2 , convergence and evaluation hold precisely on k > −1 2 . In (50) the prefactor mΓ(m(k + 1 2)) (α 2 + β2)−m(k+1/2)/2 is strictly positive for k > −1 2 , so Jm,q(k, α, β) = 0 ⇐⇒ cos ( π(k + 1 2) ) = 0 ⇐⇒ k + 1 2 ∈ Z ⇐⇒ k ∈ Z. Intersecting with k > −1 2 gives k ∈ {0, 1, 2, . . . }, which is (51). Corollary 5 gives the following explicit examples, I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 12 of 16 Example 3. Set m = 4, α = 1, β = α tan (π 4 ) = 1, q = −m 2 = −2, and define, f1(x) = e−x1/4 cos ( x1/4 ) (x1/4)−2 = e−x1/4 cos ( x1/4 ) x−1/2. Then for k > −1 2 , J4,−2(k, 1, 1) = ∫ ∞ 0 f1(x)x k dx = 4Γ(4k + 2) 2−(2k+1) cos ( π(k + 1 2) ) , and therefore J4,−2(k, 1, 1) = 0 ⇐⇒ k ∈ Z≥0. Example 4. Choose m = 6, α = √ 3, β = α tan (π 6 ) = 1, q = −m 2 = −3, and define f2(x) = e− √ 3x1/6 cos ( x1/6 ) (x1/6)−3 = e− √ 3x1/6 cos ( x1/6 ) x−1/2. Then for k > −1 2 , J6,−3 ( k, √ 3, 1 ) = ∫ ∞ 0 f2(x)x k dx = 6Γ(6k + 3) 2−(6k+3) cos ( π(k + 1 2) ) , which gives J6,−3 ( k, √ 3, 1 ) = 0 ⇐⇒ k ∈ Z≥0. We also have the following integral representations of the important constants, Corollary 6. In Theorem 5, choose m = 2, α = 1, β = 1, k = 0, q = 0. Then A = mk+ q+m = 2 and θ = arctan(β/α) = arctan(1) (so sin(2θ) = 1), which gives∫ ∞ 0 e−x1/2 sin ( x1/2 ) dx = 1 (converges since A = 2 > 0). (54) Proof. By (17), I = mΓ(2)r−1 sin(2θ) with m = 2, r = α2 + β2 = 2. Thus I = 2 · 1 · (1/2) · 1 = 1. The following theorem gives integral representation of specific values of the gamma function. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 13 of 16 Theorem 7. Let A > 0, m > 0, and choose any angle θ ∈ ( 0, π2 ) . Define r = ( m sin(Aθ) ) 2 A , α = √ r cos θ > 0, β = √ r sin θ, (55) and set k = 0, q = A−m. (56) Then A = mk + q +m and, with Im,q(k, α, β) = ∫ ∞ 0 e−αx1/m sin ( βx1/m ) xk (x1/m)q dx, (57) we have A > −1 (indeed A > 0) so the integral converges, and moreover Im,q(k, α, β) = Γ(A). (58) Proof. By (56), A = mk + q + m = 0 + (A − m) + m = A, so the parameter A in Theorem 5 agrees with the present A. Using θ = arctan(β/α) ∈ (0, π2 ) and r = α2 + β2 from (55), Theorem 5 gives (for A > 0) Im,q(k, α, β) = mΓ(A) r−A/2 sin(Aθ). (59) By construction, rA/2 = ( m sin(Aθ) ) , hence r−A/2 sin(Aθ) = sin(Aθ) m sin(Aθ) = 1 m . (60) Substituting (60) into (59) yields Im,q = Γ(A), i.e. (58). Convergence is ensured because A > 0 implies the hypothesis A > −1 in Theorem 5. Remark 2. The choice of θ ∈ (0, π2 ) guarantees α > 0 and is compatible with θ = arctan(β/α). The representation (58) uses only the freely chosen numerical parameters (m, θ) and the target parameter A, the constants α, β are then determined by (55) and do not involve Γ(A). Corollary 7 (Integer case A = n ∈ N with m = 1). Let n ∈ N and m = 1. Choose any θ ∈ (0, π2 ) and define α = ( sin(nθ) )2/n cos θ, β = ( sin(nθ) )2/n sin θ, k = 0, q = n− 1. (61) Then ∫ ∞ 0 e−αx sin(βx)xn−1 dx = Γ(n) = (n− 1)!. (62) Proof. This is Theorem 7 with A = n and m = 1, note r = (sin(nθ))2/n so that rn/2 = sin(nθ) and (60) holds. I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 14 of 16 Corollary 8 (Half–integer case A = ℓ + 1 2). Let ℓ > 0 and A = ℓ + 1 2 , m = 1. Choosing any θ ∈ (0, π2 ) and α = ( sin(Aθ) )2/A cos θ, β = ( sin(Aθ) )2/A sin θ, k = 0, q = A− 1, (63) one obtains ∫ ∞ 0 e−αx sin(βx)xA−1 dx = Γ ( ℓ+ 1 2 ) = (2ℓ)! 4ℓ ℓ! √ π (ℓ ∈ 1 2N). (64) Proof. Immediate from Theorem 7 with m = 1, A = ℓ+ 1 2 . Corollary 9. Let A > 1 and θ = π 2A ∈ (0, π2 ). For any m > 0, set α = m1/A cos ( π 2A ) , β = m1/A sin ( π 2A ) , k = 0, q = A−m. (65) Then sin(Aθ) = 1 and r = α2 + β2 = m2/A, so∫ ∞ 0 e−αx1/m sin ( βx1/m ) x A m −1 dx = Γ(A). (66) Proof. Here rA/2 = m and sin(Aθ) = 1, (59) yields Im,q = mΓ(A)m−1 · 1 = Γ(A). 3. Conclusion Motivated by Stieltjes’ classical counterexample in the (indeterminate) Stieltjes mo- ment problem, we revisited the kernel f(x) = e−x1/4 sin ( x1/4 ) and evaluated its moments against continuous powers. Theorem 4 gives the exact formula I(k) = ∫ ∞ 0 e−x1/4 sin ( x1/4 ) xk dx = 2−2kΓ(4k + 4) sin ( (k + 1)π ) , from which it follows that I(k) = 0 if and only if k ∈ Z≥0. This settles, in the affirmative and with a complete characterization, the question of which non-negative real exponents yield vanishing moments for the Stieltjes kernel. We then proved a generalization (Theorem 5) to the family Im,q(k, α, β) = ∫ ∞ 0 e−αx1/m sin ( βx1/m ) xk(x1/m)q dx, A = mk + q +m, establishing: (i) sharp convergence for A > −1, (ii) the closed form Im,q = mΓ(A) (α2 + β2)−A/2 sin ( A arctan(β/α) ) ; and (iii) a necessary and sufficient vanishing criterion A arctan(β/α) ∈ I. Ayoob / Eur. J. Pure Appl. Math, 18 (4) (2025), 7023 15 of 16 πZ. A parallel cosine variant (Theorem 6) yields Jm,q = mΓ(A) (α2+β2)−A/2 cos ( A arctan(β/α) ) with the expected zero set cos ( A arctan(β/α) ) = 0. A particularly effective parameteri- zation (Corollary 5), taking θ = π/m, β = α tan θ, and q = −m/2, forces Jm,q(k, α, β) = 0 ⇐⇒ k ∈ Z≥0, thus providing a cosine analogue that vanishes exactly at the non- negative integers. Beyond vanishing criteria, we showed that this framework generates Gamma values. Theorem 7 prescribes (α, β) via r = (m sin(Aθ))2/A to give the exact representation.∫ ∞ 0 e−αx1/m sin ( βx1/m ) xA/m−1 dx = Γ(A), from which the integer and half-integer cases follow transparently, recovering factorials and √ π factors in a unified way. This paper also includes a brief discussion of Salem’s equivalence of the Riemann hypothesis, which is based on the vanishing of an integral dependent on a continuous parameter. Acknowledgements The author would like to thank the Prince Sultan University for paying the publication fees for this work through TAS LAB. References [1] I. S. Gradshteyn and I. M. Ryzhik. Table of Integrals, Series, and Products. Academic Press, New York, 7th edition, 2007. [2] Christophe Chesneau. New integral formulas inspired by an old integral result. Inter- national Journal of Open Problems in Computer Science and Mathematics, 18(2):53– 71, 2025. 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