Generalized Stieltjes constants and integrals involving the log-log function: Kummer's Theorem in action
GGENERALIZED STIELTJES CONSTANTS AND INTEGRALS INVOLVING THELOG-LOG FUNCTION: KUMMER’S THEOREM IN ACTION.
OMRAN KOUBA † Abstract.
In this note, we recall Kummer’s Fourier series expansion of the 1-periodic function thatcoincides with the logarithm of the Gamma function on the unit interval (0 , x (cid:55)→ ln ln(1 /x ). Introduction and Notation
The aim of this paper is to present an alternative proof of the reflection principle of the first ordergeneralized Stieltjes constants, and to give an alternative approach to the evaluation of some integralsinvolving the function x (cid:55)→ ln ln(1 /x ). The basic tool for this investigation is a result of Kummer recalledbelow (Theorem 1).The first order generalized Stieltjes constant γ ( a ) is defined for a ∈ (0 ,
1) by γ ( a ) = lim n →∞ (cid:32) n (cid:88) k =0 ln( a + k ) a + k −
12 ln ( n + a ) (cid:33) . From this, it is easy to show that γ ( a ) − γ (1 − a ) = lim n →∞ (cid:32) n (cid:88) k = − n ln | a + k | a + k (cid:33) def = (cid:88) (cid:48) n ∈ Z ln | a + n | a + n , where the primed sum denotes the “principal value” as shown above. For integers p and q with 0 < p < q the difference γ ( p/q ) − γ (1 − p/q ) can be expressed as follows γ ( p/q ) − γ (1 − p/q ) = − π ln(2 πqe γ ) cot (cid:18) pπq (cid:19) + 2 π q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) ln Γ (cid:18) jq (cid:19) . The formula is attributed to Almkvist and Meurman who obtained it by calculating the derivative of thefunctional equation for the Hurwitz zeta function ζ ( s, v ) with respect to s at rational v , see [2]. However,it was recently discovered that an equivalent form of this formula was already obtained by Carl Malmstenin 1846 (see [5]). An elementary proof of this formula will be presented in Proposition 2.In a recent series of articles ([9],[10],[11],[3],[14]), the authors proved some formulas from the Tableof integrals, Series, and Products , of Gradshteyn and Ryzhik [7]. Further, the monographs [12, 13] aredevoted to providing proofs for the formulas in [7]. In fact, we are particularly interested in integrals
Key words and phrases.
Gamma function, log-log integrals, Fourier series, numerical series. † Department of Mathematics, Higher Institute for Applied Sciences and Technology. a r X i v : . [ m a t h . C A ] F e b OMRAN KOUBA involving the function x (cid:55)→ ln ln(1 /x ). Indeed, entries 4 .
325 of [7] contain the following evaluations: (cid:90) ln(ln(1 /x ))1 + x = π √ π Γ(3 / / (cid:90) ln(ln(1 /x ))1 + x + x = π √ √ π Γ(2 / / (cid:90) ln(ln(1 /x ))1 + 2 x cos t + x = π t ln (2 π ) t/π Γ (cid:0) + t π (cid:1) Γ (cid:0) − t π (cid:1) These integrals can be traced back to [6]. The first of them was the object of a detailed investigationin [14], where the author says that his approach can be adapted to prove also the second one. A generalapproach that yields the first two integrals, and much more evaluations, can also be found in [2]. Thisline of investigation was completed by adapting the methods of [14] to obtain general results that includeall the above mentioned integrals in [11].Our aim is to present an alternative approach to the evaluation of these integrals. Our starting pointwill be Kummer’s Fourier expansion of Log Γ, (Theorem 1), where Γ is the well-known Eulerian gammafunction. This result is attributed to Kummer in (1847), a more accessible reference is [4, Section 1.7]:
Theorem 1 (Kummer,[8]) . For < x < , ln Γ( x ) √ π = − ln(2 sin( πx ))2 + ( γ + ln(2 π )) (cid:18) − x (cid:19) + 1 π ∞ (cid:88) k =1 ln kk sin(2 πkx ) , where γ is the Euler-Mascheroni constant. The reflection formula for the first order generalized Stieltjes constants
As we explained in the introduction, this formula relates the first order generalized Stieltjes constant γ ( a ) to its reflected value γ (1 − a ) for rational a . The presented proof is different from that of Almkvistand Meurman, and has the advantage of being elementary in the sense that it does not make use of thefunctional equation of the Hurwitz zeta function. Proposition 2. for positive integers p and q with p < q , we have (cid:88) (cid:48) n ∈ Z ln (cid:12)(cid:12)(cid:12) n + pq (cid:12)(cid:12)(cid:12) n + pq = − π ln(2 πqe γ ) cot (cid:18) pπq (cid:19) + 2 π q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) ln Γ (cid:18) jq (cid:19) . where the primed sum denotes the “principal value”, defined as follows: (cid:88) (cid:48) n ∈ Z a n = lim n →∞ (cid:32) n (cid:88) k = − n a k (cid:33) . Proof.
The statement of Theorem 1 is written as ∞ (cid:88) k =1 ln kk sin(2 πkx ) = − π π + π πx ) + π ln(2 πe γ ) (cid:18) x − (cid:19) + π ln Γ( x ) . (1)Now, consider a positive integer q with q ≥
2. For j ∈ { , , . . . , q − } we have ∞ (cid:88) k =1 ln kk sin (cid:18) πkjq (cid:19) = − π π + π (cid:18) πjq (cid:19) + π ln(2 πe γ ) (cid:18) jq − (cid:19) + π ln Γ (cid:18) jq (cid:19) . (2) UMMER’S THEOREM IN ACTION 3
Multiply both sides of (2) by sin (cid:16) πjpq (cid:17) , where p is some integer from { , . . . , q − } , and add the resultingequalities for j = 1 , . . . , q −
1, to obtain ∞ (cid:88) k =1 ln kk A p,q ( k ) = − π ln π B p,q + π C p,q + π ln(2 πe γ ) D p,q + π q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) ln Γ (cid:18) jq (cid:19) , (3)where A p,q ( k ) = q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) sin (cid:18) πjkq (cid:19) B p,q = q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) C p,q = q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) ln sin (cid:18) πjq (cid:19) D p,q = q − (cid:88) j =1 (cid:18) jq − (cid:19) sin (cid:18) πjpq (cid:19) . These sums are now simplified. Let ω q = exp (cid:16) πiq (cid:17) , and use (cid:80) q − j =0 ω njq = qχ q ( n ) where χ q ( n ) = 1 if n ≡ q and χ q ( n ) = 0 otherwise. The imaginary part of the identity gives B p,q = 0 . (4)Also, A p,q ( k ) = 12 q − (cid:88) j =1 (cid:18) cos (cid:18) πj ( p − k ) q (cid:19) − cos (cid:18) πj ( p + k ) q (cid:19)(cid:19) = 12 (cid:60) q − (cid:88) j =0 ω ( p − k ) j − q − (cid:88) j =0 ω ( p + k ) j = q χ q ( p − k ) − χ q ( p + k )) . That is A p,q ( k ) = (cid:40) q if k ≡ p mod q, − q if k ≡ − p mod q. (5)On the other hand, the change of summation index j ← q − j in the formula for C p,q shows that C p,q = q − (cid:88) j =1 sin (cid:18) πp − πjpq (cid:19) ln sin (cid:18) π − πjq (cid:19) = − C p,q . Thus, C p,q = 0 . (6)Finally, use (4) to obtain D p,q = 1 q q − (cid:88) j =1 j sin (cid:18) πjpq (cid:19) . OMRAN KOUBA
Now for, 0 < θ < π , we have1 + 2 q − (cid:88) j =1 cos(2 jθ ) = q − (cid:88) j =1 − q e ijθ = e iqθ − e i (1 − q ) θ e iθ −
1= sin((2 q − θ )sin θ = sin(2 qθ ) cot θ − cos(2 qθ ) . Taking the derivative with respect to θ and substituting θ = πp/q we get D p,q = −
12 cot (cid:18) pπq (cid:19) . (7)Replacing (4),(5),(6) and (7) in (3) we obtain ∞ (cid:88) k =0 (cid:18) ln( qk + p ) k + p/q − ln( qk + q − p ) k + 1 − p/q (cid:19) = − π ln(2 πe γ ) cot (cid:18) pπq (cid:19) + 2 π q − (cid:88) j =1 sin (cid:18) πjpq (cid:19) ln Γ (cid:18) jq (cid:19) . (8)The final step is to use the well-known cotangent partial fraction expansion: ∞ (cid:88) k =0 (cid:18) k + p/q − k + 1 − p/q (cid:19) = lim N →∞ N (cid:88) k = − N k + p/q = qp − N (cid:88) k =1 p/qk − ( p/q ) = qp + ∞ (cid:88) k =1 p/q ( p/q ) − k = π cot (cid:18) pπq (cid:19) . (9)Thus, subtracting (ln q ) times (9) from (8) we obtain the desired conclusion. (cid:3) Examples.
Taking p = 1 and q ∈ { , } we obtain (cid:88) (cid:48) n ∈ Z ln (cid:12)(cid:12) n + (cid:12)(cid:12) n + = π √ (cid:18) ( )2 π e γ (cid:19) . (cid:88) (cid:48) n ∈ Z ln (cid:12)(cid:12) n + (cid:12)(cid:12) n + = π ln (cid:18) Γ ( )2 π e γ (cid:19) . The Evaluation of some integrals involving the log-log function
In this section we use Theorem 1, to evaluate some difficult integrals.
Proposition 3.
For < x < , we have: (cid:90) ln ln(1 /u ) u − πx ) u + 1 du = π πx ) (cid:18) (1 − x ) ln(2 π ) + ln (cid:18) Γ(1 − x )Γ( x ) (cid:19)(cid:19) . And, taking the limit as x tend to / , we obtain (cid:90) ln ln(1 /u )( u + 1) du = ln √ π + Γ (cid:48) (1 / /
2) = ln (cid:114) π − γ . UMMER’S THEOREM IN ACTION 5
Proof.
Indeed, subtracting the corresponding Kummer’s Formulas, for ln Γ( x ) and ln Γ(1 − x ) we see that,for 0 < x < (cid:18) Γ( x )Γ(1 − x ) (cid:19) = ( γ + ln(2 π )) (1 − x ) + 2 π ∞ (cid:88) k =1 ln kk sin(2 πkx ) , (10)or equivalently, ln (cid:18) (2 π ) x Γ( x )(2 π ) − x Γ(1 − x ) (cid:19) = γ (1 − x ) + 2 π ∞ (cid:88) k =1 ln kk sin(2 πkx ) . (11)Now, using the fact that for (cid:60) s > k ≥ Γ( s ) k s = (cid:82) ∞ t s − e − kt dt , we conclude that for s > < x <
1, we have ∞ (cid:88) k =1 e πikx k s = 1Γ( s ) ∞ (cid:88) k =1 (cid:18)(cid:90) ∞ t s − e − kt e πikx dt (cid:19) = 1Γ( s ) (cid:90) ∞ e − t +2 πix − e − t +2 πix t s − dt. Restricting our attention to the imaginary parts we get ∞ (cid:88) k =1 sin(2 πkx ) k s = sin(2 πx )Γ( s ) (cid:90) ∞ t s − e t + e − t − πx ) dt. (12)Now, taking the derivative with respect to s at s = 1 we obtain, for 0 < x <
1, the following: ∞ (cid:88) k =1 ln kk sin(2 πkx ) = Γ (cid:48) (1)Γ (1) (cid:90) ∞ sin(2 πx ) e t + e − t − πx ) dt − sin(2 πx )Γ(1) (cid:90) ∞ ln te t + e − t − πx ) dt. (13)Taking into account the facts Γ (cid:48) (1) = − γ , Γ(1) = 1, and (cid:90) ∞ sin(2 πx ) e t + e − t − πx ) dt = π (cid:18) − x (cid:19) , for 0 < x < ∞ (cid:88) k =1 ln kk sin(2 πkx ) = − γπ − x ) − sin(2 πx ) (cid:90) ∞ ln te t + e − t − πx ) dt. (14)The change of variables t = ln(1 /u ) yields:2 π ∞ (cid:88) k =1 ln kk sin(2 πkx ) + γ (1 − x ) = − πx ) π (cid:90) ln ln(1 /u ) u − πx ) u + 1 du. (15)Finally, combining (11) and (15) we obtain the desired result. Concerning the limit as x tend to 1 /
2, weuse the well-known fact that Γ (cid:48) (1 / / = ψ (1 /
2) = − γ − (cid:3) OMRAN KOUBA
Examples.
Taking x = 1 / x = 1 / x = 1 / (cid:90) ln ln(1 /u ) u + u + 1 du = − π √ (cid:18) π Γ (cid:18) (cid:19)(cid:19) . (16) (cid:90) ln ln(1 /u ) u + 1 du = − π (cid:18) π Γ (cid:18) (cid:19)(cid:19) . (17) (cid:90) ln ln(1 /u ) u − u + 1 du = − π √ (cid:18) π ) Γ (cid:18) (cid:19)(cid:19) , = − π √ (cid:18) π Γ (cid:18) (cid:19)(cid:19) . (18)where we used freely the duplication, and the reflection formulas for the gamma function [1, 6.1.17 and6.1.18]. In particular, we used Γ( ) = √ /π √ Γ ( ) that follows readily from these formulas.The second degree polynomial in the integrand’s denominator in Proposition 3 has negative discrim-inant. In the next proposition the corresponding denominator has real roots outside the interval [0 , Proposition 4.
Let A Γ : R −→ R be the function defined by A Γ ( y ) = − ln(2 π )2 y + sinh( πy ) π (cid:90) ln ln(1 /u ) u + 2 cosh( πy ) u + 1 du. Then, for y ∈ R we have Γ (cid:18) iy (cid:19) = (cid:114) π cosh( πy/ e iA Γ ( y ) . Proof.
Let us rephrase Proposition 3, by taking x = t +12 in order to give more symmetric aspect to theformula there: ∀ t ∈ ( − , , (cid:90) ln ln(1 /u ) u + 2 cos( πt ) u + 1 du = π πt ) (cid:18) ln(2 π ) t + ln (cid:18) Γ( t )Γ( − t ) (cid:19)(cid:19) , or equivalently, for − < t <
1, we haveexp (cid:18) − ln(2 π ) t + 2 sin( πt ) π (cid:90) ln ln(1 /u ) u + 2 cos( πt ) u + 1 du (cid:19) = Γ( t )Γ( − t ) . Using analytic continuation we deduce that, for − < (cid:60) z < (cid:18) − ln(2 π ) z + 2 sin( πz ) π (cid:90) ln ln(1 /u ) u + 2 cos( πz ) u + 1 du (cid:19) = Γ( z )Γ( − z ) . In particular, setting z = iy with y ∈ R , we obtain e iA Γ ( y ) = Γ( iy )Γ( − iy ) . But, by Euler’s reflection formula [1, 6.1.17] we know that (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) iy (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = Γ (cid:18) iy (cid:19) Γ (cid:18) iy (cid:19) = Γ (cid:18) iy (cid:19) Γ (cid:18) − iy (cid:19) = π cosh( πy/ , UMMER’S THEOREM IN ACTION 7 therefore, the square of the continuous function: y (cid:55)→ (cid:114) cosh( πy/ π Γ (cid:18) iy (cid:19) e − iA Γ ( y ) is equal to 1 for every y ∈ R , hence, it must be constant and consequently identical to 1 which is its valuefor y = 0. (cid:3) Corollary 5.
Let the principal determination of the argument of a nonzero complex number z be denotedby Arg , and let α be defined by the formula α = inf (cid:26) y > (cid:18) iy (cid:19) = − (cid:114) π cosh( πy/ (cid:27) . Then, for every y ∈ ( − α, α ) we have (cid:90) ln ln(1 /u ) u + 2 cosh( πy ) u + 1 du = ln √ π · πy sinh( πy ) + π sinh( πy ) Arg Γ (cid:18) iy (cid:19) . Moreover, using
Mathematica R (cid:13) Software, we readily obtain α ≈ .
106 689 535 698 .Proof.
The definition of α implies that ∀ y ∈ ( − α, α ) , Γ (cid:18) iy (cid:19) ∈ C \ (( −∞ , × { } ) . Thus, the function y (cid:55)→ Arg(Γ( iy )) − A Γ ( y ) is continuous on ( − α, α ), takes its values in 2 π Z , andis equal to 0 for y = 0. Therefore, A Γ ( y ) = Arg(Γ( iy )), for every y ∈ ( − α, α ), which is the desiredconclusion. (cid:3) Examples. (cid:90) ln ln(1 /u ) u + 4 u + 1 du = ln(2 π ) √ (cid:32) √ √ (cid:33) + π √ (cid:32)
12 + iπ ln (cid:32) √ √ (cid:33)(cid:33) . (cid:90) ln ln(1 /u ) u + 3 u + 1 du = 2 ln(2 π ) √ φ ) + 2 π √ (cid:18)
12 + iπ ln( φ ) (cid:19) . where φ = 1 + √
52 is the golden ratio.More generally, for 2 < k < απ ) ≈ . × , the following holds (cid:90) ln ln(1 /u ) u + ku + 1 du = 2 ln(2 π ) √ k − φ k ) + 2 π √ k − (cid:18)
12 + iπ ln( φ k ) (cid:19) with φ k = √ k + 2 + √ k −
22 .It is worth mentioning that
Mathematica R (cid:13) OMRAN KOUBA
Proposition 6 ([2]) . For any complex number z with (cid:60) z > , we have F ( z ) def = (cid:90) t z − ln(ln(1 /t ))1 + t z dt = − ln 22 z Log(2 z ) where Log is the principal branch of the logarithm.Proof.
We start by evaluating F (1). Note that F (1) = (cid:90) ln(ln(1 /t ))1 + t dt = (cid:90) ∞ e − x e − x ln( x ) dx. So, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) F (1) − n (cid:88) k =1 ( − k − (cid:90) ∞ e − kx ln( x ) dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:90) ∞ | ln x | e x e − nx dx. Because x (cid:55)→ | ln x | e x is integrable on (0 , + ∞ ), we conclude using Lebesgue’s dominated convergence theoremthat lim n →∞ (cid:90) ∞ | ln x | e x e − nx dx = 0Thus, F (1) = ∞ (cid:88) k =1 ( − k − (cid:90) ∞ e − kx ln( x ) dx. A simple change of variables shows that (cid:90) ∞ e − kx ln( x ) dx = 1 k (cid:90) ∞ e − u (ln u − ln k ) du = − γk − ln kk since (cid:82) ∞ ln( u ) e − u du = Γ (cid:48) (1) = − γ . It follows that F (1) = − γ ∞ (cid:88) k =1 ( − k − k + ∞ (cid:88) k =1 ( − k ln kk = − γ ln 2 + ∞ (cid:88) k =1 ( − k ln kk . (20)Now, note that ln ( k + 1) − ln k = ln k (cid:32)(cid:18) k ln (cid:18) k (cid:19)(cid:19) − (cid:33) = ln k (cid:18) k ln k + O (cid:18) k ln k (cid:19)(cid:19) = 2 ln kk + O (cid:18) ln kk (cid:19) . This proves that the series (cid:80) (cid:0) ln ( k + 1) − ln k − ln kk (cid:1) is convergent. Consequently, if we define G n = (cid:80) nk =1 ln kk then there is a real number (cid:96) such that G n = ln n + (cid:96) + o (1). But n (cid:88) k =1 ( − k ln( k ) k = n (cid:88) k =1 ln(2 k ) k − n (cid:88) k =1 ln kk = (ln 2) H n + G n − G n = (ln 2)(ln n + γ ) + 12 (cid:0) ln ( n ) − ln (2 n ) (cid:1) + o (1)= −
12 ln γ ln 2 + o (1) , where we used H n = (cid:80) nk =1 /k = ln n + γ + o (1), (see [1, 4.1.32]). UMMER’S THEOREM IN ACTION 9
Now, let n tend to + ∞ to obtain ∞ (cid:88) k =1 ( − k ln( k ) k = −
12 ln γ ln 2 . Combining this with (20) we conclude that F (1) = − ln z ∈ (0 , + ∞ ) the change of variables t z = u shows that F ( z ) = 1 z (cid:90) ln ln(1 /u ) − ln z )1 + u du = F (1) z − ln( z ) ln(2) z = − ln(2 z ) ln(2)2 z , and the desired conclusion follows by analytic continuation. (cid:3) Acknowledgement.
The author would like to thank the anonymous reviewers for their commentsthat greatly improved the manuscript.
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Department of Mathematics, Higher Institute for Applied Sciences and Technology, P.O. Box 31983,Damascus, Syria.
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