Certain logarithmic integrals, including solution of Monthly problem #tbd, zeta values, and 2 1 expressions for the Stieltjes constants 0 2 n Mark W. Coffey a J Department of Physics 6 1 Colorado School of Mines ] Golden, CO 80401 h p (Received 2011) - h t December 31, 2011 a m [ 1 Abstract v 3 We solve problem x proposed by O. Oloa, AMM xxx 2012 119? (to ap- 9 pear), p. yyy for certain definite logarithmic integrals. A number of generating 3 3 functions are developed with certain coefficients pn, and some extensions are 1. presented. The explicit relation of pn to N¨orlund numbers Bn(n) is discussed. 0 Certain inequalities are conjectured for the {p } sequence of coefficients, in- n 2 cluding its convexity, and an upper bound is demonstrated. It is shown that 1 v: pn values may be used to express the Stieltjes constants for the Hurwitz and i Riemann zeta functions, as well as values of these zeta functions at integer X argument. Other summations with the p coefficients are presented. n r a Key words and phrases logarithmic integrals, Pochhammer symbol, generating function, digamma func- tion, Glaisher constant, N¨orlund number, Hurwitz zeta function, Stieltjes constants 2010 AMS codes 11Y60, 11Y35, 05A15 1 Solution of problem xxx O. Oloa has proposed the following problem in the Amer. Math. Monthly 119 (?), yyy (2012?). (a) Prove the formula 1 1 1 2 3 + dx = ln(2π)− . (1.1) lnx 1−x 2 Z0 (cid:18) (cid:19) (b) If σ ≥ 0, find a closed form expression for 1 1 1 2 + xσdx. lnx 1−x Z0 (cid:18) (cid:19) N.B. Although the statement (b) is restricted to σ ≥ 0, this part may be extended to Re σ > −1, and our explicit expressions reflect this fact. We first prove (1.1) and answer part (b), and then present several extensions. Proof. Let (a) = Γ(a + n)/Γ(a) be the Pochhammer symbol, where Γ is the n Gamma function, ψ = Γ′/Γ be the digamma function, γ = −ψ(1) be the Euler constant, and F the generalized hypergeometric function (e.g., [7, 1]). We introduce p q the positive constants (e.g., [3], Proposition 11, [4], Proposition 5, [5], Proposition 2) 1 1 (−1)n+1 n s(n,k) p = − (−x) dx = , (1.2) n+1 n n! n! k +1 Z0 k=1 X where s(k,ℓ) is the Stirling number of the first kind. The first few values of these are p = 1/2, p = 1/12, p = 1/24, p = 19/720, and p = 3/160. These constants enter 2 3 4 5 6 the generating function ∞ 1 1 p zn−1 = + , |z| < 1. (1.3) n+1 z ln(1−z) n=1 X 2 Multiplying (1.3) by ln(1−z) and manipulating series, one finds the recursion relation n−1 1 p j+1 p = − , n ≥ 1. n+1 n+1 (n−j +1) j=1 X We then have Lemma 1. ∞ 1 (1−z) = [(n+1)p −np ]zn +p + , |z| < 1. (1.4) ln2(1−z) n+3 n+2 3 z2 n=1 X This follows from the derivative expression ∞ 1 1 (n−1)p zn−2 = − + . (1.5) n+1 z2 (1−z)ln2(1−z) n=2 X As a consequence, we obtain 2 ∞ 1 1 1 + = + [(n+3)p −np ]zn. (1.6) n+3 n+2 ln(1−z) z 4 (cid:18) (cid:19) n=1 X Then for (a), 1 1 1 2 1 1 1 2 + dz = + dz ln(1−z) z lnz 1−z Z0 (cid:18) (cid:19) Z0 (cid:18) (cid:19) 1 1 ∞ = + [(n+3)p −np ](1−z)n dz n+3 n+2 4 Z0 ( n=1 ) X ∞ 1 1 = + [(n+3)p −np ] . (1.7) n+3 n+2 4 n+1 n=1 X The integral representation of (1.2) may now be inserted, and the sums rewritten in terms of binomial coefficients. For instance, we have ∞ ∞ ∞ 1 (n+3) (n+3) x 2 x (−x) = (−1)n = (−1)n 1+ n+2 n+1(n+2)! (n+1) n+2 (n+1) n+2 n=1 n=1 (cid:18) (cid:19) n=1 (cid:20) (cid:21)(cid:18) (cid:19) X X X 3 1 = (2−x)(x−1)−x[1−2γ +x−2ψ(x+1)]. (1.8) 2 1 Noting that ψ(x+1)dx = 0, we have 0 1 R 1 1 2 1 1 x 3 + dz = + γ − −2γx+ x2 −2xψ(x+1) dx ln(1−z) z 4 2 2 Z0 (cid:18) (cid:19) Z0 (cid:20) (cid:21) 3 = ln(2π)− . (1.9) 2 In the last step, we integrated by parts (e.g., [2]), 1 1 1 1 xψ(x+1)dx = − lnΓ(x+1)dx = − [lnx+lnΓ(x)]dx = 1− ln(2π). (1.10) 2 Z0 Z0 Z0 For (b), we let B(x,y) = Γ(x)Γ(y)/Γ(x+y) be the Beta function. Then 1 1 1 2 + xσdx lnx 1−x Z0 (cid:18) (cid:19) 1 1 ∞ = + [(n+3)p −np ](1−x)n xσdx n+3 n+2 4 Z0 ( n=1 ) X ∞ 1 = + [(n+3)p −np ]B(n+1,σ+1). (1.11) n+3 n+2 4(σ +1) n=1 X Lemma 2. For Re y > 0, ∞ 1 1 p B(n+1,y) = − −1+2y −ln(2π)+2lnΓ(y)+(1−2y)ψ(y) . n+3 2 6y n=1 (cid:20) (cid:21) X (1.12) Proof. From (1.2) we have ∞ 1 1 p B(n+1,y) = − x(x−1)[ F (1,1,2−x;3,y +1;1)−1]dx. (1.13) n+3 3 2 2y n=1 Z0 X We now use the identity (1) (1) (2) j j j = 2 − , (1.14) (3) (2) (3) j j j 4 being a special case of (a) (a) (a+1) j j j = (a+1) −a , (a+2) (a+1) (a+2) j j j to obtain ∞ (1) (2−x) j j F (1,1,2−x;3,y +1;1) = 3 2 (3) (y +1) j j j=0 X ∞ (1) (2) (2−x) (1) j j j j = 2 − (2) (3) (y +1) j! j=0 (cid:20) j j(cid:21) j X 2y = [1−x−(x+y −1)ψ(y)+(x+y −1)ψ(x+y +1)]. (1.15) x(x−1) Carrying out the integration of (1.13) gives the Lemma. Then by Proposition 2 in the next section with y = σ +1 we obtain 1 1 1 2 3 + xσdx = (σ+1)ln(σ+1)−2σ+σψ(σ+1)−2lnΓ(σ+1)+ln(2π)− . lnx 1−x 2 Z0 (cid:18) (cid:19) (1.16) Remarks. As is apparent from (1.11) and (1.16), Re σ = −1 is the ‘critical line’ for divergence of the integral. We have the following Corollary 1. For n ≥ 1, n+1 (n+3)p −np = p p . (1.17) n+3 n+2 k+1 n−k+3 k=1 X This follows from multiplication of series, using (1.6), 2 ∞ 1 1 1 + = + [(n+3)p −np ]zn n+3 n+2 ln(1−z) z 4 (cid:18) (cid:19) n=1 X 5 ∞ n+1 = p p zn. (1.18) k+1 n−k+3 n=0 k=1 XX Is there a combinatorial interpretation of identity (1.13)? The Appendix generalizes (1.14) and the following identity for ratios of Pochham- mer symbols. Extensions We may proceed similarly as above, and find for instance ∞ 1 1 = [(n+1)(n+2)p −(n+1)(2n+1)p +n2p ]zn ln3(1−z) 2 n+4 n+3 n+2 n=1 X 1 3 1 − + − , |z| < 1, (2.1) z3 2z2 2z giving, along with (1.3) and (1.4), 3 1 1 1 3 3 1 + = + + + ln(1−z) z ln3(1−z) zln2(1−z) z2ln(1−z) z3 (cid:18) (cid:19) ∞ 1 = [(n+4)(n+5)p −(n+1)(2n+7)p +n2p ]zn n+4 n+3 n+2 2 n=4 X 1 z z2 133 + + − + z3. (2.2) 8 16 24 4320 Then 1 1 1 3 1 1 1 3 + dz = + dz ln(1−z) z lnz 1−z Z0 (cid:18) (cid:19) Z0 (cid:18) (cid:19) ∞ 1 1 3073 = [(n+4)(n+5)p −(n+1)(2n+7)p +n2p ] + n+4 n+3 n+2 2 (n+1) 17280 n=4 X 31 = − +6lnA, (2.3) 24 6 wherein A is Glaisher’s constant, such that lnA = −[ζ(−1)+ζ′(−1)] = 1/12−ζ′(−1), andζ(s)istheRiemannzetafunction. Thelattercontributionentersfromtheintegral 1 1 1 t2ψ(t+1)dt = −2 tlnΓ(t+1)dt = (1−ln2π)+2lnA. (2.4) 2 Z0 Z0 This integral may be readily determined from Kummer’s Fourier series for lnΓ. Oth- erwise, it may be found through the infinite series 1 γ ∞ 1 γ ∞ (−1)k t2ψ(t+1)dt = − + (−1)kζ(k) tk+1dt = − + ζ(k). (2.5) 3 3 k +2 Z0 k=2 Z0 k=2 X X For reference, we have from (2.1) ∞ 1 1 = (n+1)(n+2)(n+3)p −3(n+1)2(n+2)p ln4(1−z) 6 n+5 n+4 n=1 X(cid:2) 1 1 7 2 1 +(n+1)(3n2 +3n+1)p −n3p zn− − + − + , |z| < 1. (2.6) n+3 n+2 720 6z 6z2 z3 z4 (cid:3) Then also using (1.3), (1.4), and (2.1), we find 4 1 1 + ln(1−z) z (cid:18) (cid:19) ∞ 1 = [(n+1)(n+2)(n+3)p −3(n+1)2(n+2)p +(n+1)(3n2+3n+1)p −n3p ]zn n+5 n+4 n+3 n+2 6 n=1 X ∞ 1 − +2 [(n+1)(n+2)p −(n+1)(2n+1)p +n2p ]zn−1 n+4 n+3 n+2 720 n=1 X ∞ ∞ +6 [(n+2)p −(n+1)p ]zn−1 +4 p zn−1. (2.7) n+4 n+3 n+4 n=1 n=1 X X By using (1.2) we obtain 1 1 1 4 1 1 53 57 143 t4 + dz = − + t− t2 + t3 + ln(1−z) z 720 18 8 36 24 Z0 (cid:18) (cid:19) Z0 (cid:18) 7 1 8 10 + − t+6t2 − t3 [ψ(t+1)+γ] dt 6 3 3 (cid:20) (cid:21) (cid:19) 49 5 = − +2lnA+ ζ(3). (2.8) 72 2π2 From (2.6) we have ∞ 1 1 = [(n+1)(n+2)(n+3)(n+4)p −2(n+1)(n+2)(n+3)(2n+3)p ln5(1−z) 24 n+6 n+5 n=1 X +(n+1)(n+2)(6n2 +12n+7)p −(n+1)(2n+1)(2n2 +2n+1)p +n4p zn n+4 n+3 n+2 1 5 25 5 1 (cid:3) − + − + − , |z| < 1, (2.9) 24z 8z2 12z3 2z4 z5 leading to 5 1 1 + ln(1−z) z (cid:18) (cid:19) ∞ 1 = [(n+1)(n+2)(n+3)(n+4)p −2(n+1)(n+2)(n+3)(2n+3)p n+6 n+5 24 n=1 X +(n+1)(n+2)(6n2 +12n+7)p −(n+1)(2n+1)(2n2 +2n+1)p +n4p zn n+4 n+3 n+2 ∞ 5 (cid:3) + [(n+1)(n+2)(n+3)p −3(n+1)2(n+2)p +(n+1)(3n2+3n+1)p −n3p ]zn−1 n+5 n+4 n+3 n+2 6 n=1 X ∞ +5 [(n+2)(n+3)p −(n+2)(2n+3)p +(n+1)2p ]zn−1 n+5 n+4 n+3 n=1 X ∞ ∞ +10 [(n+3)p −(n+2)p ]zn−1 +5 p zn−1. (2.10) n+5 n+4 n+5 n=1 n=1 X X By performing the integration and using (1.2) we determine 1 1 1 5 4367 5 15 35 + dz = − + lnA+ ζ(3)− ζ′(−3). (2.11) ln(1−z) z 8640 3 8π2 3 Z0 (cid:18) (cid:19) Here we have used 1 49 t4ψ(t)dt = −4ζ′(−1)+ζ′(0)+6ζ′(−2)−4ζ′(−3) 180 Z0 8 11 1 3 = − +4lnA− ln(2π)− ζ(3)−4ζ′(−3). (2.12) 180 2 2π2 Based upon the evaluation of 1tkψ(t)dt = −k 1tk−1lnΓ(t)dt, k > −1, we may 0 0 k+1 anticipate that 1 1 + 1 R dz evaluates inRterms of 0 ln(1−z) z (cid:16) (cid:17) R ζ(3) ζ(5) ζ(k −1) Q+QlnA+Q +Q +...+Q π2 π4 πk−2 +Qζ′(−3)+Qζ′(−5)+...+ζ′(1−k), k even, (2.13a) and ζ(3) ζ(5) ζ(k) Q+QlnA+Q +Q +...+Q π2 π4 πk−1 +Qζ′(−3)+Qζ′(−5)+...+ζ′(2−k), k odd. (2.13b) We provide some discussion of intermediate calculations in the above steps. First, we consider the integrals 1tkψ(t)dt. These may be treated by multiple integrations 0 by parts using the fact thRat −ψ(a) is the zeroth Stieltjes constant for the Hurwitz zeta function ζ(s,a). That is, we have the limit representation 1 ψ(t) = −lim ζ(z,t)− . (2.14) z→1 z −1 (cid:18) (cid:19) Thus, 1 1 1 tkψ(t)dt = −lim tk ζ(z,t)− dt, (2.15) z→1 z −1 Z0 Z0 (cid:20) (cid:21) with the interchange justified by uniform convergence of the integral. We have the 1 properties ∂ ζ(s,a) = −sζ(s+ 1,a) and ζ(s,t)dt = 0 for Re s < 1. By iteration a 0 we may then obtain the integrals R 1 1 1 tkζ(z,t)dt = − tk∂ ζ(z −1,t)dt t z −1 Z0 Z0 9 k 1 ζ(z −1) = tk−1ζ(z −1,t)dt− . (2.16) z −1 z −1 Z0 Secondly, the insertion of the integral representation of (1.2) for p into sums n+1 such as (2.10) gives certain hypergeometric summations. Again, F denotes the p q generalized hypergeometric function. As an illustration, consider a contribution from a sum such as ∞ n3p . By recalling the property (a) = a(a+1) , we have n=1 n+6 n+1 n P ∞ ∞ (−t) (−t) − n3 n+5 = − (n+1)3 n+6 (n+5)! (n+6)! n=1 n=0 X X ∞ (2)3 (6−t) = −t(t−1)(t−2)(t−3)(t−4)(t−5) n n (1)3 (n+6)! n=0 n X 1 ∞ (2)3 (6−t) 1 = −t(t−1)(t−2)(t−3)(t−4)(t−5) n n 720 (1)2 (7) n! n=0 n n X 1 = −−t(t−1)(t−2)(t−3)(t−4)(t−5) F (2,2,2,6−t;1,1,7;1). (2.17) 4 3 720 Thirdly, (i) there are relations between the F sums so obtained, and (ii) they p+1 p are often related to the digamma function. By manipulating divided difference forms of the ψ function, relations such as the following may be obtained: F (2,2,3−t;3,4;1)+ F (2,2,2,3−t;1,3,4;1) 3 2 4 3 12[3−3t+γt+tψ(t+1)] 12[1−2t+γt+tψ(t+1)] 12 = − + = . (2.18) t(t−1)(t−2) t(t−1)(t−2) t(t−1) Likewise, we have 12(4−t) F (2,2,2,2,3−t;1,1,3,4;1)+ F (2,2,2,3−t;1,3,4;1) = . (2.19) 5 4 4 3 t(t−1)(t−2) In this way, summations over the p constants may be evaluated. n 10