WATSON RESUMMATION OF A CLASS OF HAUSDORFF–TRANSFORMED POWER SERIES 6 0 0 2 E.DEMICHELIANDG.A.VIANO n a Abstract. In this paper we study a class of Hausdorff–transformed power J series whose convergence is extremely slow for large values of the argument. 3 We performaWatson–type resummationoftheseexpansions, andobtain, by 1 theuseofthePollaczekpolynomials,anewrepresentationwhoseconvergence is muchfaster. We canthus propose anew algorithm for thenumerical eval- ] uation ofthese expansions, whichincludeseriesplayingarelevantroleinthe A computation of the partition function in statistical mechanics. By the same procedureweobtainalsoasolutionoftheclassicalHausdorffmomentproblem. C . h t a m 1. Introduction [ Following Fuchs, Rogosinski[6, 13] and Hardy [9] we say that f(x) is summable 1 to the value S by the continuous Hausdorff method (or f(x) is H–summable to S) v if 6 1 g(x)= f(xt)dχ(t) (x>0) (1) 1 3 Z0 1 tends to S as x . Integral (1) is a Lebesgue–Stieltjes integral, f(x) is Borel → ∞ 0 summable and bounded in every finite interval [0,x], and χ(t) is of bounded varia- 6 tion in [0,1]. Next, in Hardy [9] the following theorem is proved. 0 / Theorem 1 (Hardy). In order that the transformation (1) should be regular, i.e., h that f(x) S (for x ) should imply g(x) S (for x ), it is necessary t a and suffici→ent that χ(0→+)∞=χ(0)=0 and χ(1)=→1. → ∞ m Letus now supposethat f(x)is a function ofxregularonthe positive realaxis, : v and so expressible in the form i X ∞ ∞ f(n)(0) r f(x)= xn = anxn. (2) a n! n=0 n=0 X X Then, we substitute expansion (2) in integral (1) and, in view of the uniform con- vergenceoftheTaylorseries,wecanexchangethesumwiththeintegralandobtain 1 ∞ ∞ 1 g(x)= a (xt)ndχ(t)= a xn tndχ(t). (3) n n Z0 n=0 n=0 Z0 X X The last integral at the r.h.s. of (3) represents the Hausdorff moment µ , i.e., n 1 1 µ = tndχ(t)= tnu(t)dt. (4) n Z0 Z0 In formula (4), χ(t) is supposed to be a real function of bounded variation in t [0,1], and the numbers µ are called moment constant, of rank n, of χ. If we n ∈ 1 2 E.DEMICHELIANDG.A.VIANO suppose,withoutlossofgenerality,thatχ(0)=0,χ(1)=1,andχ(0+)=χ(0)=0, sothatχ(t)iscontinuousattheorigin,thenµ iscalledaregular moment constant n (see Theorem 1 above). Moreover,the following theorem can be proved [9]. Theorem 2 (Hardy). Sums, differences and products of moment constants are themselves moment constants. The product of two regular moment constants is a regular moment constant. TworelevantexamplesofregularHausdorff transformations arethefollowing[9, Theorem 200]: 1 n+ℓ −1 (i) µ =ℓ tn(1 t)(ℓ−1)dt= (ℓ>0), (5) n − ℓ Z0 (cid:18) (cid:19) which corresponds to the Cesaro transformation C(ℓ); 1 1 1 (ℓ−1) 1 (ii) µ = tn log dt= (ℓ>0), (6) n Γ(ℓ) t (n+1)ℓ Z0 (cid:18) (cid:19) which corresponds to the Ho¨lder transformation H(ℓ). From (3) we are naturally led to consider expansions of the following form: ∞ f(n)(0) µ xn, (7) n n! n=0 X where the terms µ are Hausdorff moments. If we suppose that f(n)(0) = ( 1)n, n − we obtain expansions which read: ∞ ( 1)n g(x)= − µ xn. (8) n n! n=0 X These series may be slowly convergent, for values of x sufficiently large. We thus face a serious problem of numerical analysis, which is quite relevant in view of the fact that sums like expansion (8) occur in several problems, including some of physical interest. For instance, (a) The Laplace transform of the functions of compact support gives rise to sums of the form (8), if we expand in series the exponential e−xt. This case appears in statistical mechanics, where the partition function is the Laplace transform of the density of states. If the latter is a function of compact support, as in the case of harmonic crystals, then we obtain a representation of the partition function in terms of a power series of the type (8) [14]. (b) Confluent hypergeometric function of the following type: ∞ ( 1)n n+ℓ −1 Φ(1,ℓ+1; x)= − xn (ℓ>0), (9) − n! ℓ n=0 (cid:18) (cid:19) X are expansions of the form (8) since the terms µ = n+ℓ −1 form a Haus- n ℓ dorff sequence (see (5)). (cid:0) (cid:1) (c) Hausdorffsequencescanbeconstructedasfollows[19]. Considerasequence ∞ µ of(real)numbers,anddenoteby∆theforwarddifferenceoperator: { n}0 WATSON RESUMMATION OF POWER SERIES 3 ∆µ =µ µ . Then we have n n+1 n − k k ∆kµn =∆ ∆ ∆µn = ( 1)m µn+k−m (k =0,1,2,...), (10) × ×···× − m k−times mX=0 (cid:18) (cid:19) ∆0| is the{izdentity}operator, by definition. Now, suppose that there exists a positive constant M such that n p n (n+1)(p−1) ( 1)n−i∆n−iµ <M (n=0,1,2,...;p>1). (11) i i − i=0(cid:12)(cid:18) (cid:19) (cid:12) X(cid:12) (cid:12) (cid:12) (cid:12) It can be p(cid:12)roved [19] that cond(cid:12)ition (11) is necessary and sufficient to represent the sequence µ ∞ as follows: µ = 1tnu(t)dt (see formula (4)), where u Lp[0,1].{Tnh}u0s we can say thant the0set µ ∞, constrained ∈ R { n}0 by the condition (11), forms a Hausdorff sequence. The main purpose of the present paper consists in performing a Watson–type ∞ resummationofexpansionsoftype(8),wherethesetofnumbers µ isassumed { n}0 tobeaHausdorffsequencegeneratedbyafunctionu(t)(seeformulae(4)and(11)) which belongs to L(2+ǫ)[0,1] (ǫ > 0). In this case we can regard the sequence µ ∞ as the restriction to the integers of a function µ(z) (z C), which belongs t{ont}h0e Hardy space H2(C−1/2), and which is the unique Carl∈sonian interpolation ∞ [3] of the numbers µ . We can thus perform the Watson–type resummation of { n}0 expansion(8), and finally obtain another representationwhose numerical handling is much more convenient and effective. The paper is organized as follows. In Section 2 we study the Carlsonian inter- ∞ polationofthe Hausdorffmoments µ , andexpandthe function µ(iy 1/2)in { n}0 − termsofthe so–calledPollaczekfunctions. InSection3we performaWatson–type resummation of expansion (8). In Section 4 we study an appropriate truncation procedure of the new representation obtained in Section 3. In Section 5 we solve the Hausdorff moment problem by the use of the Pollaczek polynomials [2, 16], and show the connection between this problem and the Watson resummation of expansion (8). Finally, Section 6 is devoted to numerical analysis and examples. 2. Interpolation of Hausdorff moments and Hardy spaces We prove the following theorem. ∞ Theorem 3. Let the sequence µ satisfy condition (11) with p>2+ǫ (ǫ>0). { n}0 Then there exists a unique Carlsonian interpolation of the numbers µ , denoted by n µ(z) (z C,µ(n)=µ ), that satisfies the following conditions: n ∈ (i) µ(z) is holomorphic in the half–plane Rez > 1/2, continuous at Rez = − 1/2; − (ii) µ(z) belongs to L2( ,+ ) for any fixed value of Rez x> 1/2; −∞ ∞ ≡ − (iii) µ(z)tendsuniformlytozeroasz tendstoinfinityinsideanyfixedhalf–plane Rez >δ > 1/2; − Proof. If the sequence µ ∞ satisfies condition (11) with p>2+ǫ (ǫ>0), then { n}0 1 µ = tnu(t)dt, (12) n Z0 4 E.DEMICHELIANDG.A.VIANO with u L2+ǫ[0,1]. Next, set t=e−s in formula (12), and obtain ∈ ∞ µ = e−nse−su(e−s)ds (n=0,1,2,...). (13) n Z0 Therefore the numbers µ can be regarded as the restriction to the integers of the n following Laplace transform: ∞ µ(z)= e−(z+1/2)se−s/2u(e−s)ds. (14) Z0 Indeed,onehasµ(n)=µ . ByapplyingthePaley–Wienertheorem[10]toequality n (14), and recalling that the function e−s/2u(e−s) belongs to L2[0,+ ), we can concludethatµ(z)belongstotheHardyspaceH2(C−1/2),C−1/2 = z∞C, Rez > { ∈ 1/2 (seeRef. [10]). Wecanthusstatethatµ(z)isholomorphicinthehalf–plane − } Rez > 1/2, and tends uniformly to zero as z tends to infinity inside any fixed − half–plane Rez > δ > 1/2. We can then apply the Carlson theorem [3], and say − that µ(z) is the unique Carlsonian interpolation of the numbers µ . Furthermore, n in view of the fact that e−s/2u(e−s) belongs to L2[0,+ ), then µ( 1/2 + iy) ∞ − belongs to L2( ,+ ), and, consequently, property (ii) holds true for any fixed value of Rez −∞x >∞ 1/2. Finally, let us note that the function e−s/2u(e−s) belongs to L1[≡0,+ );−in fact, ∞ e−s/2u(e−s) ds = 1 u(t)/√t dt < since ∞ 0 | | 0 ∞ u L2+ǫ[0,1] (ǫ > 0). Therefore, in view of the Riemann–Lebesgue theorem app∈lied to representation(14), itRfollows that the functiRon(cid:12)(cid:12)µ( 1/2+(cid:12)(cid:12) iy) (y R) is − ∈ continuous, and thus property (i) is proved. (cid:3) Let us now introduce the following set of functions: 1 1 Ψ (y)= Γ +iy P(1/2)(y), (15) n √π 2 n (cid:18) (cid:19) where Γ() denotes the Euler gamma function, and P(1/2)() denote the Pollaczek n · · polynomialsP(λ)(), with λ=1/2(see the appendix). These polynomials (in what n · follows the superscript λ = 1/2 will be omitted) are orthonormal in ( ,+ ) −∞ ∞ with weight function [2, 16] 2 1 1 w(y)= Γ +iy . (16) π 2 (cid:12) (cid:18) (cid:19)(cid:12) (cid:12) (cid:12) Therefore the orthonormality condition(cid:12) reads: (cid:12) (cid:12) (cid:12) +∞ w(y)P (y)P (y)dy =δ . (17) n m n,m −∞ Z ∞ Itcanbeproved[11]thatthefunctions Ψ (y) formacompletebasisinthespace { n }0 L2( ,+ ). Thereforethefunctionµ( 1/2+iy),whichbelongstoL2( ,+ ) −∞ ∞ − −∞ ∞ (seeTheorem3),canbeexpandedintermsofthisbasis. Wecanstatethefollowing proposition. Proposition 1. If the sequence µ ∞ satisfies condition (11) with p > 2 + ǫ { n}0 (ǫ>0), then ∞ 1 µ +iy = c Ψ (y), (18) n n −2 (cid:18) (cid:19) n=0 X WATSON RESUMMATION OF POWER SERIES 5 A B Imz Imz γ + C - 1 0 1 2 Rez - 1 0 1 2 Rez 2 2 Figure 1: (A) Contour integration used for evaluating the coefficients cn (see Theorem 4). (B) C+ is theintegration path of the integral in (21). which converges in the L2–norm. The coefficients c are given by n 1 +∞ 1 1 c = µ +iy Γ iy P (y)dy. (19) n n √π −∞ −2 2 − Z (cid:18) (cid:19) (cid:18) (cid:19) Proof. The sequence µ ∞ is a Hausdorff sequence satisfying condition (11) with { n}0 p > 2 + ǫ (ǫ > 0), then µ( 1/2+ iy) belongs to L2( ,+ ) (statement (ii) − −∞ ∞ of Theorem 3), and expansion (18) converges in the sense of the L2–norm; the coefficientsc arethenobtainedbytheuseoftheorthonormalitycondition(17). (cid:3) n The coefficients c can be evaluated as follows. n Theorem 4. The following equality holds true: ∞ ( 1)k 1 c =2√π − µ P i k+ , (20) n k n k! − 2 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X where P () are the Pollaczek polynomials. n · Proof. Integral (19) can be evaluated by means of the method of contour integra- tion. Set in formula (19): 1/2+iy =z, and accordingly y = i(z+1/2). Then, − − performing an integration along the contour γ shown in Fig. 1A, and taking into account the asymptotic behavior of the gamma function, we obtain −21+i∞ 1 1 µ(z)Γ( z)P i z+ dz = µ(z)Γ( z)P i z+ dz, n n Z−12−i∞ − (cid:20)− (cid:18) 2(cid:19)(cid:21) ZC+ − (cid:20)− (cid:18) 2(cid:19)(cid:21) (21) where C+ is a path which encircles the real positive semi–axis of the z–plane in counterclockwise sense (see Fig. 1B). Then, using the theorem of residues we get ∞ i 1 ( 1)k 1 µ(z)Γ( z)P i z+ dz =2√π − µ P i k+ . n k n −√π − − 2 k! − 2 ZC+ (cid:20) (cid:18) (cid:19)(cid:21) k=0 (cid:20) (cid:18) (cid:19)(cid:21) X (22) 6 E.DEMICHELIANDG.A.VIANO (cid:3) 3. Watson resummation of a class of H–transformed power series We prove the following theorem. Theorem 5. Expansion (8), where the terms µ ∞ form a Hausdorff sequence { n}0 satisfying condition (11) with p > 2+ǫ (ǫ > 0), can be rewritten in the following form: ∞ ( 1)n √2 ∞ x 1 n g(x)= − µ xn = u in − (x>0), (23) n n n! x+1 x+1 n=0 n=0 (cid:18) (cid:19) X X where ∞ ( 1)k 1 u =√2 − µ P i k+ , (24) n k n k! − 2 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X P () being the Pollaczek polynomials. The convergence of expansion (23) is uni- n · form on any compact subset of the real positive axis. Proof. We start by rewriting expansion (8) in the following form: ∞ ∞ ( 1)n ( 1)n − µ xn = − µ enα (α=lnx). (25) n n n! n! n=0 n=0 X X Next, using once again the theorem of residues, we rewrite the sum (25) as the following integral: ∞ ( 1)n 1 − µ enα = Γ( z)µ(z)eαzdz, (26) n n! 2πi − n=0 ZC+ X where the path C+ encircles the real positive semi–axis of the z–plane (see Fig. 1B). Equality (26) holds true since: (i) µ(z) is the Carlsonian interpolation of the moments µ ; n { } (ii) for z = n (n = 0,1,2,...), one has µ(n) = µ , and the function Γ( z) = n − Γ( n) is singular and has simple poles with residues ( 1)n/n!. − − We cannow close the contour C+ as shownin Fig. 1A. We have,by exploiting the asymptotic behavior of the gamma function Γ( z)µ(z)eαzdz =0. (27) − Iγ From (27), and using the Stirling formula for the gamma function, −1+i∞ 2 Γ( z)µ(z)eαzdz = Γ( z)µ(z)eαzdz ZC+ − Z−21−i∞ − (28) +∞ 1 1 =i Γ iy µ iy eα(iy−1/2)dy. −∞ 2 − − 2 Z (cid:18) (cid:19) (cid:18) (cid:19) In the latter integral we use (15) and (18). Then from (26) and (28) we have ∞ ( 1)n 1 +∞ 1 1 − µ enα = Γ iy µ iy eα(iy−1/2)dy n n! 2π −∞ 2 − − 2 n=0 Z (cid:18) (cid:19) (cid:18) (cid:19) X (29) e−α/2 +∞ ∞ 1 1 = c Γ iy Γ +iy P (y)eiαydy. n n 2π√π −∞ 2 − 2 Z n=0 (cid:18) (cid:19) (cid:18) (cid:19) X WATSON RESUMMATION OF POWER SERIES 7 Using the formula 1 1 π Γ iy Γ +iy = , (30) 2 − 2 cosh(πy) (cid:18) (cid:19) (cid:18) (cid:19) from (29) it follows that +∞ ∞ P (y) ∞ P (y) c n eiαydy =2πF−1 c n , (31) n n −∞ cosh(πy) ( cosh(πy)) Z n=0 n=0 X X where F denotes the Fourierintegraloperator. Interchangingintegrationand sum- mation, we have, from formulae (29)–(31): e−α/2 ∞ +∞ P (y) e−α/2 ∞ d 1 c n eiαydy = c P i n n n 2√π −∞ cosh(πy) 2√π − dα cosh(α/2) n=0 Z n=0 (cid:18) (cid:19)(cid:20) (cid:21) X X e−α/2 ∞ α n = c in tanh . n 2√πcosh(α/2) 2 nX=0 h (cid:16) (cid:17)i (32) Substituting α=lnx in (32) yields ∞ ( 1)n √2 ∞ x 1 n − µ xn = u in − (x>0), (33) n n n! x+1 x+1 n=0 n=0 (cid:18) (cid:19) X X where ∞ ( 1)k 1 u =√2 − µ P i k+ . (34) n k n k! − 2 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X Itremainsto provethatthe seriesatthe r.h.s. offormula(33)convergesuniformly on any compact subset of the positive real axis. Using the Schwarz inequality, ∞ x 1 n ∞ 1/2 ∞ x 1 2n 1/2 u in − 6 u 2 − . (35) n n The sum (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)nX∞n==00|un|2(cid:18)cxan+b1e(cid:19)sh(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)own tnXo=0b|e fi|ni!te by unXs=in0g(cid:12)(cid:12)(cid:12)(cid:12)xth+e1P(cid:12)(cid:12)(cid:12)(cid:12)ars!eval equality on expansionP(18). Thesum ∞ x−1 2ncanbeeasilyreducedtotheseries ∞ yn, n=0 x+1 n=0 y =(x−1)2, which is uniformly(cid:12)conv(cid:12)ergenton any compact set y 6y <1. (cid:3) x+1 P (cid:12) (cid:12) 0 P (cid:12) (cid:12) 4. Truncation of the resummed expansion We hereafter assume that only a finite number of Hausdorff moments µ (see k formula(34))aregiven,and,furthermore,wesupposethattheycanalsobeaffected by noise, being typically round–off numerical errors. Accordingly, they will be denotedbyµ(η),ηdenotingtheorderofmagnitudeofthenumericalnoise. Precisely, k we state: µ µ(η) 6 η (k = 0,1,2,...,k ; η > 0); (k + 1) is the number k− k 0 0 of Hausdor(cid:12)ff moment(cid:12)s which are supposed to be known. Next, we introduce the (cid:12) (cid:12) following fin(cid:12)ite sums:(cid:12) k0 ( 1)k 1 u(η,k0) =√2 − µ(η)P i k+ . (36) n k! k n − 2 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X 8 E.DEMICHELIANDG.A.VIANO With obvious notation we write: u(0,∞) = u . Then, the following two auxiliary n n lemmas can be proved. Lemma 1. The following statements hold true: ∞ 2 1 +∞ 1 2 (i) u(0,∞) = µ +iy dy =C (C =const); (37) nX=0(cid:12)(cid:12) n (cid:12)(cid:12) π Z−∞ (cid:12)(cid:12) (cid:18)−2 (cid:19)(cid:12)(cid:12) (cid:12) (cid:12) (cid:12)(cid:12) ∞ (cid:12)(cid:12) 2 (ii) u(η,k0) =+ ; (38) n ∞ nX=0(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (iii) lim u(η,k0) =u(0,∞) =u (n=0,1,2,...); (39) k0→∞ n n n η→0 (iv) If m (η,k ) is defined as 0 0 m 2 m (η,k )=max m N : u(η,k0) 6C , (40) 0 0 ( ∈ n ) nX=0(cid:12) (cid:12) (cid:12) (cid:12) then (cid:12) (cid:12) lim m (η,k )=+ ; (41) 0 0 k0→∞ ∞ η→0 (v) The sum m 2 M(η,k0) = u(η,k0) (m N), (42) m n ∈ nX=0(cid:12) (cid:12) satisfies the following properties: (cid:12) (cid:12) (cid:12) (cid:12) (a) It increases for increasing values of m; (b) the following relationship holds true: Mm(η,k0) > um(η,k0) 2 m→∼+∞(k1!)2 (2m)2k0 (k0 fixed). (43) 0 (cid:12) (cid:12) Proof. The proof is give(cid:12)n, with(cid:12) minor modifications, in Ref. [4]. (cid:3) (cid:12) (cid:12) Lemma 2. The following equality holds true: m0(η,k0) 2 lim u(η,k0) u(0,∞) =0. (44) k0→∞ n − n η→0 nX=0 (cid:12) (cid:12) (cid:12) (cid:12) Proof. See Ref. [4]. (cid:12) (cid:12) (cid:3) Next, we introduce the following notations: ∞ ( 1)n √2 ∞ x 1 n g(x) g(0,∞) = − µ xn = u(0,∞)in − , (45) ≡ n! n x+1 n x+1 n=0 n=0 (cid:18) (cid:19) X X √2 m0(η,k0) x 1 n g(η,k0) = u(η,k0)in − , (46) x+1 n x+1 n=0 (cid:18) (cid:19) X k0 ( 1)k 1 u(η,k0) = √2 − µ(η)P i k+ . (47) n k! k n − 2 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X WATSON RESUMMATION OF POWER SERIES 9 Then the following is true. Theorem 6. The following equality holds: lim g(η,k0) g(0,∞) =0. (48) k0→∞ − η→0 (cid:12) (cid:12) (cid:12) (cid:12) Proof. We have (cid:12) (cid:12) g(η,k0) g(0,∞) − (cid:12) (cid:12) (cid:12) √2 m0(η,(cid:12)k0) x 1 n ∞ x 1 n (cid:12)= (cid:12) u(η,k0)in − u(0,∞)in − (cid:12)x+1 n x+1 − n x+1 (cid:12) (cid:12) n=0 (cid:18) (cid:19) n=0 (cid:18) (cid:19) (cid:12) (cid:12) X X (cid:12) (cid:12)  (cid:12) (cid:12) √2 ∞ x 1 n (cid:12) (49) 6(cid:12) u(0,∞) − (cid:12) (cid:12)(cid:12)x+1(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)n=m0X(η,k0)+1 n (cid:18)x+1(cid:19) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) + m(cid:12)(cid:12)(cid:12)0(η(cid:12)(cid:12)(cid:12),k0) u(η,k0) u(0,∞) x−1 (cid:12)(cid:12)(cid:12)n . (cid:12) n − n x+1 (cid:12) (cid:12)(cid:12) nX=0 (cid:16) (cid:17)(cid:18) (cid:19) (cid:12)(cid:12) Now, using th(cid:12)(cid:12)e Schwarz inequality, (cid:12)(cid:12) (cid:12) (cid:12) ∞ x 1 n u(0,∞) − (cid:12) n x+1 (cid:12)(cid:12)n=m0X(η,k0)+1 (cid:18) (cid:19) (cid:12)(cid:12) (50) (cid:12) 1/2(cid:12) 1/2 (cid:12)(cid:12)6 ∞ u(0,∞) 2 (cid:12) ∞ x−1 2n .  n   x+1  n=m0X(η,k0)+1(cid:12)(cid:12) (cid:12)(cid:12) n=m0X(η,k0)+1(cid:12)(cid:12) (cid:12)(cid:12) Since  (cid:12) (cid:12)   (cid:12) (cid:12)  ∞ (cid:12) (cid:12) 2 m (η,k ) + , u(0,∞) <+ , 0 0 −k−0−→−∞→ ∞ n ∞ η→0 nX=0(cid:12) (cid:12) (cid:12) (cid:12) and (cid:12) (cid:12) ∞ (x 1)/(x+1)2n <+ for x>0 | − | ∞ n=0 X (see Lemma 1), the r.h.s. of formula(50) tends to zeroas k + , η 0. Now, 0 → ∞ → m0(η,k0) x 1 n u(η,k0) u(0,∞) − (cid:12) n − n x+1 (cid:12) (cid:12)(cid:12) nX=0 (cid:16) (cid:17)(cid:18) (cid:19) (cid:12)(cid:12) (cid:12) (cid:12) (cid:12) m0(η,k0) x (cid:12)1 n (cid:12) 6 u(η,k0) u(0,∞) −(cid:12) (51) n − n x+1 nmX=00(η,k(cid:12)(cid:12)(cid:12)0) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) 2 1/2(cid:12)(cid:12)(cid:12)(cid:12) m0(η,k0) x 1 2n 1/2 6 u(η,k0) u(0,∞) − .  n − n   x+1  nX=0 (cid:12)(cid:12) (cid:12)(cid:12) nX=0 (cid:12)(cid:12) (cid:12)(cid:12) From Lemma 2 it follows(cid:12)that (cid:12)   (cid:12)(cid:12) (cid:12)(cid:12)  m0(η,k0) 2 lim u(η,k0) u(0,∞) =0, (52) k0→∞ n − n η→0 nX=0 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 10 E.DEMICHELIANDG.A.VIANO while limkη0→→∞0 mn=0(0η,k0)|(x−1)/(x+1)|2n is finite for x>0. (cid:3) P 5. Connection with the Hausdorff moment problem The classical Hausdorff moment problem can be formulated as follows [1, 15]: Problem. Given a sequence of real numbers µ ∞, find a function u(t) such { n}0 that 1 µ = tnu(t)dt (n=0,1,2,...). (53) n Z0 This problem is ill-posed in the sense of Hadamard [8]: Suppose, for instance, that we are looking for a solution in the space X = L2[0,1], and assume that a solution in this space exists and is unique, but it does not depend continuously on the data. Further, in practical cases only a finite number of moments µ N are { n}0 known. Wemustthenlookforasolutioninafinite–dimensionalsubspaceX of N+1 X. Therefore,any function which is orthogonalto X cannot be recovered: the N+1 solutionisnotunique. Fromthenumericalpointofview,weareledtotheinversion of matrices which are severely ill–conditioned. We shall return on these questions ∞ later. For now we assume that a countable set of noiseless moments µ are { n}0 given, and prove the following theorem. Theorem 7. Suppose that the real sequence µ ∞ of Hausdorff moments satisfy { n}0 condition (11) with p>2+ǫ (ǫ>0). Then the function u(t) can be represented by the following expansion, which converges in the L2–norm: ∞ ∞ ( 1)k 1 u(t)= u Φ (t), u =√2 − µ P i k+ , (54) n n n k n k! − 2 n=0 k=0 (cid:20) (cid:18) (cid:19)(cid:21) X X where Φ (t)=in√2e−tL (2t), (55) n n and P () and L () are the Pollaczek and the Laguerre polynomials, respectively. n n · · Proof. Informula(14)setz = 1/2+iy. Recallingthatthesupportofthefunction f(s)=e−s/2u(e−s) belongs to−R+, 1 +∞ µ +iy = e−iyse−s/2u(e−s)ds=F e−s/2u(e−s) , (56) −2 −∞ (cid:18) (cid:19) Z n o whereFdenotestheFouriertransformoperator. Letusnowreturntotheexpansion (18) and to formula (15), which gives the expression of the functions Ψ (y). In n particular,in the integral representationof the Euler gamma function Γ(1/2+iy): i.e, Γ(1/2+iy)= +∞e−tt(iy−1/2)dt, we set t=e−s: 0 Γ 1 +Riy = +∞e−e−se−s/2e−isyds=F e−e−se−s/2 . (57) 2 −∞ (cid:18) (cid:19) Z n o Since the function e−e−se−s/2 belongs to the Schwartz space S∞, 1 1 F−1 Γ +iy P (y) F−1 Ψ (y) n n √π 2 ≡ { } (cid:26) (cid:18) (cid:19) (cid:27) (58) =P i d 1 e−e−se−s/2 , n − ds √π (cid:18) (cid:19)(cid:20) (cid:21)