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Associated Stieltjes-Carlitz polynomials and

arXiv:math/9307204v1 [math.CA] 8 Jul 1993Associated Stieltjes-Carlitz polynomials anda generalization of Heun’s differentialequation.Galliano VALENT∗COMMUNICATION AT THE 4th INTERNATIONAL CONFERENCEON ORTHOGONAL POLYNOMIALS. EVIAN.

OCTOBER 1992.AbstractThe generating function of Stieltjes-Carlitz polynomials is a solution of Heun’s differ-ential equation and using this relation Carlitz was the first to get exact closed formsfor some Heun functions. Similarly the associated Stieltjes-Carlitz polynomials leadto a new differential equation which we call associated Heun.

Thanks to the linkwith orthogonal polynomials we are able to deduce two integral relations connectingassociated Heun functions with different parameters and to exhibit the set of asso-ciated Heun functions which generalize Carlitz’s. Part of these results were used bythe author to derive the Stieltjes transform of the measure of orthogonality for theassociated Stieltjes-Carlitz polynomials using asymptotic analysis; here we presenta new derivation of this result.1IntroductionThe Stieltjes-Carlitz polynomials are othogonal polynomials with a three term recurrencerelation(λn + µn −x)Fn(x) = µn+1Fn+1 + λn−1Fn−1,n ≥0,F−1(x) = 0,F0(x) = 1,(1)with the ratesλn = k2(2n + 1)2,µn = (2n)2λn = (2n + 1)2,µn = k2(2n)2(2)and 0 < k2 < 1.

Their orthogonality measure, given for instance in [3, p. 194], wasfirst derived by Stieltjes [13] using continued fraction techniques.Later on Carlitz [2]obtained a generating function for the Fn(x) from which he was able to derive anew theorthogonality measure. The most striking fact, which does not seem to have been realizedby Carlitz himself, is that the generating functions he had obtained give a finite set of∗Laboratoire de Physique Th´eorique et Hautes Energies, Unit´e associ´ee au CNRS UA 280, Universit´ePARIS 7, 2 Place Jussieu, F-75251 CEDEX 05.1

2G. Valentexact solutions of Heun’s differential equation.

The complete list of Carlitz results weregathered in [14], and an integral transformation connecting Heun functions with differentparameters was derived. This last tool, combined with Carlitz results, led to an enlargedset of Heun functions.More recently we have obtained a generating function for the associated Stieltjes-Carlitzpolynomials with ratesλn = k2(2n + 2c + 1)2,µn = 4(n + c)2 + µδn0(3)λn = (2n + 2c + 1)2,µn = 4k2(n + c)2 + k2µδn0(4)This result, combined with asymptotic analysis, has led to the Stieltjes transform of theassociated Stieltjes-Carlitz polynomials [17]Our aim is to show that these results lead to a finite set of exact solutions of whatcould be called the “associated Heun” differential equation which emerges as an equationsatisfied by the generating functions of the polynomials Fn(x) with the rates (3), (4).This link with orthogonal polynomials is even more fruitful since it gives a convenienttool to derive two new integral transformations relating associated Heun functions withdifferent parameters.The plan of this communication is the following.Section 2 is devoted to a short summary on Heun’s differential equation and its relationwith orthogonal polynomials.In section 3 we present the associated Heun differential equation and derive two integralconnection relations for its solutions.In section 4 we give a finite set of exact solutions of the associated Heun differentialequation which generalize Carlitz results to non-vanishing (c, µ).In section 5, in order to cross-check the Stieltjes transform obtained in [17], we presenta completely different derivation which uses Karlin and MacGregor representation theoremand the link between orthogonal polynomials and birth and death processes.

Solving theKolmogorov equations leads to a Stieltjes transform which is in perfect agreement withasymptotic analysis.2Heun differential equationThis equation is the most general second order differential equation with four regularsingular pointsw = 0, 1, 1/k2, +∞0 ≤k2 ≤1and is described by the array of parametersP = {α, β; γ, δ, ǫ; s}α + β = γ + δ + ǫ −1.The accessory parameter s is unconstrained. We shall follow the standard notations of[19, p.576],[4, p.57-62] for Heun’s differential equationw(1 −w)(1 −k2w)D2F + [γ(1 −w)(1 −k2w) −δw (1 −k2w) −ǫk2w(1 −w)]DF+ (αβk2w + s)F = 0(5)

Generalized Heun differential equation3with D = ddw.We shall denote by Hn(P, w) the unique solution of (5) which is analytic for |w| < 1and is normalized according toHn(P, w = 0) = 1. (6)This equation, for arbitrary values of s, can be solved in terms of hypergeometricfunctions only for two values of the parameter k2:1) if k2 = 0 the solution analytic around w = 0 is2F1 r+, r−γ; w!r± = a ±√a2 + sa = γ + δ −122) if k2 = 1 the solution analytic around w = 0 was shown in [12] to be(1 −w)r2F1 r + α, r + βγ; w!r = a +qa2 −αβ −s,a = γ −α −β2For some particular values of the accessory parameter s Heun’s functions degenerateinto hypergeometric functions of the variable R(w), where R(w) is a polynomial of seconddegree in w. These values of s are listed in [10].In all what follows we shall not consider these particular cases.Since Hn(P, w) is analytic for |w| < 1 we can consider it as the generating function ofthe polynomials Fn(P, s), with variable s, such thatHn(P, w) =Xn≥0Fn(P, s)wn|w| < 1(7)Relations (5,7) imply routinely the three term recurrence relation for the Fn(λn + µn + γn −s −αβk2)Fn = µn+1Fn+1 + λn−1Fn−1,n ≥0F−1 = 0,F0 = 1(8)withλn = k2(n + α)(n + β),µn = n(n + γ −1),γn = (1 −k2)δn.This recurrence exhibits the polynomial character of Fn with respect to either the variables or the more familiar x = s + αβk2.One should observe on (8) that only for δ = 0 do we have a true birth and deathprocess with birth rate λn and death rate µn (see [5] for an introduction).

For δ ̸= 0 wehave killing in the sense of Karlin and Tavar´e [8],[9] with rate γn.The finite set of exact solutions of (5) obtained by Carlitz can be found in [14, p.692]as well as as the following integral transform, quoted here for convenience.Let us define P ′ = {α′ = γ, β′ = β; γ′ = α, δ′ = δ + γ −α, ǫ′ = ǫ + γ −α; s}, thenprovided that Re γ > Re α > 0, we haveHn(P, w) =1B(α, γ −α)Z 10 dt tα−1(1 −t)γ−α−1Hn(P ′; wt)(9)for w in C\[1, +∞[. This relation, combined with the set of Carlitz solutions gives anotherfinite set of solutions described in [14, p.693].

4G. Valent3Integral connection relations for associated HeunfunctionsLet us turn ourselves to the associated polynomials with the recurrence(λn + µn + γn −s −(α + c)(β + c)k2 + k2δc) Fn = µn+1Fn+1 + λn−1Fn−1, n ≥0F−1 = 0,F0 = 1(10)withλn = k2(n + c + α)(n + c + β)µn = (n + c)(n + c + γ −1) + µδn0γn = (1 −k2)δ(n + c)Two new parameters appear: c which is an association parameter, and µ which is a co-recursivity parameter.

Clearly, for vanishing (c, µ) the recurrence (10) reduces to (8). Theconstant terms added to s were chosen for notational convenience reasons.We defineHn(c, µ, P; w) =Xn≥0Fn(c, µ, P)wn|w| < 1(11)and from (10) it is easy to obtainw(1 −w) (1 −k2w)D2F + [(γ + 2c)(1 −w)(1 −k2w) −δw(1 −k2w) −ǫk2w(1 −w)]DF+"(α + c)(β + c)k2w + c(c + γ −1)w(1 −w) + s −δc#F = c(c + γ −1)w+ µ (12)In view of its origin it is natural to call this equation the associated Heun differentialequation.

This departs from the usual terminology where associated differential equationsrefer to homogeneous extensions of a given differential equation whereas here we have aninhomogeneous extension.At any rate it reduces to Heun equation in the particular cases (c = 0, µ = 0) and(c = 1 −γ, µ = 0), according to the relationsHn(0, 0, P; w) = Hn(P; w)Hn(1 −γ, 0, P; w) = Hn( ˜P; w)(13)with the array ˜P = {1 −γ + α, 1 −γ + β; 2 −γ, δ, ǫ; s −(1 −γ)δ}.Let us now derive two integral transformations relating associated Heun functions. Todo this we switch from the Fn to the Gn defined byG0 = F0,Gn = µ1 · · · µnFn = (1 + c)n(γ + c)nFnn ≥1(14)whose recurrence(λn + µn + γn −s −(α + c)(β + c)k2 + k2δc) Gn = Gn+1 + λn−1µnGn−1,n ≥0G−1 = 0,G0 = 1(15)

Generalized Heun differential equation5reveals that (−1)nGn is monic in the variable x = s + k2(α + c)(β + c) −k2δc.The basic technique to get an integral transform is to look for a mapping of the pa-rameters P which leaves invariant the recurrence (15).A first possibility is the mapping P ′αα′ = γ, β′ = βγ′ = α, δ′ = δ + γ −α, ǫ′ = ǫ + γ −αs′ = s, c′ = c, µ′ = µfor which we haveGn(P ′α) = Gn(P)n ≥0.Using (14) givesFn(P) = (c + α)n(c + γ)nFn(P ′α). (16)If Re γ > Re α > −Re c we can write(c + α)n(c + γ)n=1B(γ −α, α + c)Z 10 dt tn+c+α−1(1 −t)γ−α−1and inserting this in (16), multiplying each term by wn and summing n from zero to infinitygivesHn(c, µ, P, w) =1B(γ −α, α + c)Z 10 dt tc+α−1(1 −t)γ−α−1Hn(c, µ, P ′α; wt)(17)valid for |w| < 1.

The term by term integration is allowed since the right hand side powerseries is absolutely and uniformly convergent for |w| ≤R < 1. Analytic continuationextends this relation to C\[1, +∞[.

Clearly for c = µ = 0 we recover (9) and there isanother integral transformation P ′β obtained from P ′α by the exchange of the couples (α, α′)and (β, β′).A second possibility which leaves invariant the recurrence (15) is P ′′α withα′′ = 2 −α, β′′ = β + 1 −αγ′′ = γ + 1 −α, δ′′ = δ + 1 −α, ǫ′′ = ǫ + 1 −αs′′ = s + (α −1)(γ + δ −α), c′′ = c + α −1, µ′′ = µwhich leads toFn(P) = (c + α)n(c + 1)nFn(P ′′α).The corresponding integral transform follows analogously to (17)Hn(c, µ, P, w) =1B(1 −α, c + α)Z 10 dt tc+α−1(1 −t)−αHn(c + α −1, µ, P ′′α; wt)and is valid for 1 > Re α > −Re c and w ∈C\[1, +∞[.This second integral relation is a genuinely new result, since it changes the value ofthe association parameter from c to c + α −1. For this reason it could not appear in theprevious analyses where c = 0.

Here too the interchange of the couples (α, α′′) and (β, β′′)leads to another mapping P ′′β .

6G. Valent4Exact solutions of the associated Heun equationBefore giving the set of associated Heun functions which generalize Carlitz ones we shallexplain, on the first of them, how they can be constructed.We first make the change of functionG = wcF(18)which brings (12) tow(1 −w)(1 −k2w)D2G + [γ(1 −w)(1 −k2w) −δw(1 −k2w) −ǫk2w(1 −w)]DG+ [s + k2c(c + ǫ + γ −1) + αβk2w] G = c(c + γ −1)wc−1 + µwc.The first exact solution will correspond to the parametersP =α = 0, β = 12; γ = 12, δ = 12, ǫ = 12; s = σ −k2c2c > 0.The change of variable√w = sn (θ; k2)(19)(in what follows, concerning elliptic functions we stick to the notations of [19]; here √w isthe square root which is positive for real positive w) reduces the differential equation to∂2θG + 4σG = 2c(2c −1)(sn 2θ)c−1 + 4µ(sn 2θ)c = J(θ)Its solution isG(θ) =Z θ0 du sin 2√σ(θ −u)2√σJ(u)(20)and is meaningful provided that c > 1/2.

In order to extend this integral representationto c > 0 an integration by parts of (sn 2u)c−1 is needed with the final resultwc Hn(c, µ, P; σ −k2c2; w) =Z θ(w)0du cos 2√σ(θ(w) −u)2c(sn 2u)c−1/2cn udn u+Z θ(w)0du sin 2√σ(θ(w) −u)2√σh4(c2 + c2k2 + µ)(sn 2u)c −2c(2c + 1)k2(sn 2u)c+1i(21)The variable θ(w) is obtained through the inversion of relation (19)θ(w) =Z √w0dtq(1 −t2)(1 −k2t2).It is analytic for |w| < 1; its analytic extension to the whole complex w plane has beendescribed in [1, p.122] and is analytic but for the branch points w = 1, 1/k2.We take for (sn 2u)c−1/2 and wc one and the same principal branch in order to securethe analyticity of Hn(w) for |w| < 1.It is lengthy, even if straightforward, to check that (21) is indeed a solution of (12)in the complex plane deprived with the points w = 1, 1/k2 and that the normalizationcondition (6) does hold.

Generalized Heun differential equation7As mentioned in the previous section, this associated Heun function should reduce toa Heun function either if (c = µ = 0) or if (c = 1/2, µ = 0). In the first case a limitingprocedure which makes use oflimc→0Z θ0 du f(u) 2c(sn 2u)c−1/2 = f(0)giveslimc→0 Hn(c, µ, P; σ; w) = cos 2√σθ(w) + µσ1 −cos 2√σθ(w).In the second limiting case, using relation (20) we getlimc→1/2√wHn c, µ, P; σ −k24 ; w!= sin 2√σθ(w)2√σ+ 4µZ θ(w)0du sin 2√σ(θ(w) −u)2√σsn uFor µ = 0, using the relations (13) we recover Carlitz results:Hn0, 12; 12, 12, 12; s = σ; w= cos(2√σθ(w))√wHn 12, 1; 32, 12, 12; s = σ −1 + k24; w!= sin 2√σθ(w)2√σIn order to get the remaining set of exact solutions one has to change (18) intoG = wc+λ(1 −w)µ(1 −k2w)νFwhere λ, µ, ν take the values 0 or 1/2.

We get in this way seven more solutions to be listedbelow.• P =12, 1; 12, 32, 12; s = σ −14 −k2c2wc√1 −wHn(c, µ, P; w) =Z θ(w)0du cos 2√σ(θ(w) −u) 2c(sn 2u)c−1/2dn u+4(k2c2 + µ)Z θ(w)0du sin 2√σ(θ(w) −u)2√σ(sn 2u)ccn u(22)which reduces for (c = µ = 0) and (c = 1/2, µ = 0) to√1 −wHn12, 1; 12, 32, 12; s = σ −14; w= cos(2√σθ(w))qw(1 −w)Hn 1, 32; 32, 32, 12; s = σ −1 −k24 ; w!= sin 2√σθ(w)2√σ• P =12, 1; 12, 12, 32; s = σ −k2(c + 12)2wc√1 −k2wHn(c, µ, P; w) =Z θ(w)0du cos 2√σ(θ(w) −u) 2c(sn 2u)ccn u+4(c2 + µ)Z θ(w)0du sin 2√σ(θ(w) −u)2√σ(sn 2u)cdn u(23)

8G. Valentwhich reduces for (c = µ = 0) and (c = 1/2, µ = 0) to√1 −k2wHn 12, 1; 12, 12, 32; s = σ −k24 ; w!= cos(2√σθ(w))qw(1 −k2w)Hn1, 32; 32, 12, 32; s = σ −14 −k2; w= sin 2√σθ(w)2√σ• P =1, 32; 12, 32, 32; s = σ −14 −k2(c + 12)2q(1 −w)(1 −k2w)Hn(c, µ, P; w) =Z θ(w)0du cos 2√σ(θ(w) −u) 2c(sn 2u)c−1/2+4µZ θ(w)0du sin 2√σ(θ(w) −u)2√σ(sn 2u)ccn udn uwhich reduces for (c = µ = 0) and (c = 1/2, µ = 0) toq(1 −w)(1 −k2w)Hn 1, 32; 12, 32, 32; s = σ −1 + k24; w!= cos(2√σθ(w))qw(1 −w)(1 −k2w)Hn32, 2; 32, 32, 32; s = σ −1 −k2; w= sin 2√σθ(w)2√σThe eight particular cases for which either (c = µ = 0) or (c = 1/2, µ = 0) reproducethe results collected in the table [14, p.692].

There are four other solutions:• P =12, 1; 32, 12, 12; s = σ −14 −k2(c + 12)2Let us defineF(u) = 2c(2c + 1)(sn 2u)c−1/2 + 4µ(sn 2u)c+1/2then we have√wHn(c, µ; w) =Z θ(w)0du sin 2√σ(θ(w) −u)2√σF(u)• P =1, 32; 32, 32, 12; s = σ −1 −k2(c + 12)2qw(1 −w)Hn(c, µ; w) =Z θ(w)0du sin 2√σ(θ(w) −u)2√σF(u)cn u• P =1, 32; 32, 12, 32; s = σ −14 −k2(c + 1)2qw(1 −k2w)Hn(c, µ; w) =Z θ(w)0du sin 2√σ(θ(w) −u)2√σF(u)dn u• P =32, 2; 32, 32, 32; s = σ −1 −k2(c + 1)2qw(1 −w)(1 −k2w)Hn(c, µ; w) =Z θ(w)0du sin 2√σ(θ(w) −u)2√σF(u)cn udn u

Generalized Heun differential equation9For these last cases the limiting cases (c = 0, µ = 0) and (c = 1/2, µ = 0) do notgive anything new. Furthermore the reader can check that all the solutions given here arecorrectly normalized at w = 0.The results (22,23) were first derived in [17] whilst all the other are new.

Obviouslythis set of solutions can be further enlarged using the two integral transforms of section 3.5Associated Stieltjes-Carlitz polynomialsIn [17] the Stieltjes transform of the orthogonality measure of the associated Stieltjes-Carlitz polynomials has been derived for the the first time. The main tool used in thiswork is Markov theorem and asymptotic analysis.

It is of some interest to check this resultusing a completely different approach and this is the aim of this section.The strategy used here was already applied to the Stieltjes-Carlitz polynomials in[16] and provided for a new derivation of the orthogonality measures. Its generalization,described in [15] and in [16], led to a one parameter family of orthogonality measures forthe indeterminate moment problem corresponding to the ratesλn = (4n + 1)(4n + 2)2(4n + 3),µn = (4n −1)(4n)2(4n + 1).What is basic in this approach is the connection between orthogonal polynomials andbirth and death processes; the whole problem to get the orthogonality measure is reducedto the resolution of a linear partial differential equation.Let us first recall that the first family of associated Stieltjes-Carlitz polynomials aredefined by the recurrence relation(λn + µn −x)Fn = µn+1Fn+1 + λn−1Fn−1,n ≥0F−1 = 0,F0 = 1(24)with the ratesλn = k2(2n + 2c + 1)2,µn = 4(n + c)2 + µδn0c ≥0In order to obtain the Stieltjes transform of their orthogonality measure we shall relatethem to the birth and death process whose Kolmogorov equation isddtPm,n(t) = λn−1Pm,n−1(t) + µn+1Pm,n+1(t) −(λn + µn)Pm,n(t)Pm,n(0) = δmn.

(25)Pm,n(t) is the probability of a population n at time t provided that it was m at time t = 0.It is therefore positive and bounded by one. The link between (24) and (25) is providedby Karlin and McGregor representation theorem [6],[7]Pm,n(t) = 1πmZ ∞0dΨ(x)Fm(x)Fn(x)e−txwithπ0 = 1,πm = λ0 .

. .

λm−1µ1 . .

. µm,m = 1, 2, .

. .

10G. ValentFrom this representation theorem it follows that a possible way to get Ψ is to compute theLaplace transform ofP00(t) =Z ∞0dΨ(x) e−txwhich we shall write˜P00(p) =Z ∞0dt e−ptP00(t).For Re p > 0 this is nothing but˜P00(p) =Z ∞0dΨ(x)p + xclosely related to the Stieltjes transform of the orthogonality measure since we have˜P00(p) = −S(−p).In this approach no recourse to asymptotic analysis is needed to get S(z): we just require˜P00(p).

We shall describe in the following how this can be worked out just by solvinglinear partial differential equations.As a first step we consider the change of basis Pm,n(t) →Pmn(t) such thatPmn(t) =(1 + c)n(1/2 + c)nPm,n(t)(26)Kolmogorov equation becomesddtPm,n(t) = ˜λn−1Pm,n−1(t) + ˜µn+1Pm,n+1(t) −(λn + µn)Pm,n(t)Pm,n(0) =(1 + c)m(1/2 + c)mδmn. (27)with˜λn = k2(2n + 2c + 1)(2n + 2c + 2),˜µn = (2n + 2c −1)(2n + 2c)whilst λn, µn are given by (3).In order to solve (27) we introduce the generating functionHm(t, w) =√1 −k2wXn≥0Pmn(t) wn+c|w| < 1.The Kolmogorov equation becomes a linear partial differential equation for Hm∂tHm(t, w) = {4w(1 −w)(1 −k2w)∂2w + 2[(1 −w)(1 −k2w) −w(1 −k2w)−k2w(1 −w)]∂w} Hm(t, w) −µ + 2c(2c−1)wwc√1 −k2wPm0(t)(28)with the boundary conditionHm(0, w) =(1 + c)m(1/2 + c)mwm+c√1 −k2w.

Generalized Heun differential equation11It is convenient, from a notational point of view, to keep Pm0(t); however this is relatedto the generating function Hm(t, w) byPm0(t) = limw→0 w−cHm(t, w)From now on we shall restrict ourselves to m = 0, and in order to simplify (28) we changethe variable to w = sn 2(θ, k2) which maps [0, 1] into [0, K]. Deleting the subscript m = 0in H0 we are led to∂tH(t, θ) = ∂2θH(t, θ) − µ + 2c(2c −1)sn 2θ!

(sn 2θ)cdn θP00(t)(29)withH(0, θ) = dn θ(sn 2θ)c.Let us stress that since the moment problem for the associated Stieltjes-Carlitz polynomialsis determined the measure Ψ is unique and therefore the solution of the Kolmogorovequations is unique [11].In order to get it we extend H(t, θ), a priori defined for θ ∈[0, K], to the intervalθ ∈[−K, +K] by using the symmetry θ ↔−θ of the equation and of the boundary value,and further to all real values of θ . This last step is possible since (29) and the boundaryconditions are periodic with period 2K.We shall use a Laplace transform in the variable tH(t, θ) −→˜H(p, θ) =Z +∞0dt e−ptH(t, θ)to solve equation (29).

We shall first examine what can be said on general grounds on˜H(p, θ).Firstly since the Pmn are probabilities we have the bounds0 ≤Pmn(t) ≤(1 + c)n(1/2 + c)nn = 0, 1 · · ·t ≥0which imply0 ≤H(t, θ) ≤dn θ(sn 2θ)c2F1 1, 1 + c1/2 + c ; sn 2θ!for any real θ. From theorem 2.1 of [18, p.38] it follows that ˜H(p, θ) is analytic in thedomain Re p > 0 uniformly for real θ.Secondly the small time behaviour of the transition probabilities is given bylimt→0 Pmn(t) = δmnfrom this and theorem 1 of [18, p.181] we conclude tolimp→+∞˜H(p, θ) = limt→0 H(t, θ) = dn θ(sn 2θ)c.

12G. ValentThirdly ˜H(p, θ) must be periodic in θ, with period 2K, and for P00(t) to exist it isnecessary thatlimθ→0 H(t, θ) = 0t ≥0.

(30)Having stated the most useful properties of ˜H(p, θ) let us now take the Laplace trans-form of equation (29). We get∂2θ ˜H(p, θ) −p ˜H(p, θ) = −H(0, θ) + A(θ) ˜P00(p) = J(p, θ)withA(θ) = µ + 2c(2c −1)sn 2θ!dn θ(sn 2θ)c.This equation has for general solution even in θ˜H(p, θ) = e√pθ2√pZ θ0 dφ e−√pφJ(p, φ) + (θ ↔−θ) + C(p) cosh(√pθ)The integral over φ is convergent at 0 provided that we take c > 1/2.The necessary condition (30) implies C(p) = 0 and the periodicity of ˜H(p, θ) in thevariable θ impliesZ 2K0dφ e−√pφJ(p, φ) = 0c > 1/2from which we deduce˜P00(p) =R 2K0dφ e−√pφ H(0, φ)R 2K0dφ e−√pφ A(φ) .Using the periodicity of A and ˜H brings this ratio to˜P00(p) =R K0 dφ cosh √p(K −φ) H(0, φ)R K0 dφ cosh √p(K −φ) A(φ)and the change of variable p = −z gives eventually the Stieltjes transformZ +∞0dΨz −s = −R K0 du cos √z(K −u) H(0, u)R K0 du cos √z(K −u) A(u)so that if we defineD(c, µ; z) =Z K0du cos √z(K −u)2c(2c −1) + µ sn 2udn u (sn 2u)c−1Γ(2c + 1)we end up withZ +∞0dΨ(s)z −s = −D(c + 1, 0; z)D(c, µ; z)c > 1/2(31)in perfect agreement with the result derived in [17].

In this reference the polynomials withratesλn = (2n + 2c + 1)2,µn = 4k2(n + c)2 + µk2δn00 < k2 < 1have been shown to follow from the result (31) using the transformation theory for Jacobianelliptic functions.

Generalized Heun differential equation13References[1] N. I. Akhiezer, Elements of the theory of elliptic functions. Translations of the Amer.Math.

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