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  • 8/9/2019 Kadell K W J - A Simple Proof of an Aomoto-Type Extension of Askey's Last Conjectured Selberg Q-Integral - J. Math.

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    Journal of Mathematical Analysis and Applications 261, 419440 (2001)

    doi:10.1006/jmaa.2000.7180, available online at http://www.idealibrary.com on

    A Simple Proof of an Aomoto-Type Extension of

    Askeys Last Conjectured Selberg q-Integral

    Kevin W. J. Kadell

    Department of Mathematics, Arizona State University, Tempe, Arizona 85287-1804

    Submitted by Bruce C. Berndt

    Received June 1, 1998

    We establish an Aomoto-type extension of Askeys last conjectured Selbergq-integral, which was recently proved by Evans. We follow the lines of our proofsof Aomoto-type extensions of the Morris constant term q-identity and Gustafsons

    AskeyWilson Selberg q-integral. We require integral forms of the q-transportationtheory and its alternative for the root system An1, which are related to the simplereflections and minuscule weight of An1. We use an elementary symmetry whichis related to the minuscule weight of An1 to lift a proof of the one dimensionalq-integral to the multivariable setting. 2001 Academic Press

    Key Words: Selbergs integral; Aomoto-type extension; q-transportation theory forthe root system An1.

    1. INTRODUCTION AND SUMMARY

    Throughout this paper, we let n 1, k 0, N 0, a 0, b 0, m, andv be integers with 0 m n and 2 v n, we let x and y be complex

    with positive real parts, and we let q be real with 0 < q < 1. In 1944,Selberg [27] evaluated the multivariable beta integral

    1

    0

    1

    0

    ni=1

    tx1i 1 tiy12kn t1 t ndt1 dtn

    =n

    i=1

    x + n iky+ n ik1 + ik

    x + y+ 2n i 1k 1 + k (1.1)

    where nt1 t n =

    1i

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    420 kevin w. j. kadell

    q-beta integrals given by Ramanujan [25, 26]; see also Askey [4] andHardy [13]. Gustafson [10, 11] has proven many extensions of Selbergsintegral including extensions based upon MellinBarnes type integrals and

    the AskeyWilson integral with five parameters.Let z qa =

    a1i=0 1 q

    iz be the q-Pockhammer symbol. FollowingAskey [5], we set

    q2kn t1 t n =

    1i

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    askeys last selberg q-integral 421

    by the asymptotic relation

    qxn2k

    qIkn x n 1k + 0 C1 q

    n (1.9)

    where the constant C is independent ofx and y. Kadells proof generalizedAomotos proof [3] of an extension of Selbergs integral. See Evans [7] fora proof using Andersons argument [1].

    Let w f denote the coefficient of the monomial w in the Laurent expan-sion of f. Habsieger [12] and Kadell [17] independently observed that theSelberg q-integral (1.7) is equivalent to the Morris constant term q-identity

    1n

    i=1

    ti qaq/ti qb qakn1t1 t n

    =n

    i=1

    q qa+b+nikq qik

    q qa+nikq qb+nikq qk(1.10)

    A natural extension of the Jackson q-integral (1.5) is given by

    dc

    ftdqt = c

    0ftdqt+

    d0

    ftdqt

    = c1 q

    i=0

    qifqic + d1 q

    i=0

    qifqid (1.11)

    which is the Riemann sum using the endpoints which are furthest from zero

    in the partition

    c dq = qic i 0 qid i 0 (1.12)

    Andrews and Askey [2] have given the q-beta integral

    qIx y c d =

    d

    c

    qt/c qqt/d qqxt/c qq

    yt/d qdqt

    =cd

    c + d

    d/c qc/d qqxd/c qq

    yc/d q

    qxqy

    qx + y (1.13)

    Observe that the integrand has no poles with t c dq.Omitting m as a subscript when m = 0, we set

    qaskn mx y c d t1 t n

    =n

    i=1

    qti/c qqti/d qqx+ni+1mti/c qq

    yti/d qq

    2kn t1 t n (1.14)

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    422 kevin w. j. kadell

    where A is one or zero according to whether A is true or false, respec-tively, and we use capital letters to denote the q-integral

    qASk

    n mx y c d

    =d

    c

    dc

    qaskn mx y c d t1 t ndqt1 dqtn (1.15)

    Askeys last conjectured Selberg q-integral [5], which has been provenby Evans [8], reduces to (1.13) when n = 1. This is the m = 0 case of thefollowing theorem, which is our main result.

    Theorem 1.

    qASkn mx y c d = q

    n3k2 n2

    k2cd

    n2k

    n

    i=1

    cdd/c qc/d qc + dqx+nik+imd/c q q

    y+nikc/d q

    n

    i=1

    qx + n ik + i mqy+ n ikq1 + ik

    q

    x+

    y+ 2

    n

    i 1

    k+

    i

    m

    q1 +

    k

    (1.16)

    Observe that the integrand has no poles with t1 t n c dnq, that the

    integrand is bounded on c dnq, and that the case cd = 0 of (1.16) is Askeysfirst conjecture (1.7).

    Krattenthaler [22] used the case m = 0, k = 1 of Theorem 1 toobtain exact enumeration formulas for perfect matchings of holey Aztecrectangles.

    Observe by (1.2) and (1.14) that we have the symmetry

    qaskn x y c d t1 t n = qas

    kn y x d c t1 tn (1.17)

    and that dc

    d

    cft1 t ndqt1 dqtn

    =

    d

    c

    d

    cf t1 tndqt1 dqtn (1.18)

    Using (1.15), (1.17), and (1.18), we have the symmetry

    qASkn x y c d = qAS

    kn y x d c (1.19)

    We follow the lines of our proofs [19, 20] of Aomoto-type extensions ofthe Morris constant term q-identity (1.10) and Gustafsons AskeyWilsonSelberg q-integral. We require integral forms of the q-transportation the-ory for the root system An1 and its alternative. These are related to the

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    askeys last selberg q-integral 423

    properties of the simple reflections and the minuscule weight of An1. Weuse the elementary symmetry

    qakn1qtn t1 t n1 = qa

    kn1t1 t n (1.20)

    See Carter [6], Grove and Benson [9], and Kadell [1820] for details on theproperties of root systems. The q-transportation theory for root systems hasa long history in [1620] and, in various forms, in many of the papers onconstant term q-identities associated with root systems.

    In Section 2, we use the substitution t = qs to give a simple relation and,using a boundary condition, we prove the AndrewsAskey q-beta integral

    (1.13) when x is a positive integer.In Section 3, we give an integral form of the q-transportation theory for

    the root system An1 which we express in terms ofqaskn x y c d t1 t n.

    In Section 4, we lift the simple relation of Section 2 to the multivari-able setting, obtaining an integral form of the alternative q-transportationtheory for the root system An1 which we express in terms of qas

    kn x y

    c d t1 t n.

    In Section 5, we establish the dependence of qAS

    k

    n m x y c d on theparameters m and x.In Section 6, we recall the global form [16, Sect. 4] of the q-transportation

    theory for the root system An1 and give an application.In Section 7, we follow Askey [4] and lift the boundary condition of Sec-

    tion 2 to the multivariable setting, obtaining a recurrence relation involvingthe parameters x, c, and n.

    In Section 8, we use induction on n, x, and m to evaluate qASkn m

    x y c d when x is a positive integer and, using Ismails argument [15],we extend to complex x and complete the proof of Theorem 1.

    2. A PROOF OF THE ANDREWSASKEYq-BETA INTEGRAL (1.13)

    In this section, we use the substitution t = qs to give a simple relationand, using a boundary condition, we prove the AndrewsAskey q-beta inte-gral (1.13) when x is a positive integer.

    Observe that the q-differential

    dqft = ft fqt (2.1)

    satisfies the scale invariance property

    dqct

    ct=

    dqt

    t= 1 q (2.2)

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    The q-derivative is given by the ratio of q-differentials

    dq

    dqtft = dqft

    dqt

    =1

    t1 qft fqt (2.3)

    We define

    qN

    ftdqt = 1 qN1i=0

    qifqi (2.4)

    to be the finite sum which results when we cancel the possibly divergent

    infinite series in the definition (1.11). We have the principal value of theq-integral

    PV

    dc

    ftdqt

    = lim

    N

    qNcc

    ftdqtqNd

    dftdqt

    = limN

    c1 qN1

    i=0qifqic

    + d1 qN1i=0

    qifqid (2.5)

    which extends the definition (1.11).The following lemma gives an integral form of the alternative q-

    transportation theory for the root system An1.

    Lemma 2.If gt has no poles with t c dq, then we have

    PV

    dc

    1 + t/c1 t/dqt/c qqt/d q

    qxt/c qqyt/d q

    gtdqt

    t

    = PV

    dc

    1 + qxt/c1 qyt/dqt/c qqt/d q

    qxt/c qqyt/d q

    gqt

    dqt

    t

    (2.6)

    Proof. Observe that

    ft = 1 + t/c1 t/dqt/c qqt/d q

    qxt/c qqyt/d q

    gt (2.7)

    has no poles with t c dq and

    fqt = 1 + qxt/c1 qyt/dqt/c qqt/d q

    qxt/c qqyt/d q

    gqt (2.8)

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    Hence we may write the result (2.6) as

    PV d

    cft

    dqt

    t = PV d

    cfqt

    dqt

    t (2.9)Observe by (2.7) that

    0 = f c = fd (2.10)

    Replacing ft by ft/t in (2.5), we obtain

    PV

    dc

    ftdqt

    t

    = 1 q lim

    N

    N1i=0

    fqic + fqid (2.11)

    Using (2.10) and observing that we may shift both of the sums on the rightside of (2.11), we have

    PV

    dc

    ftdqt

    t

    = PV

    qdqc

    ftdqt

    t

    (2.12)

    Substituting (2.12) into (2.9), we see that the result (2.6) becomes

    PV

    qd

    qcft

    dqt

    t

    = PV

    d

    cfqt

    dqt

    t

    (2.13)

    which follows using the substitution t = qs, the scale invariance (2.2) ofdqt/t, and replacing s by t.

    Observe that1 + t/c1 t/d 1 + qxt/c1 qyt/d

    qxdt

    = 1 + t1/c 1/d t2/cd 1 tqx/c qy/d + qx+yt2/cd

    qxd

    t

    = qxd/c qx qxt/c q2xd/c + qx+y + q2x+yt/c

    = 1 qx + qxd/c q2xd/c 1 + qx+y qxt/c + q2x+yt/c

    = 1 qx1 + qxd/c 1 qx+y1 + qxt/c (2.14)

    Setting gt = qxd in Lemma 2 (2.6), moving both q-integrals to thesame side of the equation, using (2.14), and observing that the principal

    value extends the definition (1.11), we obtain

    0 = PV

    dc

    1 qx1 + qxd/c 1 qx+y1 + qxt/c

    qt/c q qt/d qqxt/c qq

    yt/d q

    dqt

    t

    = d

    c 1 qx1 + qxd/c 1 qx+y1 + qxt/c

    qt/c q qt/d qqxt/c qq

    yt/d qdqt (2.15)

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    which we may rearrange as

    1 qx1 + qxd/cqIx y c d = 1 qx+yqIx + 1 y c d (2.16)

    Using the functional equation

    qx + 1 =1 qx

    1 qqx (2.17)

    for the q-gamma function, we see that the product on the far right sideof (1.13) satisfies (2.16).

    We require the following q-analogue of [16, Lemma 3].

    Lemma 3. If gxt is bounded on c dq and gxc is continuous at

    x = 0 from the right, then we have

    limx0+

    1 qx

    1 q

    dc

    qt/c qqt/d qqxt/c qq

    yt/d qgxtdqt

    =cd

    c + d

    c/d qqyc/d q

    g0c (2.18)

    Proof. Using the definition (1.11) and cancelling the factor 1 q, we

    obtain

    limx0+

    1 qx

    1 q

    dc

    qt/c qqt/d qqxt/c qq

    yt/d qgxtdqt

    = limx0+

    1 qx

    c

    i=0

    qiqi+1 qq

    i+1c/d qqi+x qq

    i+xc/d qgxq

    ic

    + d

    i=0

    qiqi+1d/c qq

    i+1 q

    qi+xd/c qqi+x qgxq

    id= c

    qc/d qqyc/d q

    g0c (2.19)

    which readily simplifies to the result (2.18).

    Taking gxt = 1 in Lemma 3 (2.18), we see that qIx y c d satisfiesthe boundary condition

    limx0+

    1 qx1 q q

    Ix y c d = cdc + d

    c/d qqyc/d q

    (2.20)

    Using (1.4) and (2.17), we have

    limx0+

    1 qx

    1 qqx = lim

    x0+qx + 1 = q1 = 1 (2.21)

    Hence that the product on the far right side of (1.13) also satisfies theboundary condition (2.20).

    Combining our results (2.16) and (2.20), we may establish the AndrewsAskey q-beta integral (1.13) by induction on x.

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    3. THE q-TRANSPORTATION THEORY FORTHE ROOT SYSTEM An1

    In this section, we give an integral form of the q-transportation theory forthe root system An1 which we express in terms ofqas

    kn x y c d t1 t n.

    Using the identities

    z qa = zaq

    a2q1a/z qa (3.1)

    z qa+b = z qaqaz qb (3.2)

    for reversing and splitting the q-Pockhammer symbol, respectively, and thefact that k1 k +

    k2

    =

    k2

    , we have

    s2kq1kt/s q2k = s2kq1kt/s qkqt/s qk

    = s2k q1kt/skqk2s/t qkqt/s qk

    = stkqk2s/t qkqt/s qk (3.3)

    The following lemma gives an integral form of the q-transportation theoryfor the root system An1.

    Lemma 4. If F is a distribution function on the domain and Ys t issymmetric,

    Ys t = Yt s (3.4)

    in s and t, then we have 2

    ts ts QtYs tdFsdFt

    = Q

    2ss ts QtYs tdFsdFt (3.5)

    Proof. Since Qs ts ts Qt is antisymmetric under the substi-tution s t, we have

    0 =

    2Qs ts ts QtYs tdFsdFt (3.6)

    The result (3.5) follows by expanding the factor Qs t in (3.6) and rear-ranging the result.

    The following lemma expresses Lemma 4 (3.5) in terms of qaskn x y c d

    t1 t n.

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    Lemma 5. If t1 t n is symmetric,

    t1 t n = t1 t v2 tv tv1 tv+1 t n (3.7)

    in tv1 and tv, then we have

    PV

    dc

    d

    ctvt1 t n qas

    kn x y c d t1 t ndqt1 dqtn

    = qk PV

    dc

    d

    ctv1t1 t nqas

    kn

    x y c d t1 t ndqt1 dqtn

    (3.8)

    PV

    dc

    d

    c

    1

    tv1t1 t n qas

    kn x y c d t1 t ndqt1 dqtn

    = qk PVd

    c

    d

    c

    1

    tv

    t1 t n qaskn

    x y c d t1 t n dqt1 dqtn

    (3.9)

    Proof. Define q2kn v t1 t n by

    q2kn t1 t n = tv1 tvtv1 q

    ktv q2kn v t1 t n (3.10)

    Using (3.3) and the fact that st1 s/t1 qk

    t/s = s ts qk

    t, wehave

    s2kq1kt/s q2k = s ts qkt stk1

    qk2qs/t qk1qt/s qk1 (3.11)

    Using (3.11) with s = tv1 and t = tv, we see by (1.2) and (3.10) that

    q2kn v t1 t n = tv1tvk1qk2qtv1/tv qk1qtv/tv1 qk1

    v2i=1

    t2ki q1ktv1/ti q2kt

    2ki q

    1ktv/ti q2k

    n

    j=v+1t2kv1q

    1ktj/tv1 q2k t2kv q

    1ktj/tv q2k

    1i

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    Using (1.14), (3.7), (3.10), and (3.12), we obtain

    t1 t n qaskn x y c d t1 t n

    = tv1 tvtv1 qktv Ytv1 tv (3.13)

    where

    Ytv1 tv = t1 t nn

    i=1

    qti/c qqxti/c q

    qti/d q

    qyti/d q

    q2kn v t1 t n (3.14)

    is symmetric in tv1 and tv. The result Lemma 5 (3.8) now follows by apply-ing Lemma 4 (3.5) with Ytv1 tv given by (3.14) and s = tv1, t = tv,Q = qk.

    The result Lemma 5 (3.9) now follows by incorporating 1/tv1tv intot1 t n.

    4. THE ALTERNATIVE Q-TRANSPORTATION THEORYFOR THE ROOT SYSTEM An1

    In this section, we lift the simple relation Lemma 2 (2.6) to themultivariable setting, obtaining an integral form of the alternativeq-transportation theory for the root system An1 which we express interms of qas

    kn x y c d t1 t n.

    Setting s = ti and t = tj for 1 i < j n in (3.3), we see by (1.2) that

    q2kn t1 t n =

    1i

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    Lemma 6. If gt1 t n has no poles on c dnq, then we have

    PVd

    c

    d

    c

    1 + t1/c1 t1/d qaskn x y c d t1 t n

    gt1 t ndqt1

    t1dqt2 dqtn

    = qn1k PV

    dc

    d

    c1 + qxtn/c1 q

    ytn/dqaskn

    x y c d t1 t ngqtn t1 t n1 dqt1 dqtn1dqtn

    tn (4.3)

    Proof. Observe that

    ft1 t n = 1 + t1/c1 t1/dn

    i=1

    qti/c qqxti/c q

    qti/d qqyti/d q

    q2kn t1 t n gt1 t n (4.4)

    has no poles with t1 t n c dn

    q and

    fqtnt1tn1 = 1+qxtn/c1q

    ytn/dn

    i=1

    qti/cqqxti/cq

    qti/dqqyti/dq

    q2kn qtnt1tn1gqtnt1tn1 (4.5)

    Using (1.14), (4.2), (4.4), and (4.5), we see that we may write the result (4.3)as

    PV

    dc

    d

    cft1 t n

    dqt1

    t1dqt2 dqtn

    = PV

    dc

    d

    cfqtn t1 t n1 dqt1 dqtn1

    dqtn

    tn

    (4.6)

    Observe by (4.3) that

    0 = fc t2 t n = fd t2 t n (4.7)

    where t2 t n c dn1q . Using (4.7) and observing that we may shift

    both of the sums on the right side of (2.11), we have

    PV

    d

    c

    d

    cft1 t n

    dqt1

    t1dqt2 dqtn

    = PV

    qdqc

    dc

    d

    cft1 t n

    dqt1

    t1dqt2 dqtn

    (4.8)

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    Substituting (4.8) into (4.6), we see that the result (4.3) becomes

    PVqd

    qc

    d

    c

    d

    c

    ft1 t ndqt1

    t1dqt2 dqtn

    = PV

    dc

    d

    cfqtn t1 t n1 dqt1 dqtn1

    dqtn

    tn

    (4.9)

    which follows using the substitution t1 t n = qsn s1 sn1, thescale invariance (2.2) of dqt1/t1, and replacing si by ti, 1 i n.

    5. THE DEPENDENCE OF qASkn mx y c d ON THE

    PARAMETERS m AND x

    In this section, we establish the dependence of qASkn mx y c d on the

    parameters m and x. Throughout this section, we let 1 m n.We set

    = PVd

    c

    dc

    1 + t1/c1 t1/d

    ni=nm+2

    1 + qxti/c

    qaskn x y c d t1 t n

    dqt1

    t1dqt2 dqtn

    (5.1)

    Taking gt1 t n =n

    i=nm+21 + qxti/c in Lemma 6 (4.3), we obtain

    = qn1k

    PVd

    c d

    c1 qy

    tn/d

    ni=nm+1

    1 + qx

    ti/c

    qaskn x y c d t1 t ndqt1 dqtn1

    dqtn

    tn

    (5.2)

    Applying Lemma 5 (3.9) m 1 times to the term 1/tn in

    1 qytn

    /d1

    tn= 1/t

    n

    qy/d (5.3)

    and the q-integral on the right side of (5.2) with v running from n 1to n m + 1 and t1 t n =

    ni=nm+11 + q

    xti/c and rearrangingt1 t n, we obtain

    = PV

    dc

    d

    cqnmk/tnm+1 q

    y+n1k/d1 + qxtnm+1/c

    n

    i=nm+2

    1 + qxti/c qaskn x y c d t1 t n dqt1 dqtn

    (5.4)

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    Applying Lemma 5 (3.8) and (3.9) n m times to the terms t1/cd and1/t1, respectively, in

    1 + t1/c1 t1/d 1t1= 1/t1 + 1/c 1/d t1/cd (5.5)

    and the q-integral on the right side of (5.1) with v running from twoto n m + 1 and t1 t n =

    ni=nm+21 + q

    xti/c, and factoring theintegrand, we obtain

    = PVd

    c

    d

    c qnmk/tnm+1 + 1/c 1/d qnmktnm+1/cd

    ni=nm+2

    1 + qxti/c qaskn x y c d t1 t n dqt1 dqtn

    (5.6)

    Substituting

    xycdt x + n mk y+ n 1k

    q

    nmk

    c q

    nmk

    d tnm+1 (5.7)into (2.14) and rearranging the left side, we obtain

    qnmk/tnm+1 + 1/c1 qnmktnm+1/d

    1 + qxtnm+1/cqnmk/tnm+1 q

    y+n1k/d

    qx+nmkd

    = 1 q

    x+nmk

    1 + q

    x+nmk

    d/c 1 qx+y+2nm1k1 + qxtnm+1/c (5.8)

    Equating (5.4) and (5.6), multiplying by qx+nmkd, factoring theintegrand of (5.6), moving both q-integrals to the same side of theequation, using (5.8), and observing that the principal value extendsthe definition (1.11), we obtain

    0 = 1 qx+nmk1 + qx+nmkd/c

    d

    c

    d

    c

    n

    i=nm+2

    1 + qxti/c qaskn x y c d t1 t ndqt1 dqtn

    1 qx+y+2nm1k

    d

    c

    d

    c1 + qxtnm+1/c

    n

    i=nm+2

    1 + qxti/c Qaskn x y c d t1 t ndqt1 dqtn (5.9)

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    Using (1.14) and (1.15), we may rearrange (5.9) as

    qASkn mx y c d = 1 + q

    x+nmkd/c

    1 qx+nmk

    1 qx+2nm1kqAS

    kn m1x y c d (5.10)

    which gives the dependence ofqASkn mx y c d on the parameter m.

    Using (5.10), we see by induction on m that

    qASkn mx y c d =

    m

    i=1

    1 + qx+nikd

    1 qx+nik

    1 qx+2ni1k qASkn x y c d (5.11)

    Setting m = n in (5.11) and using the fact that

    qASknnx y c d = qAS

    kn x + 1 y c d (5.12)

    we obtain

    qASkn x + 1 y c d =

    ni=1

    1 + qx+nikd

    1 qx+nik

    1 qx+2ni1kqAS

    kn x y c d (5.13)

    which gives the dependence ofqASkn x y c d on the parameter x.

    6. THE GLOBAL FORM OF THE q-TRANSPORTATION THEORYFOR THE ROOT SYSTEM An1

    In this section, we recall the global form [16, Sect. 4] of the q-transportation theory for the root system An1 and give an application.

    Throughout this section, we follow the notation and recall results of[16, Sect. 4].

    We set

    Qijnt1 t n =

    1i

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    The case m = 0 of [16, Lemma 4 (4.4)] is

    nQi j

    nt1 t n = Sn

    1i

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    askeys last selberg q-integral 435

    which is symmetric,

    qhknx y c d t1 t n

    = qhknx y c d t1 t n Sn (6.11)

    in t1 t n, and we use capital letters to denote the q-integral

    qHkn x y c d

    =d

    c

    dc

    qhknx y c d t1 t ndqt1 dqtn (6.12)

    Setting Qij = Q, 1 i < j n, in (6.3) and using (6.9), we obtain

    n

    1i

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    436 kevin w. j. kadell

    and, using the antisymmetry (6.4),

    q2k1n t1 t n1 c =

    n1

    i=1

    t2k1i q1kc/ti q2k1 q

    2k1n1 t1 t n1

    = c2k1n1n1i=1

    q1kti/c q2k1

    q2k1n1 t1 t n1 (7.4)

    Observe that

    n1i=1

    1 + ti/cqti/c q

    qxti/c q

    x=0

    = 1 (7.5)

    Substituting (7.3) and (7.4) into (7.2), setting x = 0 and tn = c, andusing (7.5) and the fact that q2k1 = q/q

    2k1 q, we obtain

    g0c = c2n1k

    d

    c

    d

    c

    n1

    i=1q1kti/c qqti/d q qkti/c qq

    yti/d q

    n1t1 t n1q2k1n1 t1 t n1dqt1 dqtn1 (7.6)

    Since the integrand on the right side of (7.6) vanishes for ti = qjc where

    1 i n 1 and 0 j k 1, we may write (7.6) as

    g0c = c2kn1

    d

    qkc

    d

    qkc

    n1

    i=1q1kti/c qqti/d qqkti/c qq

    yti/d q

    n1t1 t n1q2k1n1 t1 t n1dqt1 dqtn1

    = c2kn1qHkn12k y q

    kc d (7.7)

    Since gx satisfies the hypotheses of Lemma 3, we see using (7.1) and (7.7)that Lemma 3 (2.18) gives

    limx0

    +

    1 qx

    1 qqH

    kn x y c d

    = ncd

    c + d

    c/d qqyc/d q

    c2kn1qHkn12k y q

    kc d (7.8)

    which is [16, (9.17)]. Using (6.14), we see that (7.8) becomes the recurrencerelation

    lim

    x0+

    1 qx

    1 q

    qASkn x y c d =

    1 qnk

    1 qk

    cd

    c + d

    c/d q

    qy

    c/d q c2n1kqAS

    kn12k y q

    kc d (7.9)

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    8. A PROOF OF THEOREM 1

    In this section, we use induction on n, x, and m to evaluate qASkn mx y

    c d when x is a positive integer and, using Ismails argument [15], weextend to complex x and complete the proof of Theorem 1.

    Using the fact that cdd/c q/c + d = dqd/c q, we may writethe function on the right side of (1.16) as

    qRkn mx y c d = q

    kn c dq

    kn mx y c dq

    kn mx y (8.1)

    where

    qkn c d = q

    n3k2 n2

    k2cd

    n2k (8.2)

    qkn mx y c d

    = dnn

    i=1

    qd/c qc/d qqx+nik+imd/c qq

    y+nikc/d q (8.3)

    qkn mx y

    =n

    i=1

    qx + n ik + i mqy+ n ikq1 + ik

    qx + y+ 2n i 1k + i mq1 + k (8.4)

    Using (2.17), we see that qRkn mx y c d satisfies (5.10) and (5.12).

    Observe that

    qkn1q

    kc d = qn1

    3 k2 n12

    k2qkcd

    n12 k

    = qn3k

    2 n12 k2cd

    n12 k (8.5)

    Comparing (8.2) and (8.5), we have

    qkn c d = q

    n1k2cdn1kqkn1q

    kc d (8.6)

    Observe that (8.3) gives

    qkn 0 y c d = d

    nn

    i=1

    qd/c qc/d qqnikd/c qq

    y+nikc/d q (8.7)

    qkn12k y q

    kc d

    = dn1n1i=1

    q1kd/c qqkc/d qqnikd/c qq

    y+nikc/d q (8.8)

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    Using (3.1) and (3.2), we obtain

    qd/c qc/d q

    q1k

    d/c qqk

    c/d q

    =c/d qk

    q1k

    d/c qk

    = qk2c/dk (8.9)

    Using (8.9) and the fact that dqd/c q/d/c q = cd/c + d tocompare (8.7) and (8.8), we obtain

    qkn 0 y c d = q

    n1k2c/dn1kcd

    c + d

    c/d qqyc/d q

    qkn12k y q

    kc d (8.10)

    Using (2.17) and (2.21), we obtain

    limx0+

    1qx

    1q qkn xy

    =n1i=1

    qnikn

    i=1

    qy+nikq1+ik

    qy+2ni1kq1+k

    = q1+nkq1+k

    n1i=1

    qnikqy+n1ikq1+ikqy+2ni1kq1+k

    =1qnk

    1qk

    n1i=1

    qn+1ikqy+n1ikq1+ik

    qy+2ni1kq1+k (8.11)

    qk

    n12ky =

    n1i=1

    qn+1ikqy+n1ik

    qy+2ni1k

    q1+ik

    q1+k (8.12)

    Comparing (8.11) and (8.12), we have

    limx0+

    1 qx

    1 q qkn x y =

    1 qnk

    1 qk qkn12k y (8.13)

    Multiplying (8.6), (8.10), and (8.13), we see that qRknx y c d satisfies the

    recurrence relation (7.9).Using (5.10), (5.12), and (7.9), and proceeding by induction on n, x, and

    m, we may evaluate qASkn x y c d when x is a positive integer.

    Recall the identity theorem by Hille [14, Sect. 8.1] that two functionswhich are analytic in the domain and agree at infinitely many points of which include an accumulation point in agree throughout .

    Let w and z be complex numbers. Observe that wz q is an entirefunction of z and that if w 1 then 1/wz q is an analytic function of

    z in the unit disc around zero. Observe that

    s2kq1kt/s q2k = s q1kt s qkt (8.14)

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    askeys last selberg q-integral 439

    Setting s = ti and t = tj for 1 i < j n in (8.14), we see by (1.2) that

    q2kn t1 t n is a polynomial. Since finite sums and products of analytic

    functions are analytic, we see that qASkn mx y c d and qR

    kn mx y c d

    are analytic functions of z = qx in the unit disc around zero. Since theyhave the same values at z = qx where x is a positive integer and theyare both analytic at z = 0, we see by the identity theorem that they areequal for z in the unit disc around zero. Since q < 1, this is the halfplane where the real part of x is positive. This completes the evaluation of

    qASkn mx y c d and establishes Theorem 1.

    Our argument was introduced by Ismail [15] who gave a simple proof

    of Ramanujans 11 summation formula. Kaneko [21] used Ismails argu-ment to give a multivariable 11 summation theorem using the Macdonaldpolynomials.

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