Kyushu J. Math. Vol . 61, 2007, pp. 551–563 JACKSON–MELLIN’S TRANSFORM OF MODULAR FORMS AND q-ZETA FUNCTIONS Shai HARAN, Nobushige KUROKAWA and Masato WAKAYAMA (Received 24 July 2006) Dedicated to Leonhard Euler on the 300th anniversary of his birthday Abstract. This note studies a q-analogue of Mellin’s transform of a modular form via Jackson’s integral. 1. Introduction The Mellin transform sends modular forms to L-functions and it is basic for the study of L-functions. The main purpose of the present note is to develop an elementary study of a q-analogue of this situation. In fact, a q-Mellin transform which we give here via the Jackson integral (see, e.g., [KC]) can indeed be applied to various modular forms for obtaining a natural approximation when q → 1 to the classical (completed) L-functions having functional equation and yet are periodic mod 2πi/log q. Let 0 <q<1. Let f :[0 ∞) −→ C be a good function. Then, we define the Jackson–Mellin (or q-Mellin) transform of f by ∞ s−1 n(s−1) n n n+1 Mq (f )(s) := x f(x)dqx = q f(q )(q − q ) 0 n∈Z ns n = (1 − q) q f(q ). n∈Z It is immediate to see that Mq (f )(s) becomes the usual Mellin transform when q → m (f )(s) := 1. To simplify the description we sometimes use the notation q ns n n∈Z q f(q ),thatis,Mq (f ) = (1 − q)mq(f ). 2000 Mathematics Subject Classification: Primary 11M36. Keywords and Phrases: Mellin’s transform; Jackson’s integral; modular forms; Riemann’s zeta function; Ramanujan’s delta function; q-series. This article is an invited contribution to a special issue of the Kyushu Journal of Mathematics commemorating the sixtieth volume. 552 S. Haran et al The first aim is to show that mq (f )(s) has a functional equation when f is a modular form. For motivation and guidance, we give here a very simple example as a −a follows. Consider the function fa(x) = 1/(x + x ) for a>0. Then it is clear that − fa(x 1) = fa(x).Putφa(s) = mq (fa)(s). Then we can calculate φa(s) as ns n ns n −ns −n φa(s) = q fa(q ) = fa(1) + q fa(q ) + q fa(q ) n∈Z n≥1 n≥1 ns −ns n = fa(1) + (q + q )fa(q ). n≥1 The series obviously converges when |Re s| <aand the functional equation φa(−s) = φa(s) follows immediately from the expression. Moreover, φa(s) can be meromorphically continued to the entire complex plane C. To see this, for instance, we assume that 0 < |q| < 1. Then we have 1 ns −ns 1 φa(s) = + (q + q ) 2 qna + q−na n≥1 = 1 + (qns + q−ns)qna (−q2an)m 2 n≥1 m≥0 1 m n(s+a+ am) n(−s+a+ am) = + (−1) (q 2 + q 2 ) 2 m≥0 n≥1 s+a+2am −s+a+2am 1 m q q = + (−1) + . 2 − qs+a+2am − q−s+a+2am m≥0 1 1 Hence, φa(s) is meromorphic in C and has a simple pole at s =±{a(1 + 2m) + (2πi/log q)} (m, ∈ Z,m ≥ 0) with residue (−1)m−1/log q. The location of the zeros φa(s), however, seems difficult to detect. Throughout the paper, we assume that 0 <q<1. 2. Jackson–Mellin’s transform of modular forms We prove the following basic theorem. THEOREM 2.1. Let F be a holomorphic modular form of weight k (k must be a non- negative even integer) for SL(2, Z); in particular, F(−z−1) = zkF(z) for z ∈ H, H being the upper half plane. Put f(y) = F(iy) for y>0. Define a q-analogue of Jackson–Mellin’s transform of modular forms and q-zeta functions 553 the completed L-function q (s, F ) by q (s, F ) :=mq (f − F(i∞))(s) − F(i) ns n = q (f (q ) − a0) − f(1). (2.1) n∈Z Here a0 is the constant term of the Fourier expansion of F(z): 2πinz F(z)= ane . n≥0 Then q (s, F ) can be analytically continued to a meromorphic function on C and 1 i−k q (s, F ) + a + (2.2) 0 1 − qs 1 − qk−s is entire. In particular, if F isacuspform,thenq (s, F ) is an entire function. Furthermore, q (s, F ) has a functional equation: −k q (s, F ) = i q (k − s, F). (2.3) Proof. Since f(q−n) = (iqn)kf(qn) = ikqnkf(qn),wehave ns n −ns −n q (s, F ) = f(1) − a0 + q (f (q ) − a0) + q (f (q ) − a0) − f(1) n≥1 n≥1 ns −k −nk −n −ns −n =−a0 + q (i q f(q ) − a0) + q (f (q ) − a0) n≥1 n≥1 ns −k −nk −n −k −nk =−a0 + q {i q (f (q ) − a0) + a0(i q − 1)} n≥1 −ns −n + q (f (q ) − a0) n≥1 −ns −k −n(k−s) −n =−a0 + (q + i q )(f (q ) − a0) n≥ 1 i−kqs−k qs + a − 0 1 − qs−k 1 − qs for Re s>k. It follows that −k −ns −k −n(k−s) −n 1 i q (s, F ) = (q + i q )(f (q ) − a ) − a + . 0 0 1 − qs 1 − qk−s n≥1 (2.4) 554 S. Haran et al f(q−n) − a = ∞ a e−2πmq−n <q< Since 0 m=1 m and 0 1, the series converges absolutely and uniformly for all s in any compact subset of C, whence it defines an entire function. Therefore, q (s, F ) is meromorphic in C and the first claim follows. Also, the functional equation follows immediately from the expression. 2 Remark 1. Similarly, we have the following expression of q (s, F ): k ns k n(k−s) n 1 i q (s, F ) = (q + i q )(f (q ) − a ) − a + . 0 0 1 − q−s 1 − q−k+s n≥1 (2.5) We notice that, however, the series in the first term may also involve (actually involves) the poles located on the lines Re s = 0andRes = k because 0 <q<1. Moreover, adding two expressions (2.4) and (2.5), we obtain ns k n(k−s) n k 2q (s, F ) = (q + i q )(f (q ) − a0) − a0(1 + i ). n=0 Now we suppose k ≡ 2 mod 4. Then, for s = k/2 + it,wehave k 1 n(k/2+it) n(k/2−it) n q + it,F = (q − q )(f (q ) − a0) 2 2 n= 0 nk/2 n = i q sin(nt log q)(f (q ) − a0). n=0 Each term in the series vanishes obviously when t log q ∈ πZ. Hence, it follows that s = k/2+i(Nπ/log q)(N ∈ Z) gives a zero of q (s, F ). One can also check this fact directly from (2.4). Note that these zeros, however, may not be expected to exhaust all of the zeros of q (s, F ) and actually disappear when q → 1 except for the point s = k/2. Remark 2. Note that as q → 1 ∞ s−1 −s (1 − q)q (s, F ) −→ F (s) := y (F (iy) − a0)dy = (2π) (s)LF (s), 0 L (s) = ∞ a n−s s>k where F n=1 n for Re . (z) = e2πiz ∞ ( − e2πinz)24 Example 1. Let n=1 1 be the Ramanujan delta function. It has a Fourier expansion ∞ (z) = τ(n)e2πinz, n=1 Jackson–Mellin’s transform of modular forms and q-zeta functions 555 that is, is a cusp form of weight 12 for SL2(Z). Hence, by Theorem 2.1 we have ns n −ns −n(12−s) −n q (s, ) := q (iq ) − (i) = (q + q )(iq ). (2.6) n∈Z n≥1 (i) = (1 )24/ 24π18 q (s, ) ( 4 2 .) It follows that is entire and satisfies the functional equation q (s, ) = q (12 − s, ). Note also that the same procedure used in Remark 1 yields 6n n q (6 + it,) = q cos(nt log q)(iq ), (2.7) n=0 because q6n(iqn) = q−6n(iq−n). The proof of the previous theorem provides the following variant. − k COROLLARY 2.2. Let f be a function satisfying f(x 1) = x f(x) for x>0. Suppose f has the following series expansion: −2πnx f(x)= ane . n≥0 Define Lq (s, f ) by ns n/k Lq (s, f ) := q (f (q ) − a0). (2.8) n∈Z Then Lq (s, f ) can be analytically continued to a meromorphic function on C. Furthermore, we have Lq (s, f ) = Lq (1 − s, f ). (2.9) Proof. Since f(q−n/k) = (qn/k)kf(qn/k) = qnf(qn/k), in a similar way as in the preceding proof, we obtain ns n/k −ns −n/k Lq (s, f ) = f(1) − a0 + q (f (q ) − a0) + q (f (q ) − a0) n≥1 n≥1 ns −n −n/k −ns −n/k = f(1) − a0 + q (q f(q ) − a0) + q (f (q ) − a0) n≥1 n≥1 556 S. Haran et al ns −n −n/k −n = f(1) − a0 + q {q (f (q ) − a0) + a0(q − 1)} n≥1 −ns −n/k + q (f (q ) − a0) n≥1 −ns −n(1−s) −n/k = f(1) − a0 + (q + q )(f (q ) − a0) n≥ 1 qs−1 qs + a0 − 1 − qs−1 1 − qs for Re s>1.