INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK. 1. Introduction

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1 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Dedicated with appreciation thanks to Richard Askey on his 65 th birthday Abstract. On pages 5 53 of his lost notebook, S. Ramanujan expressed several integrals of products of Dedekind eta-functions in terms of incomplete elliptic integrals of the first kind. In this paper, we prove these identities using only results found in Ramanujan s notebooks. We then construct several new elliptic integrals of this type using modular identities associated with certain Hauptmoduls.. Introduction On pages 5 53 in his lost notebook [7], Ramanujan recorded several identities involving integrals of theta-functions incomplete elliptic integrals of the first kind. We offer here one typical example, proved in Theorem 7.5 below. Let in Ramanujan s notation f q = q; q. Detailed notation is given in Section. The function f is essentially the Dedekind eta-function; see.4. Let. v := vq := q f 3 qf 3 q 5 f 3 q 3 f 3 q 5. Then. q f tf t 3 f t 5 f t 5 dt = 5 tan / 5 r tan v v 5 +v v. 9 5 sin ϕ The reader will immediately realize that these are rather uncommon integrals. Indeed, we have never seen identities like. in the literature. In a wonderful paper [3], all of these integral identities were proved by S. Raghavan S. S. Rangachari. However, in almost all of their proofs, they used results with which Ramanujan would have been unfamiliar. In particular, they relied heavily on results from the theory of modular forms, evidently not known to Ramanujan. For example, for four identities, including., Raghavan Rangachari appealed to differential equations 99 Mathematics Subject Classification. Primary: 33E5; Secondary F. Key words phrases. Incomplete elliptic integrals, modular equations, Dedekind eta-function, theta functions, Rogers Ramanujan continued fraction.

2 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG satisfied by certain quotients of eta-functions, such as., which can be found in R. Fricke s text [9]. In an effort to discern Ramanujan s methods to better underst the origins of identities like., the present authors have devised proofs independent of the theory of modular forms other ideas with which Ramanujan would have been unfamiliar. In particular, we have relied exclusively on results found in his ordinary notebooks [5] his lost notebook [7]. It should be emphasized that at the time of the publication of Raghavan Rangachari s paper [3] a decade ago, many of these results had not yet been proved. Particularly troublesome for us were the aforementioned four differential equations for quotients of eta-functions. To prove these, we used identities for Eisenstein series found in Chapter of Ramanujan s second notebook several eta-function identities scattered among the unorganized pages of his second notebook [, Chap. 5]. We have also utilized several results in the lost notebook found on pages in close proximity to the elliptic integral identities. The authors owe a huge debt to Raghavan Rangachari s paper [3]. In many cases, we have incorporated large portions of their proofs, while in other instances we have employed different lines of attack. This paper could have been made shorter by referring to their paper for large portions of certain proofs, but considerable readability would have been lost in doing so. In Section 3, we prove two identities for integrals of theta-functions of forms unlike.. The first proof is virtually the same as that given by Raghavan Rangachari, while the latter proof is completely different. In Sections 4 6, we prove several integral identities associated with modular equations of degree 5. Here some transformations of incomplete elliptic integrals due to J. Len Ramanujan play key roles. In Section 7, several identities of order 5 are established. Here two of the aforementioned differential equations are crucial. Differential equations are also central in Sections 8 9, where identities of orders 4 35, respectively, are proved. Since differential equations for quotients of eta-functions are of such paramount importance in proving identities akin to., we have systematically derived several new differential equations for eta-function quotients in Section. We have used two of these new differential equations to derive two new formulas in the spirit of.. In Section, we also point out the connection of such integrals with elliptic curves. We plan to return to these matters in a future paper.

3 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 3. Preliminary Results As usual, set, for each nonnegative integer n, n a; q n = aq k k= a; q = lim a; q n, q <. n Ramanujan s general theta-function fa, b is defined by fa, b = a nn+/ b nn /, ab <. n= Theta-functions satisfy the very important useful Jacobi triple product identity [, p. 35, Entry 9],. fa, b = a; ab b; ab ab; ab. The most important special cases are given by ϕq :=fq, q = q n = q; q q ; q. q ; q q; q,.3.4 ψq :=fq, q 3 = n= n= f q :=f q, q = q nn+/ = q ; q q; q, n= n q n3n / = q; q = : e πiz/4 ηz, q = e πiz, Im z >. The product representations in..4 are instances of the Jacobi triple product identity.. The function ηz, defined in.4, is the Dedekind eta-function. It has the transfomation formula.5 η /z = z/iηz. The functions ϕ, ψ, f in..4 can be expressed in terms of the modulus k the hypergeometric function z := F, ; ; k. For a catalogue of formulas of this type, see [, pp. 4]. We will need two such formulas in the sequel. If α = k F q = exp, ; ; α F,, ; ; α then.6 ψ q = z {α α/q}/8

4 4 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG.7 f q = z /3 {α α/q} /. The Eisenstein series P q, Qq, Rq are defined by nq n.8 P q := 4 q n,.9. Qq := + 4 Rq := 54 n= n= n= n 3 q n q n, n 5 q n q n. This is the notation used by Ramanujan in his lost notebook paper [4], [6, pp. 36 6], but in his ordinary notebooks, P, Q, R are replaced by L, M, N, respectively. The Rogers Ramanujan continued fraction uq is defined by. u := uq := q/5 q, q < With f q defined by.4, two of the most important properties of uq are given by [, p. 67, eqs..5,.6]..3 q q 3 uq uq = f q/5 q /5 f q 5 u 5 q u5 q = f 6 q qf 6 q 5. A common generalization of..3 was recorded by Ramanujan in his lost notebook proved by S. H. Son [8]. Lastly, it can be shown that, with the use of the Rogers Ramanujan identities [, p. 79],.4 uq = q /5 q; q5 q 4 ; q 5 q ; q 5 q 3 ; q Two Simpler Integrals Theorem 3. p. 5. Let P q, Qq, Rq be the Eisenstein series defined by.8.. Then q dt Qt e π t = log Q 3/ q Rq Q 3/. q + Rq

5 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 5 Proof. Following Ramanujan s suggestion, let z = R t/q 3 t. Then 3. dz z dq = dr R dq 3 dq Q dq. Using Ramanujan s differential equations [4, eq. 3], [7, p. 4], [, p. 33] q dr dq = P R Q q dq dq = P Q R, 3 in 3., we find that 3. q dz z dq = R Q 3 RQ. Hence, by 3., q d dq log Q 3/ R Q 3/ =q d z + R dq log + z =q d z dz dz log + z dq = q dz zz dq = Q. It follows that q dt q Qt e π t = d e π dt log Q 3/ R Q 3/ dt + R Q 3/ q Rq Q 3/ e π Re π = log Q 3/ log q + Rq Q 3/ e π + Re π. But it is well-known that Re π = [8, p. 88], so Theorem 3. follows. Theorem 3. p. 53. Let uq denote the Rogers Ramanujan continued fraction, defined by., set v = uq. Recall that ψq is defined by.3. Then ψ 5 q dq ψq 5 q = logu v log + 5 uv. 5 + uv Proof. Let k := kq := uv. Then from page 53 of Ramanujan s lost notebook [7], or from page 36 of his second notebook [3, pp. 3], k + k 3.4 u 5 = k v 5 = k. + k k

6 6 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG See also S. Y. Kang s paper []. It follows that 3.5 logu v 3 = 5 log k 8 k. + k If we set ɛ = 5 + /, we readily find that ɛ 3 = 5 + ɛ 3 = 5. Then, with the use of 3.5, we see that 3.3 is equivalent to the equality 8 ψ 5 q dq ψq 5 q = 5 log k 8 k + + ɛ 3 k 5 log + k ɛ 3. k Now from Entry 9vi in Chapter 9 of Ramanujan s second notebook [, p. 58], 3.7 ψ 5 q ψq 5 = 5q ψqψ 3 q 5 + 5q d dq log fq, q 3 fq, q 4. By the Jacobi triple product identity., 3.8 fq, q 3 fq, q 4 = q ; q 5 q 3 ; q 5 q; q 5 q 4 ; q 5 = q; q5 q 4 ; q 5 q 4 ; q q 6 ; q q ; q 5 q 3 ; q 5 q ; q q 8 ; q /5 uq =q vq, by.4. Using 3.8 in 3.7, we find that 8 ψ 5 q dq 8 5 ψq 5 q =4 qψqψ 3 q 5 dq + 5q dq 8 =4 qψqψ 3 q 5 dq 8 logu/v 3.9 =4 qψqψ 3 q 5 dq log k 4 5 log k + k, d dq log q /5 u/v dq where 3.4 has been employed. Comparing 3.9 with 3.6, we now see that it suffices to prove that 3. 8 qψqψ 3 q 5 dq = log k + k + + ɛ 3 k log 5 ɛ 3. k Upon differentiation of both sides of 3. simplification, we find that 3. is equivalent to 3. qψqψ 3 q 5 = We now prove 3.. By 3.4 again, 3. kqk q k q 4kq k q. v u = + k k.

7 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 7 Taking the logarithmic derivative of both sides of 3., we find that 3.3 k q k q = v q vq u q uq. By the logarithmic differentiation of.4, u q uq = 5q n nq n 5 q n n= v q vq = 5q n nq n 5 q n, n= where n 5 denotes the Legendre symbol. Using these derivatives in 3.3, we see that 3.4 k q k q = n nq n 5 q n. n= However, from Entry 8i in Chapter 9 of Ramanujan s second notebook [, p. 49], n nq n 5 n= so that, by 3.4, q n = qψ3 qψq 5 5q ψqψ 3 q 5, 3.5 k q k q = ψ3 qψq 5 5qψqψ 3 q 5. From page 56 in Ramanujan s lost notebook [7], 3.6 ψ q qψ q 5 = k q +, kq which has been proved by Kang [, Thm. 4.]. Putting 3.6 in 3.5, we deduce that k q k 3.7 k q = q 4 qψqψ 3 q 5. kq It is easily seen that 3.7 is equivalent to 3., so the proof of 3.3 is complete.

8 8 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG 4. Elliptic Integrals of Order 5 I Theorem 4. p. 5. With f q, ψq, uq defined by.4,.3,., respectively, with ɛ = 5 + /, /4 q f tf t 5 π/ dt = t = ɛ 5 5 3/ sin ϕ cos ɛu 5/ tan 5 3/4 qf 3 q 5 /f 3 q = 5 tan 5 /4 qψq 5 /ψq To prove 4., we need the following lemma. Lemma 4.. Let uq be defined by.. Then u q = uq f 5 q 5q f q 5. Proof. By.4 the Jacobi triple product identity., uq = q /5 q; q5 q 4 ; q 5 q ; q 5 q 3 ; q 5 = q /5 f q, q4 f q, q 3. ɛ 5 5 3/ sin ϕ ɛ5 / sin ϕ. By logarithmic differentiation the use of Entry 9v in Chapter 9 of Ramanujan s second notebook [, p. 58], which completes the proof. Proof of 4.. Let u q uq = 5q + d dq log f q, q4 f q, q 3 = 5q + + f 5 q 5q f q 5 = f 5 q 5q f q 5, 4.4 cos ϕ = ɛ 5 u 5 t. If t =, then ϕ = π/; if t = q, then ϕ = cos ɛu 5/. Upon differentiation the use of Lemma 4., 4.5 cos ϕ sin ϕ dt =5ɛ5 u 4 tu t =ɛ 5 u5 t t f 5 t f t 5 = cos ϕ f 5 t tf t 5.

9 4.6 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 9 Hence, by 4.5,.3, 4.4, q 5 3/4 f tf t 5 dt t cos =5 3/4 ɛu 5/ f tf t 5 tf t 5 sin ϕ t f 5 t cos ϕ π/ = 5 3/4 π/ = 5 3/4 π/ = 5 3/4 π/ cos ɛu 5/ cos ɛu 5/ cos ɛu 5/ Since ɛ ±5 = 5 5 ± /, t f 3 t 5 f 3 t sin ϕ cos ϕ sin ϕ /u 5 t u 5 t cos ϕ sin ϕ ɛ 5 cos ϕ ɛ 5 cos 4 ϕ. ɛ 5 cos ϕ ɛ 5 cos 4 ϕ =ɛ 5 sin ϕ ɛ 5 cos 4 ϕ Thus, from 4.6, 5 3/4 q which is 4.. f tf t 5 t dt = =ɛ 5 + sin ϕ ɛ 5 cos 4 ϕ =ɛ 5 cos ϕ + cos ϕ + sin ϕ =ɛ 5 sin ϕ sin ϕ + sin ϕ = sin ϕɛ 5 + ɛ 5 sin ϕ = sin ϕ5 5 ɛ 5 sin ϕ =5 5 sin ϕ ɛ 5 5 3/ sin ϕ. π/ cos ɛu 5/ ɛ 5 5 3/ sin ϕ, To prove 4., we need two transformations for incomplete elliptic integrals found in Chapter 7 of Ramanujan s second notebook [, pp. 5 6, Entries 7ii, vi]. Lemma 4.3. If tan γ = x tan α, then 4.7 α If cot α tanβ/ = x sin α, then 4.8 α γ x sin ϕ = x cos ϕ. β x sin ϕ = x sin ϕ.

10 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Proof of 4.. In 4.7, replace ϕ by π/ ϕ combine the result with 4.8 to deduce that 4.9 β π/ x sin ϕ = π/ γ x sin ϕ, provided that i cot α tanβ/ = x sin α, ii tan γ = x tan α. Examining 4. 4., we see that we want to set x = ɛ 5 5 3/ γ = π cos ɛu 5/. We also see that, to prove 4., we will need to show that i ii imply that 4. β = tan 5 3/4 qf 3 q 5 /f 3 q. Since ɛ ±5 = 5 5 ± /, a short calculation gives ɛ 5 5 3/ = ɛ 5 5 3/. Thus, from ii elementary trigonometry, tan α = ɛ 5 5 cot cos ɛu 5/ 3/ 4. =ɛ 5/ 5 3/4 ɛu 5/ ɛu 5 = 53/4 u 5/ ɛu 5. Thus, by i, 4. tanβ/ = From 4. elementary trigonometry, Using this in 4., we deduce that tanβ/ = ɛ 5 5 3/ sin α 53/4 u 5/ ɛu 5. x sin α = ɛ 5 u 5 + ɛ 5 u 5. ɛ 5 u 5 + ɛ 5 u 5 5 3/4 u 5/ ɛu 5 5 3/4 u 5/ = + ɛ 5 u 5 ɛ 5 u 5 5 3/4 u 5/ = u 5 u 5 3/4 = /u 5 u 5 =5 3/4 qf 3 q 5 /f 3 q,

11 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK by.3. Clearly, the last equality is equivalent to 4., so the proof of 4. is complete. For the proof of 4.3, we need another transformation for incomplete elliptic integrals. Lemma 4.4. If < p < 4.3 tan A B then + p A = p tan B, + p B = 3. p 3 +p +p sin 3 ϕ p +p +p sin ϕ This lemma is Entry 6iv in Chapter 9 in Ramanujan s second notebook is a consequence of a Theorem of Jacobi; see [, pp. 38 4] for a proof. Proof of 4.3. We apply Lemma 4.4 with where ɛ = 5 + /. Then p = ɛ 5, p = p = 3ɛ 5, so + p + p 3 p 3 = ɛ 5 5 3/ p = ɛ. + p + p 5 If we substitute these quantities in Lemma 4.4, if we set 4.5 A = tan 5 3/4 qf 3 q 5 /f 3 q 4.6 B = tan 5 /4 qψq 5 /ψq, we shall be finished with the proof of 4.3 if we can prove 4.3. Using the subtraction formula for the tangent function, 4.5, 4.6, we deduce that 4.7 tan A B = 53/4 qf 3 q 5 /f 3 q 5 /4 qψq 5 /ψq + 5q f 3 q 5 ψq 5. f 3 qψq It will be convenient to use some results from the lost notebook proved by Kang []. Set 4.8 t = q /6 q5 ; q 5 q; q s = ϕ q ϕ q 5,

12 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG where ϕq is defined by.. Then 4.9 f q q /6 f q 5 = s t ψq qψq 5 = s t 3. Employing 4.9 in 4.7, we readily deduce that 5t 4. 5 /4 3 s t 3 s 3 tan A B = s 4 + 5t 6. Next, a simple calculation shows that 4. p = 3 ɛ 5. Hence, by 4.4, 4., 4.6, the double angle formula, 5 /4 p + p tan B =5 /4 ɛ tan B =5 /4 ɛ tan tan 5 /4 qψq 5 /ψq = ɛ qψq 5 /ψq 5 qψ q 5 /ψ q = ɛ t 3 4. s s 5 t. 6 Comparing 4. 4., in view of 4.3, we must prove that ɛ t 3 s 5 t 3 s 5 t = s t 3 s 3 6 s 4 + 5t 6. After considerable simplification, the last equality is seen to be equivalent to 4.3 s 4 + 5t 6 = s + s t 6. Now, from 4.8.3, we find that t = tq = q/6 ψq 5 f q ψqf q. Replacing q by q employing.6.7, we find that ψ q t 6 5 f q 6 β β /4 q = q ψ qf q =, α α where β has degree 5 over α. On the other h, from 4.8, s q = ϕq ϕq 5 =: m, where m is the multiplier of degree 5. Hence, replacing q by q in 4.3, we see that this equality is equivalent to β β /4 β β /4 4.4 m 5 = m m. α α α α

13 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 3 Using formulas for m 5/m given in Entry 3xii of Chapter 9 in Ramanujan s second notebook [, pp. 8 8], namely, m = 5 m = β /4 + α α /4 + β β /4 α α /4 β β β /4 α α α α /4, β β we may easily verify that 4.4 does hold to complete the proof. 5. Elliptic Integrals of Order 5 II Theorem 5. p. 5. As before, let ɛ = 5+/, let uq f q be defined by..4, respectively. Then 5. π/ q 5 3/4 f 5 t dt f t /5 f t 5 t 9/ = tan 5 /4 q / f q 5 /f q /5 5. = 5.3 = 5 tan Proof of 5.. Let 5 3/4 q / f q /5 +q /5 f q 5 f q /5 +5q /5 f q 5 cos ɛu r 5.4 cos ϕ = ɛut. ɛ 5 / sin ϕ ɛ 5 / sin ϕ f q 5 f q /5 ɛ 5 5 3/ sin ϕ. Thus, if t =, then ϕ = π/; if t = q, then ϕ = cos ɛu. Upon differentiation the use of Lemma 4., 5.5 cos ϕ sin ϕ dt = ɛu t = ɛ ut f 5 t 5t f t 5.

14 4 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Therefore, by 5.5,., 5.4, 5.6 Now, q 5 3/4 f 5 t f t /5 f t 5 π/ = 5 /4 = 5 /4 π/ = 5 /4 π/ cos ɛu cos ɛu cos ɛu dt t 9/ t /5 f t 5 f t /5 sin ϕ cos ϕ ɛut sin ϕ /ut ut cos ϕ sin ϕ ɛ cos ϕ ɛ cos 4 ϕ. ɛ cos ϕ ɛ cos 4 ϕ =ɛ sin ϕ ɛ cos 4 ϕ =ɛ + sin ϕ ɛ cos 4 ϕ =ɛ cos ϕ + cos ϕ + sin ϕ =ɛ sin ϕ sin ϕ + sin ϕ = sin ϕɛ ɛ sin ϕ + = sin ϕ 5 ɛ sin ϕ. Using this calculation in 5.6, we find that q 5 3/4 f 5 t f t /5 f t 5 from which 5. is immediate. dt = 5/4 t9/ π/ cos ɛu 5 ɛ sin ϕ, Proof of 5.. The proof is similar to that of 4.. We begin with 4.9, set x = ɛ 5 /, put γ = π cos ɛu. Thus, 5.7 tan γ = cot cos ɛu ɛu = ɛu. As with the proof of 4., we want to show that conditions i ii imply that 5.8 β = tan 5 /4 q f q / 5 /f q /5. From condition ii 5.7, 5.9 tan α = 5 tan γ u ɛ 5 = / ɛu 5. sin α = tan α + tan α = 5 u + ɛ u.

15 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 5 Using in conjunction with condition ii, we arrive at 5 u tanβ/ =tan α x sin α = ɛ u ɛu + ɛ u =5 /4 /u u = 5/4 q / f q 5 f q /5. Hence, 5.8 follows, so the proof of 5. is finished. To prove 5.3, we need another transformation for incomplete elliptic integrals from Chapter 9 in Ramanujan s second notebook [, p. 38, Entry 6iii]. Lemma 5.. If tan α + β = + p tan α, where < p <, then α β + p =. p 3 +p +p sin 3 ϕ p +p +p sin ϕ Proof of 5.3. We apply Lemma 5. with p = /ɛ. Thus, + p = 5 + p = ɛ. Hence, p 3 + p + p = ɛ 5 We abbreviate notation by setting + p 3 p = ɛ5 + p 5 5. A = f q /5, B = f q 5, C = q /5. Put α = tan 5 /4 CB A β = tan 3/4 A + CB CB 5. A + 5CB A Examining in relation to Lemma 5., we see that we will be finished with the proof if we can show that 5. tan α + β = ɛ tan α.

16 6 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG First, by the addition formula for the tangent function, tan α + β CB = tan tan 5 /4 + tan 3/4 A + CB 5 A A + 5CB 5. CB 5 /4 A = 5 CB A A + CB + 53/4 A + 5CB A + CB A + 5CB CB A = 5/4 A + 5CB + 5A + CB ABC A 5B C. CB A On the other h, by the double angle formula for the tangent function, BC 5.3 ɛ tan α = ɛ tan tan 5 /4 = ɛ5/4 ABC A A 5 CB. Comparing , we are required to prove that ɛ A 5 CB = A + 5CB + 5A + CB A 5B C. This can be established by elementary algebra, so the proof is complete. 6. Elliptic Integrals of Order 5 III Theorem 6. p. 5. Recall that Ramanujan s continued fraction uq is defined by.. Then there exists a constant C such that 6. u 5 + u 5 = q f 3 q f 3 q 5 C + q f 8 t dt f 4 t t3/ q f 8 t 5 tdt f 4. t To prove Theorem 6., we need to establish a differential equation for a certain quotient of eta-functions. Lemma 6.. Let 6. λ := λq := q f 6 q 5 f 6 q. Then 6.3 q d dq λq = qf qf q 5 5λ 3 + λ + λ.

17 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 7 Proof. By logarithmic differentiation, dλ λ dq = q 3 nq 5n q 5n + 6 nq n q n. n= n= We now apply Entry 4i in Chapter of Ramanujan s second notebook [, p. 463]. Accordingly, q dλ f λ dq = q + qf 6 qf 6 q 5 + 5q f q 5 f qf q 5 = qf qf q λ, λ or q dλ dq = qf qf q 5 λ + λ + 5λ 3, the proof is complete. Proof of Theorem 6.. From.3, in the notation 6., 6.4 u 5 u5 = λ. Considering 6.4 as a quadratic equation in x := u 5, we find upon solving it that x = u 5 = λ + + 5λ λ + λ +. The other root of this quadratic equation is easily seen to be u 5 = λ + 5λ λ + λ +. Hence, Thus, u 5 + u 5 = λ 5λ + λ Gq := λu 5 + u 5 = Thus, by , 5λ + + λ. dg dq = 5 /λ dλ 5λ + + /λ dq 5 /λ f qf q 5 = 5λ 3 + λ + λ 5λ + + /λ q = 5λ /λ q f qf q 5 = 5 q f 8 q 5 f 4 q f 8 q q 3/ f 4 q 5, upon the use of 6. again.

18 8 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Thus, for any q such that < q <, Gq Gq = q or, by , q dg dt = 5 dt q f 8 t 5 q f 4 t q f 8 t dt tdt f 4 t 5 t 3/, u 5 + u 5 = f 3 q q f 8 t 5 q f 8 t dt qf 3 q 5 Gq + 5 tdt q f 4 t q f 4 t 5 t 3/ = f 3 q q f 8 t 5 q f 8 t 5 qf 3 q 5 Gq + 5 tdt 5 tdt f 4 t f 4 t f 8 t dt + q f 4 t 5 t 3/ f 8 t dt q f 4 t 5 t 3/ = f 3 q q f 8 t 5 f 8 t dt qf 3 q 5 C + 5 tdt + f 4 t f 4 t 5, where q f 8 t 5 f 8 t dt 6.6 C = Gq 5 tdt f 4 t q f 4 t 5 t 3/. Thus, we have completed the proof of Theorem 6. have furthermore shown that the constant C is given by 6.6. Now set q = e π/θ. Ramanujan calculated Ge π/θ for three values of θ. Theorem 6.3 p. 5. We have i Ge π/5 = 4 q 5/ 5 +, ii Ge π = Ge π/5 = 6 5 / Ramanujan erroneously claimed that Ge π = Ge π/5 = 6 5 / Raghavan Rangachari [3] used a different method to prove Theorem 6.3. Proof. To prove i, we need to evaluate 6.7 λe π/5 = e π/ 5 f 6 e π 5 q f 6 e π/5 =: L 5. To evaluate L 5, we will use [5, Theorem.3i]. Ramanujan Weber class invariant. Set 6.8 V = G 5n G n t 3/ Let G n denote the

19 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 9 n 6.9 A = e π n/6 f e π f e πn. Then 6. A 5V 5V A = V 3 V 3. 5 Let n = /5. It is well known that G n = G /n, so, from 6.8, V =. Thus, from 6., A 5 =. 5 A Hence, A = 5, since, from , L 5 = A 6, we conclude that L 5 = 5 3/. Thus, from 6.5, Ge π/5 = 5 5 3/ / = 5 3/ + / 5/ 5 + = 4. We now prove ii. First, by 6.,.4,.5, 5λe π/5 = 5 η6 i η 6 i/5 = η6 i η 6 5i = λe π. Hence, by 6.5, Ge π/5 = 5λe π/5 + + /λe π/5 = /λe π + + 5λe π = Ge π. Thus, it suffices to evaluate Ge π/5, so we need to determine 6. λe π/5 = e π/5 f 6 e π f 6 e π/5 =: L5. Again, we shall employ 6.. Set n = /5. Then, by 6.8 the relation G n = G /n, 6. V = G = 5 =. G /5 G 5 See, e.g. [3, p. 9] for the value of G 5. Set ɛ = 5 + /. Then from 6. 6., ɛa 5 5 ɛa = 4, 5

20 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG from which we easily find that A = ɛ. Hence, from , L5 = A 6 = ɛ 3. Lastly, we then conclude from 6.5 that Ge π/5 = 5ɛ ɛ 3 = which completes the proof of ii. = 6 5 / = 6 5 / , Ramanujan claimed that 6.3 C =5 3/4 π + 4 π/ ɛ 5 5 3/ sin ϕ π/, ɛ 5 5 3/ sin ϕ which is quite different from 6.6. Now in Theorem 4., let q tend to. Then u tends to ɛ, so cos ɛu 5/ tends to cos =. Thus, 4. yields 5 3/4 f tf t 5 π/ dt = t ɛ 5 5 3/ sin ϕ. Thus, one of the integrals in 6.3 can be identified as an integral of etafunctions. But this is the only progress we have made in identifying 6.3 with 6.6. Numerically, do not agree. First, by 6.3, 6.4 C = 5 3/4 π = To calculate C via 6.6, we set q = e π use Theorem 6.3. Accordingly, 6.5 C = 6 5 / π η 8 5ix η 4 dx π ix η 8 ix η 4 5ix dx. We used Mathematica to calculate the integrals in 6.5 found that 6.6 C = π π = Thus, show that Ramanujan s claim 6.3 is erroneous. Nonetheless, we are haunted by the possibility that a corrected version of 6.3 exists, for Ramanujan very rarely made a serious error.

21 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 7. Elliptic Integrals of Order 5 The entries in this the following two sections depend upon remarkable differential equations satisfied by certain quotients of eta-functions. For the first series of results, that quotient is defined by f qf q v := vq := q f q 3 f q 5 We need three ancillary lemmas. The first third are found in Ramanujan s notebooks [5]. Lemma 7.. Let v be defined by 7., let Then R = q f qf q 5 f q 3 f q 5. R R = v v. For a proof of Lemma 7., see Berndt s book [, p., Entry 6]. Lemma 7.. Let R be given above, let P = f q q f q 5 Then 6 Q = q 3 P + 5 P = R R R + 79 R 3 Q + 5 Q = R3 + 6R + 5R R. f q 3 6 f q 5. Proof. From Berndt s book [, p. 3, Entry 63; p. 6, Entry 64], we have, respectively, 5 7. P Q + = K + 4K 9, P Q 7.3 P Q 5 P Q = K 4 K K +, where 7.4 K = v v, where v is given by 7.. The forms of Entries in [] are slightly different from those in , respectively, but their equivalences

22 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG are easily demonstrated by elementary algebra. Multiplying 7. by v +v 7.3 by K in 7.4, we deduce that 7.5 P + 5 P + Q + 5 Q = K + 4K P 5 P + Q + 5 Q = KK 4 K K +. From Lemma 7., we know that 7.7 K = R R. Hence, from 7.5, 7.6, 7.7, we deduce that 7.8 P + 5 P + Q + 5 Q = R3 + 6R + 6R R R + 79 R P 5 P + Q + 5 Q = R3 + 6R + 4R 6 R 486 R 79 R 3. Solving yields Lemma 7.. Lemma 7.3. We have + 6 k= kq k q k 3 kq 5k q 5k k= f q + qf 6 qf 6 q 5 + 5q f q 5 = f qf q 5. For a proof of Lemma 7.3, see Berndt s book [, p. 463, Entry 4]. Lemma 7.4. Let v be defined by 7.. Then dv dq = f qf q3 f q 5 f q 5 v 3v + v 3 + v 4.

23 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 3 Proof. From the definition 7. of v, we find that 7. dv v dq = d log v dq = 3 q + 9 = d log{q3/ f 3 q5 f 3 q 3 } dq n= q + 5 = 3 q q nq 3n q 3n 45 nq 5n q 5n n= n= nq 5n q 5n 3 n= + d log{q / f 3 q f 3 q 5 } dq nq n q n f q 3 + q 3 f 6 q 3 f 6 q 5 + 5q 6 f q 5 f q 3 f q 5 f q + qf 6 qf 6 q 5 + 5q f q 5 f qf q 5, by Lemma 7.3. Simplifying 7. by using the definitions of P, Q, R from Lemmas 7. 7., as well as Lemmas 7., 7. themselves, we find that dv 3 dq = vf qf q3 f q 5 f q 5 f q/ q3 f q 5 f qf q 5 Q Q f qf q / q5 f q 5 f q 3 P P 3 =vf qf q 3 f q 5 f q 5 Q R Q R P P =vf qf q 3 f q 5 f q 5 3 R 3 + 6R + 5R R R R R R R + 79 R 3 =vf qf q 3 f q 5 f q 5 3 R R R R

24 4 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG R =vf qf q 3 f q 5 f q 5 3 R 3 + R R =vf qf q 3 f q 5 f q 5 v v v v + =f qf q 3 f q 5 f q 5 v 3v + v 3 + v 4. This completes the proof. Theorem 7.5 p. 5. Let v be defined by 7., let ɛ = 5 + /. Then 7. q f tf t 3 f t 5 f t 5 dt = 5 Proof of. Let = 9 = 4 π/ tan vɛ 3 +vɛ 3 tan 3 5 tan 3 5 r r tan / 5 r tan v v 5 +v v +vɛ vɛ 5 vɛ +vɛ 5 vɛ vɛ 5 +vɛ+vɛ 5 vt v t 7.4 tanϕ/ = 5 + vt v t. 9 5 sin ϕ 8 sin ϕ. 5 6 sin ϕ Clearly, 7.5 ϕ = tan 5 ϕq = tan vq v q 5 + vq v q. Differentiating both sides of 7.4 with respect to t, we find, after a modest calculation, that 7.6 tanϕ/ sec ϕ/ dt = + v t dv 5 + vt v t dt. From 7.4 elementary trigonometry, with the argument t deleted for brevity, 7.7 tanϕ/ sec ϕ/ = 6 v v v v 5 + v v 3/.

25 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 5 From , it follows that /dt dv/dt = 5 + v v v v v + v v 7.8 = From further elementary trigonometry, so 5 + v v v v 3v + v 3 + v 4. sin ϕ = 4 sin ϕ/ cos ϕ/ = 5 v v + v v 9 v v, sin ϕ = 4 + v 5 v v. From , we deduce that /dt 7. dv/dt = sin ϕ v 3v + v 3 + v. 4 Using , we find that, with v = vq, q f tf t 3 f t 5 f t 5 dt tan / 5 r tan v v 5 = f tf t 3 f t 5 f t 5 5 +v v v 3v + v 3 + v 4 dv. 9 dt 5 sin ϕ Invoking Lemma 7.4, we complete the proof of 7.. Lemma 7.6 First version of Len s transformation. If α, β π/, < x <, sinβ α = x sin α, then α x sin ϕ = + x β. 4x sin ϕ +x Lemma 7.6 can be found as Entry 7xiii in Chapter 7 in Ramanujan s second notebook [, p. 3]. If we replace x by x / + x in Lemma 7.6 interchange the roles of α β, we obtain the following second version of Len s transformation. Lemma 7.7 Second version of Len s transformation. If α, β π/, < x <, tanβ α = x tan α, then α x sin ϕ = + x β. x + x sin ϕ

26 6 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Proof of 7.. We apply Lemma 7.7 with x = 3 5. Then tanβ α = tan α. Suppose that α = tan vq v q 5 + vq v q. If q =, then α = tan / 5. In comparing 7. 7., we must prove that, with the agrument q deleted for brevity, 7. tanβ/ = vɛ 3 + vɛ vɛ 5 + vɛ 3 vɛ + vɛ 5, for if q =, then β = π/. Set t = tanα/ t = tan β α/. Then t t = tanβ α = 4 5 tan α = 8t 5 t. If we consider the extremal equality as a quadratic equation in t, a routine calculation gives 7.3 t = 5 t + 5 t 8t 6t + 4, since t >. Using 7. the definition of t, we find that 7.4 t = 4 + 4v v 5 + v v t 6t + 4 = 9 + v + v v v v, after a lengthy calculation. Employing 7., 7.4, 7.5 in 7.3, we conclude that v v v t = + v v v v ɛ ɛ vɛ 5 v = vɛ + vɛ vɛ 5 + vɛ 5 =ɛ vɛ vɛ 5 + vɛ + vɛ 5.

27 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 7 Hence, by , tanβ/ = tan α/ + β α/ = t + t t t vɛ 5 + vɛ 5 5 vɛ + vɛ + vɛ vɛ 5 ɛ + vɛ + vɛ 5 = ɛ vɛ vɛ vɛ 5 + vɛ + vɛ 5 + ɛ 5 vɛ = vɛ + vɛ vɛ ɛ vɛ 5 vɛ 5 + vɛ 3 = 5 3 vɛ 3 vɛ + vɛ vɛ3. Thus, 7. has been established, the proof of 7. is complete. Proof of 7.3. We apply Lemma 7.6 with x = 3 5, let α be given by 7.. Comparing , we see that it suffices to prove that, with the argument q deleted for brevity, 7.7 t := tan β = ɛ vɛ vɛ 5 + vɛ + vɛ 5, for if q =, then t = ɛ = 3 5. Now the hypothesis sinβ α = 3 5 sin α in Lemma 7.6, by the addition formula for the sine function the double angle formulas for both the sine cosine functions, easily translates to the condition 7.8 tan α = sinβ 5 tan β = + cosβ 4 tan β. Using 7.8, 7.7, 7., we have t tanα/ = tan α = 4 t tan α/ 5 v v = + v v + 4v v. Considering 7.9 as a quadratic equation in t, we solve it to deduce that t = 5 + 4v v v v + v v + 4v v + v v + v v + 6,

28 8 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG since t >. Simplifying, we find that 5 + 4v v t = v v + v v v v v + v v =ɛ vɛ vɛ 5 + vɛ vɛ 5, by identically the same calculation that we used in 7.6. Thus, 7.7 has been proved, the proof of 7.3 is complete. For the remainder of this section, set f q 3 f q v = q f qf q 5. Theorem 7.8 p. 53. If v is defined by 7.3, then 7.3 q f tf t 3 f t 5 f t 5 dt = tan / 5 5 tan 3v. 5+3v 9 5 sin ϕ Proof. Because of the conflict in notation between , for this proof only, we set f qf q u := q f q 3 f q 5. In the notation , Lemma 7. takes the form v v = u u. By using the previous equality, we can easily verify that 3v + 3v = u u + u u. Thus, 7.3 follows immediately from Elliptic Integrals of Order 4 As in the previous section, the primary theorem in the present section depends upon a first order differential equation satisfied by a certain quotient of eta-functions established through a series of lemmas. Let f qf q v := vq := q f q f q 7.

29 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 9 Lemma 8.. If v is defined by R = q then f qf q 7 3 f q f q 4, 8.3 R R = v + v. Lemma 8. is a reformulation of Entry 9ix in Chapter 9 of Ramanujan s second notebook [, p. 35]. To see this, replace q by q in the definitions of Then use Entries i, iii in Chapter 7 of the second notebook [, p. 4] to convert 8.3 into a modular equation, which is readily seen to be the same as Entry 9ix. Lemma 8.. Let 8.4 P = f q 4 q f q 7 Q = f q 4 q f q 4. Then 8.5 P + 49 P = R + 48 R + 64 R 8.6 Q + 49 Q = R + 6R + 8 R. Proof. In the notation 8.4, Ramanujan discovered the eta-function identity [, p. 9, Entry 55] P Q + 49 P Q =v 3/ 8v / 8v / + v 3/ = v + 3 v v + v 8.7 =KK, where 8.8 K = v + v. Letting c = KK solving 8.7 for P Q, we find that c + c 96 P Q =,

30 3 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG where the correct root was found by an examination of P Q in a neighborhood of q =. A brief calculation now gives P Q 49 P Q = c = K 6 K 4 + K 96 = K 4K 4 8K Multiplying 8.7 by K given by by v v = K 4, using 8.3, we deduce that, respectively, P + 49 P + Q + 49 Q =K K = R + 8R R R 8. P 49 P + Q + 49 Q = R + 8R + R R + 8 R =K 4 K 4 8K + 49 = R + 8R + 5 R + 8 R R + 8R = R + 5R + 8R 8 R. Solving 8. 8., we deduce Lemma 8.3. We have + 4 k= kq k q k 8 k= kq 7k q 7k { f 8 q + 3qf 4 qf 4 q q f 8 q 7 } /3 = f qf q 7. Lemma 8.3 is part of Entry 5i in Chapter of Ramanujan s second notebook [, p. 467]. Lemma 8.4. If v is defined by 8., then dv dq = f qf q f q 7 f q 4 4v + 9v 4v 3 + v 4.

31 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 3 Proof. From the definition 8. of v, Lemma 8.3, 8.4, 8., Lemma 8., 8. q dv v dq = q d log{q f 4 q4 f 4 q } dq d log v =q dq { =q q + 8 n= { + q q + 8 nq n q n 56 n= n= + q d log{q f 4 q f 4 q 7 } dq } nq 4n q 4n nq 7n q 7n 4 n= nq n q n f 8 q + 3q f 4 q f 4 q q 4 f 8 q 4 /3 = f q f q 4 f 8 q + 3qf 4 qf 4 q q f 8 q 7 /3 f qf q 7 { =qf qf q f q 7 f q 4 q /3 f q f q 4 f qf q 7 Q + 49 /3 Q + 3 q /3 f qf q7 f q f q 4 P + 49 } /3 P + 3 { =qf qf q f q 7 f q 4 R /3 R + 6R /3 R R /3 R R + 64 } /3 R =qf qf q f q 7 f q 4 R + 4 R /3 6 /3 } R 4 =qf qf q f q 7 f q 4 { R /3 R 8. R } R + 6 R /3 Now, by using Lemma 8., we can easily verify that R R 8 = R + R R R R R = R R 8 4 R R

32 3 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG 8.3 = v + 4 v v v = 4v + 9v 4v 3 + v 4 v. Taking the square roots of both sides of 8.3 substituting in 8., we complete the proof. Theorem 8.5 p. 5. If v is defined by 8. if then 8.4 q c =, 7 f tf t f t 7 f t 4 dt = 8 cos c Proof. Let 8.5 cos ϕ = c + vt vt, cos c +v v sin ϕ so that at t =, q we obtain the upper lower limits, respectively, in the integral on the right side of 8.4. Differentiating 8.5, we find that 8.6 sin ϕ dt = By elementary trigonometry, c vt dv dt. 8.7 sin ϕ = v c + v. v Putting 8.7 in 8.6, we arrive at 8.8 /dt dv/dt = c v v c + v. Next, by 8.7, sin ϕ = 3 v 6 3 { v c + v } 3 v = v v 3 v.

33 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 33 Thus, by , dv/dt 6 3 /dt 3 sin ϕ v = c + v v v c 8 49 v v v v = v + 9v = 4v 3 + v 4, 8 after a calculation via Mathematica. Thus, q f tf t f t 7 f t 4 dt = cos c 8 cos c +v v f tf t f t 7 f t 4 4v + 9v 4v 3 + v 4 dv dt sin ϕ = 8 cos c upon the use of Lemma 8.4 cos c +v v, sin ϕ 9. An Elliptic Integral of Order 35 To avoid square roots, we have modestly reformulated Ramanujan s integral equality Theorem 9.5 below. Throughout this section, set 9. v := vq := q f qf q35 f q 5 f q 7. Ramanujan defined v by the square of the right side of 9.. Ramanujan s theorem depends upon a differential equation for v which we prove through a series of lemmas. Lemma 9.. Let Then R = f qf q 5 q 3/ f q 7 f q 35. R R = v 3 5 v 5v v 3.

34 34 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Lemma 9. can be found on page 33 of Ramanujan s second notebook [5]; a proof is given in [, pp. 36 4]. Lemma 9.. Let Then P = P Q P Q 3 = f q q /6 f q 5 Q = f q7 q 7/6 f q 35. v 4 v4 7 v 3 + v3 + 7 v v + 4 v + v. This eta-function identity is not found in Ramanujan s ordinary notebooks [5], but it is recorded in his lost notebook [7, p. 55]. Berndt s paper [4] contains a proof. Lemma 9.3. We have + 6 k= kq k q k 3 = k= kq 5k q 5k f q + qf 6 qf 6 q 5 + 5q f q 5 f qf q 5. For a proof of this result from Chapter of Ramanujan s second notebook, see Berndt s book [, p. 463, Entry 4i]. Lemma 9.4. If v is defined by 9., then v dv dq = qf qf q5 f q 7 f q 35 + v v 5v 9v 3 5v 5 v 6. Proof. Set 9. K = v v. Then Lemma 9. can be reformulated as 9.3 P Q P Q 3 = K3 7K + 9K + 7 K + 4. Considering 9.3 as a quadratic equation in P Q 3, we solve it. Then after a tedious, but elementary, calculation, we find that 9.4 P Q 3 5 P Q 3 = K K 4 K + K 3 5K + 3K 9. Now multiply both sides of 9.3 by v 3 + v3 = K + K + 4 both sides of 9.4 by v 3 v3 = KK + 3,

35 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 35 use the observation v = P/Q to deduce that, respectively, 9.5 P P 6 + Q6 + 5 Q 6 = U 9.6 P 6 5 P 6 + Q6 + 5 Q 6 = U, where 9.7 U := K + 4K + K 3 7K + 9K U := KK K 4K + 3 K + K 3 5K + 3K 9. Solving , we deduce that 9.9 P P 6 = U U 9. Q Q 6 = U + U. Using the definition of K in 9., we can rewrite Lemma 9. in the form 9. R + 49 R = K3 5K + 3K 5. Considering 9. as a quadratic equation in R, we solve it find that 9. R = V + V 49 R = V V, where 9.3 V := K 3 5K + 3K V := K 3 K + K 3 5K + 3K 9.

36 36 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Now, by Lemma 9.3, we find that 9.5 dv v dq =d log v dq = 7 6q + 7 = 7 6q = d log{q7/6 f q35 f q 7 } dq n= 6q + 5 nq 7n q 7n 35 nq 35n q 35n n= n= nq 5n q 5n n= + d log{q /6 f q f q 5 } dq nq n q n f q 7 + q 7 f 6 q 7 f 6 q q 4 f q 35 f q 7 f q 35 6q f q + qf 6 qf 6 q 5 + 5q f q 5 f qf q 5 =qf qf q 5 f q 7 f q 35 7 Q 6R Q R P P 6 +. where we have used the definitions of P, Q, R in Lemmas Squaring both sides of 9.5 simplifying with the use of 9., 9.9, 9., 9.7, 9.8, 9.3, 9.4, we find that dv 9.6 qf qf q 5 f q 7 f q 35 dq { = v R Q Q R P P Q Q 6 + P } P 6 + { = v V V U + U + + V + V U U 36 U 4 + U } 4 U = v 36 U V U V + V 4 + U 4 = v K 4 4K 3 K 6K 9, + where the last step involves a considerable amount of algebra. Lastly, by 9., we substitute K = v v into 9.6. Upon simplification, factorization, taking the square roots of both sides, we complete the proof.

37 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 37 Theorem 9.5 p. 53. If v is defined by 9., then = q v t f tf t 5 f t 7 f t 35 dt t dt + t t 5t 9t 3 5t 5 t 6. Proof. Let vt be defined by 9.. Then the limits t =, q are transformed into, v = vq, respectively. Thus, = = q vq vq t f tf t 5 f t 7 f t 35 dt t f tf t 5 f t 7 f t 35 dv dv/dt v dv + v v 5v 9v 3 5v 5 v 6, upon the employment of Lemma 9.4. Thus, the proof is complete.. Constructions of new incomplete elliptic integral identities It is clear from the previous sections that Ramanujan s incomplete elliptic integrals arise from differential equations satisfied by quotients of etafunctions. In this section, we first list three differential equations derive their corresponding elliptic integrals of order 6. We will then briefly describe why such differential equations exist construct several new examples, as well as their corresponding incomplete elliptic integrals. Theorem.. Let Then v := q f 4 q 3 f 4 q 6 f 4 qf 4 q. dv dq = q / f qf q f q 3 f q 6 8v 3 + 4v + v. To prove Theorem., we need the following lemmas: Lemma.. Let Then [, p. 4, Entry 5] P := q /6 f q f q 3, Q := q /3 f q f q 6, R := Q/P, S := q /6 f qf q3 f q f q 6.. P Q + 9 P Q = R3 + R 3

38 38 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG [, p. 5, Entry 5]. P Q 9 P Q = S3 8 S 3. Lemma Q Q 3 P P 3 = S =, S S Lemma.3 appears to be new. We will present the proof of this lemma using ideas illustrated in Sections 7 9 Lemma.. Proof. By cubing both sides of., we deduce that P Q P Q P Q + 9 = R P Q R 3. This implies that.5 P Q P Q 3 = S 4 R R 3 7 R 3 + R 3, by.. Similarly, by cubing. simplifying, we deduce that.6 P Q 3 93 P Q 3 = S S S 3 8S 3. Multiplying.5 by R 3 + /R 3.6 by R 3 /R 3, we find that.7 Q Q 6 + P P 6 =.8 Q Q 6 P 6 93 P 6 = Next, from.., we have R R 3 7 R 3 + R 3 R 3 R { 3 S S S 3 8 } S 3. Therefore,.9 R 3 + R 3 = P Q + 9 P Q = S 3 8 S R 3 + R 3 = S S 6 +.

39 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 39 From.9, we also obtain. R 3 R 3 = S3 + 8 S 3. Substituting.9. into.7.8, we find that. Q Q 6 + P P 6 =. Q Q 6 P 6 93 P 6 = Adding.., we have S S S S 6 + S 3 + 8S { 3 S S S 3 8 } S 3. Q Q 6 = S + S S 6, which implies.3. Similarly, by subtracting. from., we have P P 6 = S S S, which implies.4. This completes the proof of Lemma.3 Lemma.4. [, p. 46, Entry 3i] nq n + q n 36 nq 3n f q 3n = q + 7qf q 3 /3 f 3 qf 3 q 3. n= n= Proof of Theorem.. By logarithmically differentiating v, we deduce that dv v dq = nq n + 3q q n 36 nq 3n q 3n + n= n= nq n +4 q n 7 nq 6n q 6n. n= By Lemmas.,.3,.4, we may rewrite the last equality in the form dv v dq = 3q / f qf q f q 3 f q 6 { S P /3 P 3 + Q } /3.3 S Q 3 = q / f qf q f q 3 f q 6 S 3 + 8S 3. n=

40 4 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Now, it is clear that v = /P Q hence, by., we deduce that dv dq = q / f qf q f q 3 f q 6 8v 3 + 4v + v. Now, from., we observe that if v = R 6 then v = + v. v v This implies that v v 9 v From.4, we find that v 9 v = dv dq = v v + v 36 = v v dv dq. v + v 34. The last two equalities imply that dv v dq = + v 34 dv v v v v dq = q / f qf q f q 3 f q 6 + v 34, v by..3. Hence, we deduce the following differential equation. Theorem.5. Let Then v := q f qf q 6 f q f q 3. dv dq = q / f qf q f q 3 f q 6 v 3 34v + v. Recall that v = /P Q, define v = /S 6. Then. takes the form 9 v = 8 v, v v which implies that v + 9 v v dv dq = + 8 dv v v v dq. By the previous two equalities,.,.3, we find that dv v dq = dv v + 9 v + 8 v v dq v = q / f qf q f q 3 f q v +. v

41 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 4 Hence we obtain the following differential equation. Theorem.6. Recall that.5 v := q f 6 q f 6 q 6 f 6 qf 6 q 3. Then dv dq = q / f qf q f q 3 f q 6 v 4v + 6v +. Using Theorem.6, we now derive an identity associated with an incomplete elliptic integral of order 6. Theorem.7. If v is defined by.5, then q t f tf t f t 3 f t 6 dt = π sin q 4v sin ϕ.6 = cot 4 v. 3 4 sin ϕ We remark here that the upper limit of the second integral may be expressed in terms of v or v by using.., namely, = + 8v v + 4 v v v = + 64v v + 6. v v This yields incomplete elliptic integrals associated with v or v. Of course, these integrals are by no means unique representations of the left h side of Theorem.7 in terms of v v. Proof of Theorem.7. By Theorem.6, we find that.7 q v f tf t f t 3 f t 6 q dv dt = t v 4v + 6v +. Set.8 sin ϕ = 4v +. Then a routine calculation gives = v 4v + 3 6v + 4 sin ϕ = 44v +,

42 4 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG which yield.9 v q dv v 4v + 6v + = π q sin 4v sin ϕ Combining.7.9, we deduce the first equality in.6. The second integral follows from the transformation formula [, p. 6] where α π x sin ϕ = cot α cot β = x. β x sin ϕ, The key to the proof of Theorem.7 is the substitution.8. We briefly describe here how we arrive at this substitution. Consider the equation y = x4x + 6x + for an elliptic curve E. By substituting x = 4x +, we may rewrite the equation in the form y = x x x 3 4. Next, by setting y = yx 3 x = /x [, pp. 4 43], we obtain the Legendre form of the elliptic curve E, namely, y = x 3 4 x. Our substitution.8 is obtained by letting sin ϕ = x = x = 4x +. We now describe how one can construct differential equations analogous to that of Theorem.. The quotients of eta-products which appear in Ramanujan s integrals happen to be Hauptmoduls associated with discrete groups of genus zero of the form Γ N + W p, where p N W p is an Atkin-Lehner involution of Γ N see [7] for more details. Suppose v is the Hauptmodul associated with a discrete group of genus zero Γ. Then the derivative of v with respect to q is a modular form of weight under Γ. To construct a differential equation associated with v, we search for another modular form of weight under Γ for which the quotient w dv dq is invariant under Γ. Since every modular function invariant under Γ can be expressed as a rational function of v, we can easily determine the relation between the two modular forms. Using this method, we derive the following differential equations.

43 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 43 Theorem.8. Let Then v = q f 4 q f 4 q f 4 qf 4 q 5. dv dq = q / f qf q 5 v4v + 6v + v +. Theorem.9. Let v = q f 3 q f 3 q 4 f 3 qf 3 q 7. Then dv dq = f qf q f q 7 f q 4 8v + v + 8v + 5v +. Theorem.. Let Then v = q f q f q f qf q. v dv dq = qf q f q 4v 3 + 8v + 4v + 6v 3 + 6v + 8v +. Theorem.. Let Then v dv dq =q4 f q 3 f q 33 v = q f q3 f q 33 f qf q. 3v + v + 7v v v v 3 + 8v + 7v +. We now use two of these differential equations to derive two new identities for incomplete elliptic integrals. Theorem.. If then. q v = q f 4 q f 4 q f 4 qf 4 q 5, sin f tf t 5 dt = 5 /4 t r 4 5v 6+ 5v sin ϕ Proof. We first transform the differential equation in Theorem.8 to a differential equation involving σ := /v, namely, dσ dq = q / f qf q 5 σ + 4σ + σ + 6.

44 44 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG Using the method described after the proof of Theorem.7, we then derive the substitution. sin 4 5 ϕ = σ = 4 5v 6 + 5v +. It follows that Hence, q dσ = 5 /4 σ σ /q f tf t 5 dt = t =5 /4 sin dσ σ + 4σ + σ + 6 r 4 5v 6+ 5v+ Upon simplification with the use of., we obtain.. σ σ + 4 We next describe how we derive an integral identity from Theorem.9. First, we derive a substitution similar to that of Ramanujan. Our aim is to convert the differential expression dx 8x + x + 8x + 5x + to Rdt t k t. Using the method outlined in [6, pp. 3 3], we use the substitution obtain. where x = + s s dx 8x + x + 8x + 5x + ds = ns + ms, n = m = Next, we use the substitution [6, p. 36] t = ns, simplifying, we find that.3 ds ns + ms = dt 7/4 t k t,

45 INCOMPLETE ELLIPTIC INTEGRALS IN RAMANUJAN S LOST NOTEBOOK 45 where k = Hence, we should use the substitution.4 sin ϕ = t = ns x = n, x + or.5 cos ϕ = n x + x. Using.,.3,.4, we deduce that.6 dx 8x + x + 8x + 5x + = 7/4 k sin ϕ. Using Theorem.9,.5,.6, we can deduce the following theorem with no difficulty. Theorem.3. Let Then v = q f 3 q f 3 q 4 f 3 qf 3 q 7 q f tf t f t 7 f t 4 dt c = cos = 7/4 c. cos {c x + } x sin ϕ The integral above is clearly an analogue of 8.4. It is not clear how one can obtain this integral from 8.4 vice versa using the modular equation found in [, p. Entry 9ix] or [7, Entry 3.6]. References [] B. C. Berndt. Ramanujan s Notebooks, Part III. Springer Verlag, New York, 99. [] B. C. Berndt. Ramanujan s Notebooks, Part IV. Springer Verlag, New York, 994. [3] B. C. Berndt. Ramanujan s Notebooks, Part V. Springer Verlag, New York, 998. [4] B. C. Berndt. Modular equations in Ramanujan s lost notebook. In R. P. Bambah, V. C. Dumir, R. Hans-Gill, editors, Number Theory. Hindustan Book Co., Delhi, 999. [5] B. C. Berndt, H. H. Chan, L.-C. Zhang. Explicit evaluations of the Rogers Ramanujan continued fraction. J. Reine Angew. Math., 48:4 59, 996. [6] A. Cayley. An Elementary Treatise on Elliptic Functions. Dover, New York, 96. [7] H. H. Chan M. L. Lang. Ramanujan s modular equations Atkin Lehner involutions. Israel Jour. Math., 3: 6, 998. [8] K. Chrasekharan. Elliptic Functions. Springer Verlag, Berlin, 985. [9] R. Fricke. Die Elliptic Funcktionen und Ihre Anwendungen, Bd.. B. G. Teubner, Leipzig, 9. [] S.-Y. Kang. Ramanujan s formulas for the explicit evaluation of the Rogers Ramanujan continued fraction theta-functions. Acta Arith., 9:49 68, 999.

46 46 BRUCE C. BERNDT, HENG HUAT CHAN, AND SEN SHAN HUANG [] S.-Y. Kang. Some theorems on the Rogers Ramanujan continued fraction associated theta function identities in Ramanujan s lost notebook. The Ramanujan J., 3:9, 999. [] V. Prasolov Y. Solovyev. Elliptic Functions Elliptic Integrals. American Mathematical Society, Providence, RI, 997. [3] S. Raghavan S. S. Rangachari. On Ramanujan s elliptic integrals modular identities, pages Oxford University Press, Bombay, 989. [4] S. Ramanujan. On certain arithmetical functions. Trans. Cambridge Philos. Soc., :59 84, 96. [5] S. Ramanujan. Notebooks volumes. Tata Institute of Fundamental Research, Bombay, 957. [6] S. Ramanujan. Collected Papers. Chelsea, New York, 96. [7] S. Ramanujan. The Lost Notebook Other Unpublished Papers. Narosa, New Delhi, 988. [8] S. H. Son. Some theta function identities related to the Rogers Ramanujan continued fraction. Proc. Amer. Math. Soc., 6:895 9, 998. Department of Mathematics, University of Illinois, 49 West Green Street, Urbana, IL 68, USA address: berndt@math.uiuc.edu Department of Mathematics, National University of Singapore, Kent Ridge, Singapore 96, Republic of Singapore address: chanhh@math.nus.edu.sg Department of Mathematics, National Chang Hua University of Education, Chang Hua City, Taiwan, Republic of China address: shuang@math.math.ncue.edu.tw