0 [Ref. 21, items 2.12.42.25, 2.13.22.11, 2.13.25.1]:1 Z∞ Z∞ Z∞ 2 2 3 2 [ J0 (ax)] x d x = p , , [ J0 (ax)] Y0 (ax) x d x = 0, J0 (ax)[Y0 (ax)]2 x d x = p 2 0 0 0 3π a 3 3πa2 as well as the familiar asymptotic behavior of Bessel functions for large arguments x ≫ 1:
J0 ( x) ∼ which allows us to verify Z∞
s
³ π´ 2 , cos x − πx 4
Y0 ( x) ∼
s
³ π´ 2 , sin x − πx 4
½
µµ ¶ ¶ ¶ ¶ ¾ µµ ¶ ¶ µµ 1 1 1 J0 ν + x [1 + cos(νπ)] + Y0 ν + x sin(νπ) J0 ν + x × p 2 2 2 1/ 2ν+1 ¶ ¶ ¶ ¶ ¶ ¸ µ µµ · µµ 1 1 1 x cos(νπ) + Y0 ν + x sin(νπ) x d x = O × J0 ν + 2 2 (2ν + 1)9/4
via integration by parts.2 Meanwhile, using the asymptotic forms of Bessel functions and integration by parts, one can also verify that3 ¶ ¶ ¶ ¶ ¾ µµ ¶ ¶ µµ µµ 1 1 1 θ [1 + cos( νπ )] + Y θ sin( νπ ) J θ × ν + ν + ν + 0 0 p 2 2 2 1/ 2ν+1 s · µµ ¶ ¶ ¶ ¶ ¸ µµ 1 1 θ3 × J0 ν + dθ θ cos(νπ) + Y0 ν + θ sin(νπ) 2 2 sin θ µ · ¶ µµ ¸3/2 Zπ−(1/p2ν+1) ¶ ¶ µµ ¶ ¶ 1 4 1 π 1 π dθ 2 =O ∼ . cos ν + θ− cos ν + (π − θ ) − p p π(2ν + 1) 2 4 2 4 (2ν + 1)9/4 1/ 2ν+1 sin θ ½
Zπ/2
J0
1 The first integral here belongs to the generalized Weber-Schafheitlin type [Ref. 15, §13.46]. The proof of the second integral can be found in [22]. The third integral can be derived from the first one, in conjunction with Nicholson’s formula [Ref. 15, §13.73] [ J0 ( x)]2 + [Y0 ( x)]2 = 8 R∞ K (2 x sinh t) d t and a special case of the modified Weber-Schafheitlin integral [Ref. 15, §13.45] R∞ J ( x)K (2 x sinh t) x d x = (1 + 4sinh2 t)−1 . 0 0 0 2 0 0 π
2 To wit, we need to estimate
Z∞
Z∞ Z∞ sin(a(2ν + 1) x + b) 1 ∂ 1 ∂ sin b cos b dx= p sin(a y) d y − p cos(a y) d y p p p p p x a 2ν + 1 2ν+1 y ∂ y a 2ν + 1 2ν+1 y ∂ y p µ ¶ Z∞ Z∞ 1 sin b cos b cos(a 2ν + 1 + b) sin(a y) cos(a y) + p d y− p dy=O = , p p 3/4 3/2 3/2 3/4 a(2ν + 1) (2ν + 1) 2a 2ν + 1 2ν+1 y 2a 2ν + 1 2ν+1 y p 1/ 2ν+1
where the coefficients a, b ∈ R arise from the product-to-sum formulae of trigonometric functions applied to asymptotic expansions for the products of Bessel functions. 3 One might wish to compare this with the direct application of a hypergeometric asymptotic formula for Legendre functions (see [Ref. 8, p. 295] or [Ref. 18, p. 162]): s · µµ ¶ ¶ µ ¶¸ 2 1 1 π Γ(ν + 1) cos ν + θ− +O , where θ ∈ [ε, π − ε], ε > 0. Pν (cos θ) = 3 2 4 ν Γ(ν + ) π sin θ 2
11
Thus, the asymptotic expansion
Tν,ν =
¶¸ · µ 1 8[1 + cos(νπ)][1 + 2 cos(νπ)] 1+O p p 4 3 3π(1 + 2ν)2 2ν + 1
is established for ν = (1 + 4 N )/4 > 0. For an arbitrary point z satisfying Re z > −1/2, we may modify the aforementioned procedure by deforming the integral path joining the points 0 and π/2 into the sum of two line segments: Zπ/2 0
(· · · ) d θ =
Zp|2 z+1|/(2 z+1) 0
(· · · ) d θ +
Zπ/2
(· · · ) d θ , p |2 z +1|/(2 z +1)
R∞ R∞ followed by the integrations 0 (· · · ) d x and p|2 z+1|/(2 z+1) (· · · ) d x along contours that run to infinity in such a manner that (2 z + 1) x → +∞ along the positive real axis. This results in · µ ¶¸ 8[1 + cos(π z)][1 + 2 cos(π z)] 1 1 T z,z = (22a) 1+O p , z ∈ C N , Re z > − , p 4 2 2 3 3π(1 + 2 z) N
where the error bound is uniform along the contour C N , on which both | tan(π z)| and | cot(π z)| are uniformly bounded. After reflection z 7→ − z − 1, we obtain ¶¸ · µ 8[1 − cos(π z)][1 − 2 cos(π z)] 1 T z,z = , 1+O p p 4 3 3π(1 + 2 z)2 N
1 z ∈ C N , Re z < − . 2
(22b)
We may combine Eqs. 22a and 22b into a single formula ¡ ¢ ¡ ¢ · µ ¶¸ πΓ z+2 1 Γ 3 z2+2 1 1 + 2 cos(π z) , T z,z = £ ¡ 1− z ¢¤2 £ ¡ z+2 ¢¤3 ¡ 3 z+3 ¢ 1 + O p 4 3 N Γ Γ Γ 2
2
2
z ∈ CN ,
which also implies the bound estimate T( z) = O( N −9/4 ) uniformly applicable to all the points z on the rectangular contour C N , thanks to the inequalities ¶ µ p p 1 1 sup | cot(π z)| ≤ max(1, coth π) = coth π, sup = 2, ≤ max 2, sinh π z ∈C N | sin(π z )| z ∈C N for N ∈ Z>0 . By Cauchy’s integral formula, we have the identity I T( z ) d z T(ν) = lim = 0, N →∞ C N z − ν 2π i
∀ν ∈ C,
which proves Eq. 19(ν,ν) . Both Eqs. 19(2ν−1,ν) and 19(2ν+2,ν) follow directly from the recursion relation (Eq. 18).
Remark As in our proof of Eq. 19(ν,ν) , we can use Bessel functions to establish the following asymptotic behavior for | z| → ∞, Re z > −1/2: s 8 cos(ρπ z + π z) + cos(π z) 16 sin(ρπ z) sin(π z) ρ2 , Tρ z,z ∼ 0 < ρ < 2; + 2 p π − arcsin 1 − p π ρ 4 − ρ 2 (1 + 2 z )2 4 π ρ 4 − ρ 2 (1 + 2 z )2 s 8 sin(ρπ z − π z) − sin(π z) 16 sin(ρπ z) sin(π z) ρ2 −1 Tρ z,z ∼ − 1, ρ > 2. − 2 p sinh p π ρ ρ 2 − 4(1 + 2 z )2 4 π ρ ρ 2 − 4(1 + 2 z )2
12
along with an asymptotic reflection formula Tρ (− z−1),− z−1 ∼ e iπ(1−ρ ) arg z/| arg z| Tρ z,z , 0 < | arg z| < π for ρ > 0, | z| → ∞. Put differently, for large | z| and −π < arg z < π, we have an “asymptotic trigonometric modulation” of the standard ClebschGordan integral formulae: ´ ³ ´ ³ (ρ +2) z +2 1+(2−ρ ) z Γ πΓ 2 2 ´i2 h ³ ´i2 ³ ´ ³
Tρ z,z ∼ h ³ ´× 1 −ρ z 2 +ρ z (ρ +2) z +3 2+(2−ρ ) z Γ 2 Γ 2 Γ Γ 2 2 s 2 sin( ρπ z ) sin( π z ) cos( ρπ z + π z ) + cos( π z ) 2 ρ π − arcsin 1 − , × + 1 + cos(ρπ z) π 1 + cos(ρπ z ) 4 ³ ´ ³ ´ 1+(2−ρ ) z (ρ +2) z +2 πΓ Γ 2 2 Tρ z,z ∼ h ³ ´i h ³ ´i ³ ´ ³ ´× 1 −ρ z 2 2+(2−ρ ) z 2 +ρ z 2 (ρ +2) z +3 Γ 2 Γ Γ 2 Γ 2 2 s (ρ − 2)π z sin( ρπ z − π z ) − sin(π z ) 2 sin(ρπ z ) sin(π z ) ρ2 − 1 − sinh − 1 cot , × 1 + cos(ρπ z) π 1 + cos(ρπ z ) 4 2
0 < ρ < 2;
ρ > 2.
For ρ 6= 1, the right-hand side of the penultimate (resp. last) asymptotic formula diverges at z = −2/(2 + ρ ) (resp. z = −1/ρ ), so the asymptotic analysis does not result in exact forms of Tρ z,z (ρ 6= 1) for finitely sized | z|. Corollary 2.2 (Some Multiple Elliptic Integrals in Clebsch-Gordan Forms) We have the integral formulae Z1 p p π2 π3 T0,−1/2 = 2 , K( t)K( 1 − t) d t = 4 4 0 Z1 p p p p π2 T0,1/2 = 2 [2E( t) − K( t)][2E( 1 − t) − K( 1 − t)] d t = 0, 4 0 Z1 3 p p [Γ( 41 )]8 π , T−1/2,−1/2 = 2 [K( 1 − t)]2 K( t) d t = 8 192π2 0 !#2 Ãs ! " Ãs Z π3 p3 (2 + p) 27 1 (1 − p2 ) p(2 + p) p3 (2 + p) 1− T−1/3,−1/3 = K dp K p 8 2 0 1 + 2p 1 + 2p 3 + 6 p(1 + p + p2 ) !#2 Ãs ! " Ãs p Z 3 3[Γ( 31 )]9 p3 (2 + p) 27 1 (1 − p2 ) p(2 + p) p3 (2 + p) 1− K dp= , = K p 6 0 1 + 2p 1 + 2p 256π2 1 + 2 p(1 + p + p2 ) " Ãs p !#2 Ãs p ! p p Z1 p Z1 (1 − t)[K( t)]2 K( 1 − t) 1− u 2 u π3 4 K K dt T−1/4,−1/4 = p p p du = 4 2 8 (1 + t)3/2 0 0 (1 + u )3/2 1+ u 1+ u p ! p p ¶3/2 " Ãs p !#2 Ãs Z1 µ Z1 [Γ( 81 )Γ( 38 )]2 2 u 1− u 2 (1 − t)[K( 1 − t)]2 K( t) d t = , K K = p p p du = 4 24 (1 + t)3/2 0 0 1+ u 1+ u 1+ u !#2 Ãr ! p p Z Z 1 " Ãr [Γ( 41 )]4 1 − x2 27 π3 1−x 1+x 27 1 t(1 − t)[K( 1 − t)]2 K( t) K T−1/6,−1/6 = p K d x = d t = p p , 8 2 2 4 0 (3 + x2 )7/4 (1 − t + t2 )7/4 2 2 −1 8 2 3 Ãs p ! Ãs p p p ! p Z1 2E( u ) − K( u) p π3 2 u 1− u T1/2,−3/4 = 2 2 K p p K p d u = 2π , 8 0 1+ u 1+ u 1+ u Ãs p ! Ãs p p p p ! p Z π3 2 u 1− u 2π 2 2 1 8(1 − 2u)E( u) − (5 − 8u)K( u) K T3/2,−1/4 = . p p K p du = − 8 3 0 9 1+ u 1+ u 1+ u Proof Special cases of Eqs. 19(2 m,n+ 1 ) , 19(ν,ν) and 19(2ν+2,ν) lead to Eqs. 23, 24, 25, 26, 27, 28, 29 and 30. 2
(23) (24) (25)
(26)
(27)
(28)
(29)
(30)
3 Spherical Rotations and Multiple Elliptic Integrals In this section, we shall focus on the transformations of multiple elliptic integrals using variable substitutions motivated by rotations on a unit sphere. In effect, our attention will be momentarily restricted to two special Legendre functions P−1/2 and P−1/4 . 13
The transformation methods in this section not only provide evaluations of certain Cauchy principal values: s ! Ãs ! p Z1 Ã Z1 K( (1 − ξ)/2) d ξ 2dξ 1+ξ 1−ξ , −1 < x < 1; P K , −1 < x < 1 etc. K P p 2 2 π( x − ξ) −1 −1 π( x − ξ) 1 + ξ
which will be used in §4, but also pave way for further applications in our subsequent works on elliptic integrals.
3.1
Beltrami Rotations
The technique of spherical rotation, which traces back to Beltrami’s work on the Abel integral equations [Ref. 23, p. 328], has found applications in the geometrically-motivated evaluation of certain definite integrals related to Bessel functions (see [Ref. 15, §3.33, §12.12 and §12.14]). The lemma below is a simple realization of the Beltrami rotation in the case of the complete elliptic integrals K(k) and E(k). Lemma 3.1 (Beltrami Transformations) (a) We have the following integral identities p Z 2 1 K( 1 − κ2 ) d κ K(k) = , 0 < k < 1, π 0 1 − k2 κ2 p p Z 2 1 1 − ξ2 K( 1 − κ2 ) d κ K(ξ) = , 0 < ξ < 1, π 0 1 − ξ2 (1 − κ2 ) p Z 2 1 (1 + r)K( 1 − κ2 ) d κ , 0 < r < 1, K(r) = π 0 (1 + r)2 − 4rκ2 p Z 2 1 (1 − η)K( 1 − κ2 ) d κ K(η) = , 0 < η < 1. π 0 (1 − η)2 + 4ηκ2 (b) For 0 < k < 1, we have
p E( 1 − κ2 ) d κ , 1 − k2 κ2 0 p Z 4(1 − k2 ) 1 E( 1 − κ2 ) d κ , E(k) = π (1 − k2 κ2 )2 0 p Z E(k) 4(1 − k2 ) 1 E( 1 − κ2 ) d κ = . p 2 2 2 π 0 [1 − k (1 − κ )] 1 − k2
K(k) − E(k) 2 = π k2
(31) (31′ ) (31L ) (31′L )
Z1
(32) (33) (33′ )
Proof (a) We start by writing the complete elliptic integral of the first kind as an integral on the unit sphere S 2 : Z Z Z 2π Z 1 dθ 1 dσ 1 π 1 π sin θ d θ = . dφ K(k) = = p p 2 2 2 2 0 4 π sin θ (1 − k sin θ cos φ ) 2 S 0 0 1 − Z (1 − kX ) 4π 1 − k sin θ
As we rotate about the Y -axis by a right angle, effectively swapping the rôles of the X - and Z -axes, we may rewrite the integral above as Z Z Z2π Z 1 π 1 1 1 dσ dσ , K(k) = = = sin θ d θ dφp p p S2 0 S2 1 − X 2 (1 − kZ ) 4π 1 − X 2 (1 − k2 Z 2 ) 4π 4π 0 1 − sin2 θ cos2 φ(1 − k2 cos2 θ)
where we have combined the values of the integrand at points ± Z and restored the spherical coordinates. Integrating over the azimuthal angle φ, we obtain Z 1 π K(sin θ) sin θ d θ K(k) = , π 0 1 − k2 cos2 θ which is equivalent to Eq. 31. Now that we have proved the relation 2 K(k) = π
Z1 0
p K( 1 − κ2 ) d κ, 1 − k2 κ2
0 < k < 1,
which analytically continues to k ∈ C r [1, +∞), we may derive Eqs. 31′ , 31L and 31′L , respectively, from the imagip p p nary modulus transformation K(ξ) = K( i ξ/ 1 − ξ2 )/ 1 − ξ2 , Landen’s transformation K(r) = K(2 r /(1 + r))/(1 + r), and a p combination thereof K(η) = K(2 i η/(1 − η))/(1 − η). 14
(b) We adapt our derivation of part (a) to accommodate for the complete elliptic integrals of the second kind E(k), 0 < k < 1: Z Z Z2π Z p 1 − Z2 d σ 1 π K(k) − E(k) 1 π sin2 θ d θ sin θ = = sin θ d θ = d φ p 2 0 1 − k sin θ cos φ k2 S 2 1 − kX 4π 0 1 − k2 sin2 θ 4π 0 p p p Z Z Z Z2π Z 1 π 1− X2 dσ 1 − X2 dσ 1 − sin2 θ cos2 φ 1 π E(sin θ) sin θ d θ = = sin θ d θ d φ = , = 2 2 4π 0 π 0 1 − k2 cos2 θ 1 − k2 cos2 θ S 2 1 − k Z 4π 0 S 2 1 − kZ 4π which verifies Eq. 32. As we multiply both sides of Eq. 32 by k2 , and differentiate in k, we obtain the formula stated in Eq. 33: p Z1 p Z kE(k) 2 d E( 1 − κ2 )k2 d κ 4k 1 E( 1 − κ2 ) d κ d [K(k) − E(k)] = = = , 0 < k < 1. π dk 0 π 0 (1 − k2 κ2 )2 1 − k2 d k 1 − k2 κ2 Now, by analytic continuation, we have 4(1 − k2 ) E(k) = π
Z1 0
p E( 1 − κ2 ) d κ , (1 − k2 κ2 )2
k ∈ C r [1, +∞).
(33∗ )
In particular, by the imaginary modulus transformation for complete elliptic integrals of the second kind [Ref. 24, item 160.02], we obtain à ! p Z p ξ 4(1 + ξ2 ) 1 E( 1 − κ2 ) d κ 2 , ξ > 0, 1+ξ E p = E( i ξ) = π (1 + ξ2 κ2 )2 0 1 + ξ2 which is equivalent to Eq. 33′ .
Remark Admittedly, the spherical rotation is not the only road to any or all of the identities in Eqs. 31, 31′ , 31L and 31′L . For example, one may still verify Eq. 31 via the method of moments and hypergeometric summations [6]. The same can be said for several formulae that we will encounter in Proposition 3.3. However, it is our hope that the spherical rotations provide clearer geometric motivations than a purely combinatorial approach. The author thanks an anonymous referee for pointing out a recent contribution of V. Anghel [25], which generalizes the Beltrami rotation to hyperspheres embedded in Euclidean spaces of arbitrary dimensions. It is highly probable that the hyperspherical analog of Beltrami rotation can lead to geometric proofs of various integral identities related to elliptic integrals, which we hope to pursue in a separate publication. Beltrami transformations allow us to convert one multiple elliptic integral into another. For instance, the following chain of identities ¸ Z Z1 Z1 Z1 p ´ · κ arccos κ K(ξ)ξ K( 1 − κ2 ) 2 1 ³p dξ dκ = K K(ξ) d ξ = − log(2κ) d κ = 1 − κ2 p p p 1+κ 0 0 1 + 1 − ξ2 0 1 − κ2 1 − ξ2 π 0 result from two applications of Eq. 31′ , and the transformations Z1 0
2 E(k) d k = π
Z1 0
p p Z1 (2 + κ)E( 1 − κ2 ) E( 1 − κ2 ) 2 −1 [−κ + (1 + κ ) tanh κ] d κ = dκ κ3 (1 + κ)2 0
can be justified by Eqs. 33 and 33′ . More examples will be described elsewhere. In the next corollary, we mention a few consequences of Beltrami transformations that are pertinent to later developments in the current work. Corollary 3.2 (Some Applications of Beltrami Transformations) (a) For 0 < u < 1, we have the following duality relations for multiple elliptic integrals: Zp u 0
Z1 p
u
K(k) d k = p 1 − k2 u − k2
Z1
kK(k) d k , p p 0 1 − k2 1 − k2 u p Z1 kK(k) d k K( 1 − κ2 ) d κ . = p p p p 0 1 − k2 k2 − u 1 − κ2 1 − κ2 u p
15
(34) (35)
(b) We have the following Cauchy principal values for −1 < x < 1:
p p K( (1 − ξ)/2) K( (1 + x)/2) dξ = P , p p −1 π( x − ξ) 1 + ξ 1+x p p p p Z1 K( (1 − ξ)/2) − 2E( (1 − ξ)/2) 2E( (1 + x)/2) − K( (1 + x)/2) dξ = . P p p −1 1+x π( x − ξ) 1 + ξ Z1
(36) (37)
Proof (a) With Eq. 31′ , we may interchange the order of integrations to compute # p p Zpu " Z1 p Z1 Zpu 1 − ξ2 K( 1 − κ2 ) d κ K( 1 − κ2 ) d κ K(k) d k 2 dξ = = , p p p p p π 0 1 − ξ2 (1 − κ2 ) 0 0 0 1 − k2 u − k2 1 − ξ2 u − ξ2 1 − u(1 − κ2 )
p where the last expression is equivalent to the right-hand side of Eq. 34, by the correspondence κ 7→ 1 − k2 . Similarly, we may prove Eq. 35 using Eq. 31: # p Zp u " Z1 p Z1 Z1 kK(k) d k K( 1 − κ2 ) d κ kdk K( 1 − κ2 ) d κ 2 = = . p p p p p p p π 0 1 − k2 κ2 u 1 − k2 k2 − u 0 0 1 − k2 k2 − u 1 − κ2 1 − κ2 u
(b) For z > 1, one can readily compute (cf. [Ref. 26, p. 233])
Z1/p z Z1 dt dt dt K( z) = = −i p p p p p p p 0 1/ z 1 − t2 zt2 − 1 0 1 − t2 1 − zt2 1 − t2 1 − zt2 p p Z1 Zp z ds K( 1/ z) − i K( ( z − 1)/ z) i 1 ds = −p =p . q q p p p z 0 z 1 z 1 − 1z s2 1 − s2 1 − 1z s2 s2 − 1
p
Z1
(38)
This is the inverse modulus transformation for complete elliptic integrals of the first kind. Now, in the identity K(k) =
2 π
Z1 0
p K( 1 − κ2 ) d κ , 1 − k2 κ2
∀k ∈ C r [1, +∞),
we approach the limit of k ± i 0+ ∈ [1, +∞), read off the real part, and employ the inverse modulus transformation (Eq. 38), to obtain
k K(k) = P π
p K( 1 − κ2 ) d κ , k2 − κ2 −1
Z1
0 < k < 1,
which can be reformulated into Eq. 36 after the variable substitutions k = From Eq. 36, we may directly compute (cf. [Ref. 27, Eq. 11.215])
p p (1 + x)/2, κ = (1 + ξ)/2.
p p p Z1 K( (1 − ξ)/2)[1 + (ξ − x) + x] d ξ K( (1 − ξ)/2) 1 + ξ d ξ =P P p π( x − ξ) −1 −1 π( x − ξ) 1 + ξ p p Z1 Z K( (1 − ξ)/2) d ξ 1 1 K( (1 − ξ)/2) d ξ = (1 + x)P − p p π −1 −1 π( x − ξ) 1 + ξ 1+ξ ! Ãr p Z1 p K( (1 − ξ)/2) d ξ 1 1+x , −1 < x < 1 . − = 1 + xK p 2 π −1 1+ξ Z1
Here, the last integral can be simplified with the knowledge of brings us
R1 0
p K( 1 − κ2 ) d κ = π2 /4 [Ref. 13, item 6.141.2], which
Ãr ! p p K( (1 − ξ)/2) 1 + ξ d ξ p π 1+x = 1 + xK −p , P π( x − ξ) 2 −1 2 Z1
16
−1 < x < 1 .
Applying integration by parts to the Tricomi transform (cf. [Ref. 27, Eqs. 11.218 and 11.219]), one can use the equation above to deduce that # Ãr " Ãs ! " !# Z1 1−ξ p 1+x d p d 1 1 , −1 < x < 1 . 1 + ξ dξ = 1 + xK P K +p 2 dξ 2 −1 π( x − ξ) d ξ 2( x − 1) In other words, we have p p · p ¸ K( (1 − ξ)/2) − E( (1 − ξ)/2) E( (1 + x)/2) 1 1 P −p , dξ = p p 1−x −1 1+x 2 π( x − ξ)(1 − ξ) 1 + ξ Z1
−1 < x < 1 .
Carrying this further, we obtain
p p p p p Z K( (1 − ξ)/2) − E( (1 − ξ)/2) 1 1 1 K( (1 − ξ)/2) − E( (1 − ξ)/2) E( (1 + x)/2) P −p + dξ = dξ p p p −1 1+x 2 π −1 π( x − ξ) 1 + ξ (1 − ξ) 1 + ξ " Ãs ! # p p Z 1 1 1 d 1−ξ p E( (1 + x)/2) E( (1 + x)/2) K − , −1 < x < 1 . + 1 + ξ d ξ = = p p p 2 1+x 1+x 2 π −1 d ξ Z1
We may combine the relation above with Eq. 36 into a symmetric form p p p p K( (1 − ξ)/2) − 2E( (1 − ξ)/2) 2E( (1 + x)/2) − K( (1 + x)/2) P dξ = , p p −1 1+x π( x − ξ) 1 + ξ Z1
−1 < x < 1 ,
as stated in Eq. 37.
3.2
Ramanujan Rotations
The next proposition is based on reverse engineering of some formulae in Ramanujan’s notebooks that were stated without proofs. p p p Proposition 3.3 (Ramanujan Transformations) (a) For s, t ∈ [0, 1), we may represent K( s)K( t) and [K( t)]2 as integrals on the unit sphere S 2 : Z p p dσ 1 , (39) K( s)K( t) = p 8 S 2 (1 − sX 2 )(1 − tY 2 ) − Z 2 p Z p 2 2(2 − t)K( X 2 + Y 2 ) d σ [K( t)] = (40) 4π (2 − t)2 − t2 X 2 S2 p Z (1 + t)K( X 2 + Y 2 ) d σ , (40L ) = (1 + t)2 − 4 tX 2 4π S2 where X = sin θ cos φ,Y = sin θ sin φ, Z = cos θ are the Cartesian coordinates on the unit sphere S 2 . Consequently, the following identities hold for 0 < u < 1: " Ãs p !#2 1 2 u kK(k) d k , = p p p K p 2 2 0 0 1+ u 1−k 1−k u 1+ u " Ãs p !#2 Zp1−u Z1 2 K(k) d k 1− u kK(k) d k = = . p p p p K p p 0 0 1+ u 1 − k2 1 − u − k2 1 − k2 1 − k2 (1 − u) 1 + u Zp u
K(k) d k = p p 1 − k2 u − k2
Z1
(41)
(42)
(b) We have the integral identity Z
S2
K
³p
X2 +Y2
´
dσ 2 f (| X |) = 4π π
Z1 0
17
s
f (k)K
s 1+k 1−k K dk 2 2
(43)
so long as the function f (| X |), 0 ≤ | X | ≤ 1 assumes non-negative values and the surface integral is convergent. This allows us to convert Eq. 40 into the following identities for 0 < t < 1: p
2 [K( t)] = π 2
Z1
p p K( µ)K( 1 − µ) d µ
(40∗ )
, 1 − µt p p Z p 8 1 K( µ)K( 1 − µ) d µ [K( 1 − t)]2 = p p , π 0 (1 + t)2 − µ(1 − t)2 0
(40∗L )
and to deduce the following integral formulae:4 p K( 1 − κ2 )
p p [K( 1 − t) − i K( t)]2 , dκ = p 2 0 (2 t − 1)2 − (1 − κ2 ) p p p Z1 K( 1 − κ2 ) [K( 1 − t) + i K( t)]2 dκ = , p 2 0 (2 t − 1)2 − (1 − κ2 )
Z1
Moreover, we have the identity below for 0 < λ ≤ 1/2: Z1
p p kK(k) d k = K( λ)K( 1 − λ) = p p 2 2 2 1 −2 λ 1 − k k − (1 − 2λ)
0
(44)
t < 0.
(44′ )
Z1
or equivalently,
Zπ/2
1 0 1; g ∈ L q (−1, 1), q > 1 and 1p + 1q < 1. We call such a procedure “Tricomi pairing”, for which an example in the proposition below puts finishing touches on a proof for the conjectural identity stated in the introduction. Proposition 4.1 (An Application of Tricomi Pairing to Multiple Elliptic Integrals) There are several integrals that evaluate to the same number [Γ( 41 )]8 /(128π2 ): Z Z1 h ³p Z1 ´i3 10 1 K 1 − k2 [K(k)]3 d k = 5 [K(k)]3 k d k dk = 3 0 0 0 Z1 Z1 Z1 ´ ´i2 ´ ³p ³p h ³p 2 2 2 d k = 6 [K(k)]2 K 1 − k2 k d k. =3 1−k dk = 2 1−k [K(k)] K K(k) K 0
(56)
0
0
p p p p Proof If we set f ( x) = K( (1 − x)/2)/ 1 + x, −1 < x < 1 and g( x) = 2K( (1 + x)/2)K( (1 − x)/2), −1 < x < 1 in Eq. 55, while recalling the Cauchy principal values given in Eqs. 36 and 51, then we arrive at Z1 " Ã 2 K −1
s
1+x 2
!#2
K
Ãs
s ! ! " Ãs !#2 " Ãs !#2 Z1 Ã dξ 1−x 1−ξ 1+ξ 1−ξ dx =− K . − K K p p 1+ξ 2 2 2 2 −1 1+x
p p R1 R1 This instantly rearranges into 0 [K( 1 − k2 )]3 d k = 3 0 [K(k)]2 K( 1 − k2 ) d k (an identity that Wan conjectured numerically in [6] without an analytic proof), which reveals the equivalence between the leading items of the first two lines in Eq. 56. p Writing k = (1 − ξ)/(1 + ξ) and employing Landen’s transformation K(2 ξ/(1 + ξ)) = (1 + ξ)K(ξ), one has s µ ¶2 3 Z1 h ³p Z1 Z1 ´i3 1 − ξ d 1 − ξ = 2 K 1 − K dk = 1 − k2 [K(ξ)]3 (1 + ξ) d ξ; 1+ξ 1+ξ 0 0 0
23
p writing k = (1 − ξ)/(1 + ξ) and employing Landen’s transformation 2K((1 − ξ)/(1 + ξ)) = (1 + ξ)K( 1 − ξ2 ), one has ¶¸ ¶¸3 Z · µq Z Z Z1 Z1 · µ ´i3 1 1 1 1 h ³p 1−ξ 3 1−ξ 1 1 dk+ K [K(η)]3 η d η, = K d 1 − ξ2 1 − k2 (1 + ξ) d ξ = [K(k)]3 d k = K 1+ξ 1+ξ 4 0 4 0 4 0 0 0 p where the last step is a trivial substitution ξ 7→ 1 − η2R . The p two simultaneous above make it posR1 equationsRdisplayed 1 1 sible to eliminate any one among the three quantities 0 [K( 1 − k2 )]3 d k, 0 [K(k)]3 d k, 0 [K(k)]3 k d k, and determine the ratio between the two remaining numbers. Thus, we have verified the chain of identities in the first line of Eq. 56 (cf. [Ref. 6, Eq. 29]). The relations within the second line of Eq. 56 can be likewise established by successive Landen’s transformations, as detailed in the first paragraph of [Ref. 6, p. 139]. As we recall from Eq. 25 that
6
Z1
[K(k)]2 K
0
´ ³p [Γ( 41 )]8 , 1 − k2 k d k = 128π2
the verification is complete.
Remark Sometimes, the output of Tricomi pairing can also be immediately recovered by more straightforward means. For example, upon setting Ãs ! Ãs ! 1+ξ 1−ξ f (ξ) = 2K K , 2 2 Ãs ! Ãs ! Ãs ! Ãs ! Ãs ! Ãs ! 1+ξ 1−ξ 1−ξ 1+ξ 1+ξ 1−ξ g(ξ) = K K +K E −K E 2 2 2 2 2 2 in Eq. 55, one arrives at a vanishing identity: Z1 h ³p ´i2 h ³p ´ ³p ´ ³p ´i K 1 − k2 1 − k2 K(k) + K 1 − k2 E(k) − 3K(k)E 1 − k2 k d k. K 0= 0
As pointed out by an anonymous referee, the equation above is anticipated from the fact that the integrand is precisely p the derivative of k2 (1 − k2 )K(k)[K( 1 − k2 )]3 , which makes it less surprising than Eq. 56. It might be still interesting to ask if Eq. 56 can be likewise reduced into finite steps of algebraic manipulations on elliptic integrals and applications of the Newton-Leibniz formula (cf. §5).
4.2
Tricomi Transform of Pν ( x)Pν (− x)
The formula in Eq. 51 can be rewritten as Z1 2P−1/2 (ξ)P−1/2 (−ξ) P d ξ = −{[P−1/2 ( x)]2 − [P−1/2 (− x)]2 }, π( x − ξ) −1
−1 < x < 1 .
This is not accidental. In the next proposition, we will generalize such a Tricomi transform relation to Legendre functions of arbitrary degree ν. Proposition 4.2 (Tricomi Transform of Pν (ξ)Pν (−ξ)) For any ν ∈ C r Z, we have P
Z1
[Pν ( x)]2 − [Pν (− x)]2 2Pν (ξ)Pν (−ξ) dξ = , π( x − ξ) sin(νπ) −1
−1 < x < 1 .
For n ∈ Z≥0 , there is an identity Z1 Z1 [P n (ξ)]2 [P−n−1 (ξ)]2 P dξ = P d ξ = P n ( x)Q n ( x), 2( x − ξ) −1 2( x − ξ) −1
−1 < x < 1 .
Proof To verify Eq. 57, it would suffice to demonstrate that ¾ ½ Z1 Z1 2Pν (ξ)Pν (−ξ) [Pν ( x)]2 − [Pν (− x)]2 −P dξ d x 0= Pℓ ( x) sin(νπ) π( x − ξ) −1 −1 Z1 Z 1 4 1 2 2 = Pℓ ( x){[Pν ( x)] − [Pν (− x)] } d x + Q ℓ ( x)Pν( x)Pν (− x) d x, sin(νπ) −1 π −1 24
(57)
∀ℓ ∈ Z≥0 .
(58)
(59)
Here, in the last line, we have used the Parseval identity of Tricomi pairing (Eq. 55), along with the Neumann integral representation for Legendre functions of the second kind (cf. [Ref. 27, Eq. 11.269] or [Ref. 33, Table 1.12A, Eq. 12A.26]):
Q ℓ ( x) = P
Z1
−1
Pℓ (ξ) d ξ , 2( x − ξ)
∀ x ∈ (−1, 1), ∀ℓ ∈ Z≥0 .
(60)
For a non-negative even number ℓ, both addends in the last line of Eq. 59 vanish because the integrands are odd functions. We may now settle Eq. 59 for odd numbers ℓ by induction. For ℓ = 1, we use Eq. 16 to compute Z1 1 x{[Pν ( x)]2 − [Pν (− x)]2 } d x sin(νπ) −1 −1 · ¸ 1 d d{[Pν ( x)]2 − [Pν (− x)]2 } =− lim + (1 − x2 ) (1 − x2 ) 4ν(ν+1) sin(νπ) x→1−0 dx dx · ¸ 2 d d { [ P ( x )] − [Pν (− x)]2 } 4 sin(νπ) 1 ν 2 2 lim (1 − x ) + (1 − x ) = . 4ν(ν+1) sin(νπ) x→−1+0+ dx dx ν(ν + 1)π2
1 sin(νπ)
Z1
P1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x =
Meanwhile, we may check that ¶ Z1 µ 1+x x −1 + log P ν ( x ) P ν (− x ) d x 2 1−x −1 −1 ½ · ¸ ¾ Z x log 11+ 4 1 2 d d(Pν ( x)Pν (− x)) −x d =− P ν ( x ) P ν (− x ) d x − (1 − x2 ) (1 − x2 ) + 4ν(ν + 1)(1 − x2 )Pν ( x)Pν(− x) d x π −1 π −1 4ν(ν+1) d x dx dx · ¸ Z1 d 4 d(Pν ( x)Pν (− x)) 4 sin(νπ) 1 = (1 − x2 ) dx =− . π −1 4ν(ν+1) d x dx ν(ν + 1)π2
4 π
Z1
4 π Z1
Q 1 ( x ) P ν ( x ) P ν (− x ) d x =
Thus, Eq. 59 holds for ℓ = 1. Noting that the Legendre functions of the first and second kinds satisfy similar recursion relations, namely, (2µ + 1)(1 − x2 ) (2µ + 1)(1 − x2 )
d Pµ ( x) dx d Q µ ( x) dx
= µ(µ + 1)[Pµ−1 ( x) − Pµ+1 ( x)];
(2µ + 1) xPµ ( x) = (µ + 1)Pµ+1 ( x) + µPµ−1 ( x);
= µ(µ + 1)[Q µ−1 ( x) − Q µ+1 ( x)];
(2µ + 1) xQ µ ( x) = (µ + 1)Q µ+1 ( x) + µQ µ−1 ( x),
we can deduce Z 4ν(ν + 1)(ℓ + 1) 1 Q ℓ +1 ( x ) P ν ( x ) P ν ( − x ) d x sin(νπ) π −1 −1 Z Z 4ν(ν + 1)ℓ 1 ν(ν + 1)ℓ 1 Pℓ−1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x + Q ℓ −1 ( x ) P ν ( x ) P ν ( − x ) d x + sin(νπ) −1 π −1 Z Z ν(ν + 1)(2ℓ + 1) 1 4ν(ν + 1)(2ℓ + 1) 1 = xPℓ ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x + xQ ℓ ( x)Pν ( x)Pν (− x) d x, sin(νπ) π −1 −1 ν(ν + 1)(ℓ + 1)
Z1
Pℓ+1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x +
as well as Z Z1 4ν(ν + 1) 1 2 2 [(ℓ + 1)Pℓ+1 ( x) + ℓPℓ−1 ( x)]{[Pν ( x)] − [Pν (− x)] } d x = 4ν(ν + 1) xPℓ ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x 2ℓ + 1 − 1 −1 ½ · ¸ ¾ Z1 d d d{[Pν ( x)]2 − [Pν (− x)]2 } =− Pℓ ( x) (1 − x2 ) (1 − x2 ) + 4ν(ν + 1)(1 − x2 )[Pν ( x)]2 − 4ν(ν + 1)(1 − x2 )[Pν ( x)]2 d x dx dx dx −1 · ¾ Z1 ½ 2 2 ¸ d{[Pν ( x)] − [Pν (− x)] } d Pℓ ( x) d dx = (1 − x2 ) + 4ν(ν + 1)(1 − x2 )[Pν ( x)]2 − 4ν(ν + 1)(1 − x2 )[Pν ( x)]2 (1 − x2 ) dx dx dx −1
8[1 + (−1)ℓ ] sin2 (νπ) π2 Z Z1 4ℓ(ℓ + 1)ν(ν + 1) 1 d{[Pν ( x)]2 − [Pν (− x)]2 } = ℓ(ℓ + 1) dx+ [Pℓ−1 ( x) − Pℓ+1 ( x)]{[Pν ( x)]2 − [Pν (− x)]2 } d x (1 − x2 )Pℓ ( x) dx 2ℓ + 1 −1 −1 8[1 + (−1)ℓ ] sin2 (νπ) , + π2 +
25
which leads to a recursion relation 2
2
2
(ℓ + 1) [(ℓ + 1) − (2ν + 1) ] − ℓ2 [ℓ2 − (2ν + 1)2 ]
Z1
−1
Z1
−1
Pℓ+1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x
Pℓ−1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x = −
8[1 + (−1)ℓ ](2ℓ + 1) sin2 (νπ) , π2
and Z Z1 4ν(ν + 1) 1 [(ℓ + 1)Q ℓ+1 ( x) + ℓQ ℓ−1 ( x)]Pν ( x)Pν (− x) d x = 4ν(ν + 1) xQ ℓ ( x)Pν ( x)Pν(− x) d x 2ℓ + 1 − 1 −1 ½ · ¸ ¾ Z1 d 2 d 2 d( Pν ( x) Pν (− x)) 2 (1 − x ) (1 − x ) + 4ν(ν + 1)(1 − x )Pν ( x)Pν (− x) d x =− Q ℓ ( x) dx dx dx −1 · ¸ ¾ Z1 ½ d d(Pν ( x)Pν (− x)) d Q ℓ ( x) = (1 − x2 ) dx (1 − x2 ) + 4ν(ν + 1)(1 − x2 )Pν ( x)Pν(− x) dx dx dx −1 Z Z1 4ℓ(ℓ + 1)ν(ν + 1) 1 d(Pν ( x)Pν (− x)) [Q ℓ−1 ( x) − Q ℓ+1 ( x)]Pν( x)Pν (− x) d x dx+ = ℓ(ℓ + 1) (1 − x2 )Q ℓ ( x) dx 2ℓ + 1 −1 −1 2[1 + (−1)ℓ ] sin(νπ) − , π
which brings us another recursion relation (ℓ + 1)2 [(ℓ + 1)2 − (2ν + 1)2 ] − ℓ2 [ℓ2 − (2ν + 1)2 ]
Z1
−1
Z1
−1
Q ℓ +1 ( x ) P ν ( x ) P ν ( − x ) d x
Q ℓ −1 ( x ) P ν ( x ) P ν ( − x ) d x =
Here, we have used the identities · ¸ d Pν ( x) 8 sin2 (νπ) 2 2 d 2(1 − x )Pν ( x) = ; lim (1 − x ) dx dx x→−1+0+ π2
2[1 + (−1)ℓ ](2ℓ + 1) sin(νπ) . π
Pℓ (1) = (−1)ℓ Pℓ (−1) = 1, ∀ℓ ∈ Z≥0
to take care of boundary contributions to the integral concerning Pℓ ( x), and resorted to the limit behavior d Q µ ( x) d(Pν ( x)Pν(− x))
2 sin(νπ) dx dx π 2 cos(µπ) sin(νπ) 2 2 d Q µ ( x) d( Pν ( x) Pν (− x)) = lim (1 − x ) dx dx π x→−1+0+ lim (1 − x2 )2
x→1−0+
=−
for the integration by parts involving Q ℓ ( x). Therefore, the last line of Eq. 59 satisfies a homogeneous recursion ¾ ½ Z1 Z 4 1 1 Pℓ+1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x + Q ℓ +1 ( x ) P ν ( x ) P ν ( − x ) d x (ℓ + 1)2 [(ℓ + 1)2 − (2ν + 1)2 ] sin(νπ) −1 π −1 ½ ¾ Z1 Z1 1 4 = ℓ2 [ℓ2 − (2ν + 1)2 ] Pℓ−1 ( x){[Pν ( x)]2 − [Pν (− x)]2 } d x + Q ℓ −1 ( x ) P ν ( x ) P ν ( − x ) d x , ℓ ∈ Z ≥0 sin(νπ) −1 π −1 and the truthfulness of Eq. 59 for ℓ = 1 entails any scenario with a larger odd number ℓ. This completes the verification of Eq. 57 for ν ∈ C r Z. For a fixed x ∈ (−1, 1), both sides of Eq. 57 represent continuous functions of ν, so we may handle Eq. 58 by investigating the ν → n limit where n ∈ Z≥0 . Suppose that n is even and non-negative, then we have
P ν ( x ) − P ν (− x ) cos(νπ)Pν ( x) − Pν (− x) [Pν ( x)]2 − [Pν (− x)]2 = 2P n ( x) lim = 2P n ( x) lim ν→ n ν→ n ν→ n sin(νπ) sin(νπ) sin(νπ) 4 4 4 = P n ( x) lim Q ν ( x) = P n ( x)Q n ( x) = P n (− x)Q n ( x). ν→ n π π π lim
If we start from an odd and positive n instead, we will end up with
P ν ( x ) + P ν (− x ) cos(νπ)Pν ( x) − Pν (− x) [Pν ( x)]2 − [Pν (− x)]2 = 2P n ( x) lim = −2P n ( x) lim ν→ n ν→ n ν→ n sin(νπ) sin(νπ) sin(νπ) 4 4 4 = − P n ( x) lim Q ν ( x) = − P n ( x)Q n ( x) = P n (− x)Q n ( x). ν→ n π π π lim
26
Hence, we have proved Z1 Z1 4 4 2P−n−1 (ξ)P−n−1 (−ξ) 2P n (ξ)P n (−ξ) dξ = P d ξ = (−1)n P n ( x)Q n ( x) = P n (− x)Q n ( x), P π( x − ξ) π( x − ξ) π π −1 −1
n ∈ Z ≥0 ,
which is equivalent to Eq. 58. In fact, Eq. 58 is not particularly surprising. It is just a special case of the stronger statement that (cf. [Ref. 27, Eqs. 11.280 and 11.281] or [Ref. 33, Table 1.12A, Eq. 12A.27]) Z1 P n (ξ) p(ξ) d ξ = Q n ( x) p( x), deg p( x) ≤ n, (61) P −1 2( x − ξ) where p( x) is any polynomial whose degree does not exceed n.
We follow up with some additional examples involving the product of four elliptic integrals in the integrands. Corollary 4.3 (Another Application of Tricomi Pairing) We have an integral identity Z1 sin(2νπ) cos(νπ) x[Pν ( x)]3 Pν (− x) d x = , ν ∈ C r {−1/2}, (2ν + 1)2 π −1
(62)
which leads to
−
π
2
= =
−
p 9 3 = 4π
Z1
−1
32 π4 Z1
x[P−1/2 ( x)]3 P−1/2 (− x) d x
Z1
−1
0
p p 32 (1 − 2 t)[K( t)]3 K( 1 − t) d t = − 4 π
Z1 0
p p (1 − 2 t)[K( 1 − t)]3 K( t) d t,
(63)
x[P−1/3 ( x)]3 P−1/3 (− x) d x
!#3 Ãs ! · ¸ " Ãs 3 27 p2 (1 + p)2 p (2 + p) (1 − p2 ) p(2 + p) p3 (2 + p) 1−2 K dp K 1− 1 + 2p 1 + 2p 1 + 2p 4(1 + p + p2 )3 0 !#3 Ãs ! · ¸ " Ãs Z1 27 p2 (1 + p)2 72 (1 − p2 ) p(2 + p) p3 (2 + p) p3 (2 + p) 1−2 K d p, =−p K 1− 1 + 2p 1 + 2p 1 + 2p 4(1 + p + p2 )3 3π 4 0 p Z1 2 2 − = x[P−1/4 ( x)]3 P−1/4 (− x) d x π −1 " Ãs p !#3 Ãs p ! p Z 2 u 1− u 32 2 1 1 − 2u K = p p du p 2 K π4 0 (1 + u ) 1+ u 1+ u " Ãs p Z p !#3 Ãs p ! 64 2 1 1 − 2u 1− u 2 u =− K p p d u, p 2 K 4 π 0 (1 + u ) 1+ u 1+ u Z1 27 − = x[P−1/6 ( x)]3 P−1/6 (− x) d x 16π −1 Z p p 54 1 t(1 − t)(1 + t)(2 − t)(1 − 2 t) [K( t)]3 K( 1 − t) d t = 4 2 3 π 0 (1 − t + t ) Z p p 54 1 t(1 − t)(1 + t)(2 − t)(1 − 2 t) [K( 1 − t)]3 K( t) d t. =− 4 2 3 π 0 (1 − t + t )
216 =p 3π 4
Z1
(64)
(65)
(66)
Proof For integer degrees ν ∈ Z, the left-hand side of Eq. 62 represents the integration of an odd function over the interval [−1, 1], hence vanishing. This is consistent with the corresponding behavior on the right-hand side. We now turn our attention to the scenarios where ν ∈ C r (Z ∪{−1/2}). From the Tricomi transform formula in Eq. 57, one may readily deduce the following relation Z1 Z Z1 2ξPν (ξ)Pν (−ξ) [Pν ( x)]2 − [Pν (− x)]2 2 1 2[ x − ( x − ξ)]Pν (ξ)Pν (−ξ) P dξ = P dξ = x − Pν (ξ)Pν (−ξ) d ξ. π( x − ξ) π( x − ξ) sin(νπ) π −1 −1 −1 Here, according to Eq. 19(0,ν) , we have
T0,ν =
Z1
−1
Pν (ξ)Pν (−ξ) d ξ = 27
2 cos(νπ) , 2ν + 1
so we may combine the formula P
Z1
[Pν ( x)]2 − [Pν (− x)]2 4 cos(νπ) 2ξPν (ξ)Pν (−ξ) dξ = x − π( x − ξ) sin(νπ) (2ν + 1)π −1
with Eq. 57 for the implementation of the following Tricomi pairing: ¸ · Z1 Z1 Z1 2Pν (ξ)Pν (−ξ) sin(νπ) 2 2 xPν ( x)Pν (− x){[Pν ( x)] − [Pν (− x)] } d x = dξ d x xPν ( x)Pν (− x) P π( x − ξ) −1 −1 −1 ¸ Z1 · Z1 2ξPν (ξ)Pν (−ξ) sin(νπ) d ξ P ν ( x ) P ν (− x ) d x =− P π( x − ξ) −1 −1 Z1 Z 2 sin(2νπ) 1 =− xPν ( x)Pν (− x){[Pν ( x)]2 − [Pν (− x)]2 } d x + Pν ( x)Pν (− x) d x. (2ν + 1)π −1 −1 After rearrangement, we obtain 4
Z1
−1
x[Pν ( x)]3 Pν (− x) d x = 2
Z1
−1
xPν ( x)Pν (− x){[Pν ( x)]2 − [Pν (− x)]2 } d x =
2 sin(2νπ) (2ν + 1)π
Z1
−1
Pν ( x)Pν (− x) d x,
which entails the claimed identity in Eq. 62. The left-hand side of Eq. 62 extends to be a continuous function in ν ∈ C. Taking the ν → −1/2 limit, one arrives at Eq. 63. The special cases ν = −1/3, −1/4, −1/6 correspond to Eqs. 64, 65 and 66. A key step in the proof of Proposition 4.1 hinges on the identity Z1 h ³p Z1 ´i3 ´ ³p 2 1−k K d k = 3 [K(k)]2 K 1 − k2 d k, 0
Z1
[P−1/2 ( x)]3 dx =3 i.e. p −1 1+x
0
Z1
−1
P−1/2 ( x)[P−1/2 (− x)]2 d x. p 1+x
This result is actually just a special case within a family of identities satisfied by multiple elliptic integrals involving the product of three complete elliptic integrals (of the first and second kinds). To flesh out, for any integer n ∈ Z, we have Z1 Q (2 n+1)/2 ( x) P(2 n+1)/2(ξ) d ξ π P(2 n+1)/2(− x) = p P ≡ (−1)n+1 , ∀ x ∈ (−1, 1), (67) p p 2 −1 1+x 1+x 1 + ξ 2( x − ξ) which generalizes Corollary 3.2(b), and Eq. 67 entails Z1
[P(2 n+1)/2( x)]3 dx =3 p −1 1+x
Z1
−1
P(2 n+1)/2 ( x)[P(2 n+1)/2(− x)]2 dx p 1+x
(68)
after Tricomi pairing with Eq. 57. In particular, for n = 0, Eq. 68 specializes to
Z1 Z1 h ³p ´ ³p ´i3 ´ ³p h ³p ´i d k = 3 [2E(k) − K(k)]2 2E 1 − k2 − K 1 − k2 1 − k2 − K 1 − k2 d k. 2E 0
0
p p The identity stated in Eq. 67 is the Tricomi transform of P(2 n+1)/2( x)/ 1 + x ≡ P−(2 n+3)/2( x)/ 1 + x, n ∈ Z, which extends the Neumann integral representation of Legendre functions Q ℓ for ℓ ∈ Z≥0 (Eq. 60). In the next proposition, we shall prove Eq. 67 in an even broader context, which in turn, also gives rise to an independent verification of Eq. 57 in Proposition 4.2.
Proposition 4.4 (Generalized Neumann Integrals) For n ∈ Z≥0 and Re(ν − n) > −1, one has P
Z1
−1
(1 + ξ)ν−n Pν (ξ)
dξ = (1 + x)ν−n Q ν ( x), 2( x − ξ)
−1 < x < 1 .
(69)
Proof We note that there is a standard moment formula for Legendre functions [Ref. 13, item 7.127]: Z1
−1
(1 + x)σ Pν ( x) d x =
2σ+1 [Γ(σ + 1)]2 , Γ(σ + ν + 2)Γ(1 + σ − ν)
Re σ > −1.
One can verify Eq. 70 by termwise integration over the Taylor series expansion for Pν ( x) = 2 F1 to (1 − x)/2 and a reduction of the generalized hypergeometric series 3 F2 . 28
(70) ¡ −ν,ν+1 ¯ 1− x ¢ ¯ with respect 1
2
We now consider the Mellin inversion of Eq. 70: 1 2π i
Z c+ i ∞ c− i ∞
( Pν (ξ), −1 < ξ < 1 [Γ(s)]2 2s ds = s Γ(s + ν + 1)Γ(s − ν) (1 + ξ) 0, ξ>1
(71)
R c+ i ∞ R c+ iT where c > 0 and the integration c− i∞ := lim T →+∞ c− iT is carried out along a vertical line Re s = c. To facilitate analysis, we momentarily assume that Re(ν − n) > − 21 and pick c = Re(ν − n) + 21 , so that c > 0 and Re(ν − n − c) = − 21 . For −1 < x < 1, one may enlist Eq. 71 to compute Z1
1 (1 + ξ)ν−n Pν (ξ) d ξ P = 2( x − ξ ) 2 πi −1
ZRe(ν−n)+ 1 + i∞ 2
Re(ν− n)+ 21 − i∞
(1 + x)ν−n = 4i
[Γ(s)]2 2s Γ(s + ν + 1)Γ(s − ν)
ZRe(ν−n)+ 1 + i∞ 2
Re(ν− n)+ 21 − i∞
·
lim
Z∞ µ
ε→0+ −1
¶ ¸ 1 (1 + ξ)ν−n−s d ξ 1 + ds x − ξ + iε x − ξ − iε 4
[Γ(s)]2 2s cot(νπ − sπ) d s , Γ(s + ν + 1)Γ(s − ν) (1 + x)s
(72)
where the integration over ξ ∈ (−1, ∞) is an elementary exercise in complex analysis. From Eq. 72, it is straightforward to verify the Legendre differential equation: · ¸ Z1 Z1 d d (1 + ξ)ν−n Pν (ξ) d ξ (1 + ξ)ν−n Pν (ξ) d ξ (1 − x2 ) P + ν ( ν + 1) P ν − n ν− n ( x − ξ) dx dx ( x − ξ) −1 2(1 + x) −1 2(1 + x) (Z ) 1 Z 1 Re(ν− n)+ 2 + i∞ [Γ( s)]2 2 s cot(νπ − sπ) 2 s2 d s Re(ν− n)+ 2 + i∞ [Γ( s)]2 2 s cot(νπ − sπ) ( s + ν)( s − ν − 1) d s 1 − = 4 i Re(ν−n)+ 21 − i∞ Γ(s + ν + 1)Γ(s − ν) (1 + x)s+1 Γ(s + ν + 1)Γ(s − ν) (1 + x)s Re(ν− n)+ 21 − i∞ ÃZ ! Re(ν− n)+ 21 + i∞ ZRe(ν− n)− 21 + i∞ [Γ( s)]2 2 s cot(νπ − sπ) 2 s2 d s 1 = =0 − 4 i Re(ν−n)+ 21 − i∞ Γ(s + ν + 1)Γ(s − ν) (1 + x)s+1 Re(ν− n)− 21 − i∞ by noting that s = 0 and s = ν − n are both removable singularities for the integrand in the last line. Thus, we are sure that the left-hand side of Eq. 69 can be written as (1 + x)ν−n [aν,n Pν ( x) + bν,n Q ν ( x)] for some coefficients aν,n and bν,n , provided that Re(ν − n) > − 21 . Instead of directly coping with aν,n and bν,n for generic ν and n, we impose a temporary constraint that n = 0, − 21 < ν < 0. Again, by Eq. 72, we can compute two moment integrals Z1
−1
=
[aν,0 Pν ( x) + bν,0 Q ν ( x)] d x =
1 2π i
Z ν+ 1 + i ∞ 2
ν+ 21 − i∞
[Γ(s)]2 π cot(νπ − sπ) d s Γ(s + ν + 1)Γ(s − ν) 1 − s
∞ 2 νπ [Γ(m + ν)]2 cos(νπ) X + =− sin2 ; ν(ν + 1) m=1 (1 − m − ν)Γ( m)Γ( m + 2ν + 1) ν(ν + 1) 2 Z1 Z ν+ 1 + i ∞ 2 [Γ(s)]2 π cot(νπ − sπ) 2 d s 1 [aν,0 Pν ( x) + bν,0 Q ν ( x)](1 + x) d x = 2π i ν+ 12 − i∞ Γ(s + ν + 1)Γ(s − ν) 2 − s −1
=−
∞ X 2[ν2 + ν + cos(πν) − 1] 2[Γ(m + ν)]2 2 cos(νπ) + =− (ν − 1)ν(ν + 1)(ν + 2) m=1 (2 − m − ν)Γ(m)Γ(m + 2ν + 1) (ν − 1)ν(ν + 1)(ν + 2)
by closing the contours to the right. These two moment integrals reveal that aν,0 = 0, bν,0 = 1 for − 21 < ν < 0. For any fixed x ∈ (−1, 1), the integral ZRe ν+ 1 + i∞ 2
Re ν+ 12 − i∞
ds 1 [Γ(s)]2 2s cot(νπ − sπ) = P s Γ(s + ν + 1)Γ(s − ν) 4 i (1 + x) (1 + x)ν
Z1
(1 + ξ)ν Pν (ξ) d ξ 2( x − ξ) −1
is analytic for Re ν > − 21 and is equal to Q ν ( x) for − 12 < ν < 0, so it can be identified with Q ν ( x) for Re ν > − 21 . Moreover, so long as Re(ν − n) > − 12 for some n ∈ Z≥0 , we have ÃZ ! Re(ν− n)+ 21 + i∞ ZRe ν+ 12 + i∞ [Γ( s)]2 2 s cot(νπ − sπ) ds = 0, − 1 1 Γ(s + ν + 1)Γ(s − ν) (1 + x)s Re(ν− n)+ 2 − i∞ Re ν+ 2 − i∞
as the ratio cot(νπ − sπ)/Γ(s − ν) remains bounded at all the removable singularities enclosed in the contour. This allows us to simplify Eq. 72 into Eq. 69 for Re(ν − n) > − 21 , n ∈ Z≥0 . By analytic continuation in ν, one can extend the applicability to Re(ν − n) > −1, n ∈ Z≥0 . 29
Remark One may well recognize that Eq. 69 incorporates the classical result in Eq. 61 as a special case: P
Z1
−1
P n (ξ) p(ξ) d ξ = Q n ( x ) p ( x ), 2( x − ξ)
deg p( x) ≤ n ∈ Z≥0 .
To deduce Eq. 57 from Eq. 69, we need the Hardy-Poincaré-Bertrand formula (see [Ref. 31, §4.3, Eq. 4] or [Ref. 27, Eq. 11.52]): c[ f (T c g ) + g (T c f )] = (T c f )(T cg) − f g, T
which applies to the scenarios where f ∈ L p (−1, 1), p > 1; g ∈ L q (−1, 1), q > 1 and the following identities valid for −1 < ν < 0: (1 + x)−ν−1 Q −ν−1 ( x) = (1 + x)ν+1Q ν ( x) −
π
2
Z1
P
−1 Z1
2ν [Γ(ν + 1)]2 π = P Γ(2ν + 2) 2
−1
(73) 1 p
+
(1 + ξ)−ν−1 P−ν−1 (ξ) (1 + ξ)ν+1 Pν (ξ)
1 q
< 1. By Eqs. 69 and 70, we have
dξ π( x − ξ)
dξ
π( x − ξ)
.
,
(74) (75)
Taking the last pair of equations (Eqs. 74 and 75) as inputs for the Hardy-Poincaré-Bertrand formula (Eq. 73), we obtain the following identity for −1 < ν < 0 and −1 < x < 1: π
2
P
Z1
−1
Pν (ξ)[Q ν (ξ) + Q −ν−1 (ξ)]
dξ π( x − ξ)
= Q ν ( x)Q −ν−1 ( x).
According to the relation between Q ν and Pν (Eq. 3), the last equation reduces to the Tricomi transform of Pν (ξ)Pν (−ξ) (Eq. 57) upon analytic continuation in ν.
5 Discussion and Outlook In our evaluations of some generalized Clebsch-Gordan integrals (Eqs. 25-28), the expressions involving Euler’s gamma function actually certain special values of complete elliptic integrals as well. For example, p correspond to p the knowledge of K(1/ 2) = [Γ( 41 )]2 /(4 π) [Ref. 13, item 8.129.1] allows us to recast the formula [Γ( 14 )]8 /(128π2 ) = p R1 2 3 0 [K( 1 − k )] d k into the following form: 2
"Z 1 0
ds p p 2 1 − s 1 − (s2 /2)
#4
=
Z1 "Z1 0
0
p
dt
1 − t2
p 1 − (1 − k2 ) t2
#3
d k.
(76)
Here, both sides are integrations of algebraic functions over algebraic domains, which qualify them as members in the “ring of periods” defined by Kontsevich and Zagier [1]. It is generally believed that identities for periods can be proved by “algebraic means” [1], namely, relying on nothing else than additivity of the integral, algebraic change of variables, and the Newton-Leibniz-Stokes formula. However, the analytic proof we produced for the generalized Clebsch-Gordan integrals does not fall into such a category: we have invoked Bessel functions, which are “exponential periods” [1]. We would like to see purely algebraic evaluations of the generalized Clebsch-Gordan integrals when the choice of degree ν leads to special values of complete elliptic integrals. In particular, it could be interesting to search for potential connections to high degree modular equations and the Chowla-Selberg theory. After communicating the first version of this manuscript to Prof. Jonathan M. Borwein in January 2013, I was sent a preview of a forthcoming book [34] that contains a sketched proof for Eq. 76 using lattice sums and modular forms, totally independent of the methods presented in my work. Later this January, James G. Wan also wrote me about his plan to give a fuller account for the proofs of his own conjectures in a joint work with Rogers and Zucker, currently available as [35]. Some new integrals in [35] have inspired me to compose a short sequel [36] to the current work, in which there are further applications of the Hardy-Poincaré-Bertrand formula (Eq. 73) and the Tricomi transform of (1 + ξ)ν−n Pν (ξ), n ∈ Z≥0 , Re(ν − n) > −1 (Eq. 69), as well as an extension of the Hansen-Heine scaling analysis in Proposition 2.1 to the computations of other multiple elliptic integrals. The spherical methods in this work (Legendre functions and spherical rotations) enable us to handle a large variety of multiple elliptic integrals, but they are by no means a cure-all. The methods of Bessel moments [2, 4, 5], geometric transformations of Watson type [3], hypergeometric summations [6] still play fundamental rôles in our quantitative understandings for integrals over elliptic integrals.
30
With a synthesis of various techniques, it is sometimes possible to handle integrals over the product of more than three Legendre functions of the same degree ν ∈ C, such as Z1
−1
2 sin4 (π z)[ψ(2) ( z + 1) + ψ(2) (− z) + 28ζ(3)] , z →ν (2 z + 1)2 π4
x[Pν ( x)]4 d x = lim
where ψ(2) ( z) :=
d3 log Γ( z). d z3
(77)
R1 One may wish to compare Eq. 77 to the integrals −1 x[Pν ( x)]3 Pν (− x) d x, ν ∈ C (Eq. 62) that reduce to elementary functions. Clearly, special cases of Eq. 77 bring us some interesting evaluations of multiple elliptic integrals. For P −3 example, we have the following integral representations for Apéry’s constant ζ(3) = ∞ n =1 n : ζ(3) = −
π4
243
Z1
−1
x[P−1/3 ( x)]4 d x = −
as well as a critical scenario involving ζ(5) = π4
P∞
Z1
π4
168
Z1
−1
x[P−1/4 ( x)]4 d x = −
2π 4 189
Z1
−1
x[P−1/6 ( x)]4 d x
−5 n =1 n :
8 ζ(5) = − x[P−1/2 ( x)] d x = 372 −1 93 4
Z1 0
p (2 t − 1)[K( t)]4 d t.
Acknowledgements The author thanks two anonymous referees for their thoughtful comments that helped improve the presentation of this paper. This work was partly supported by the Applied Mathematics Program within the Department of Energy (DOE) Office of Advanced Scientific Computing Research (ASCR) as part of the Collaboratory on Mathematics for Mesoscopic Modeling of Materials (CM4). The author thanks Prof. Weinan E (Princeton University) for his encouragements.
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