COXETER’S INTEGRALS SOS440

Abstract. In this paper, we generalize the Ahmed’s integral and exhibit some identities involved in it.

Contents 1.

Generalized Ahmed integrals

1

1.1.

Notations

1

1.2.

Identities involving Ahmed integrals

2

2.

Calculation of Coxeter’s integrals

5

2.1.

Reduction of Coxeter’s integral to Ahmed integral

5

2.2.

Calculation of Classical Coxeter’s Integrals

6

References

8

1. Generalized Ahmed integrals 1.1. Notations. In this section, we introduce some notations that will be used throughout this paper. The function Bpxq is defined for x on p´1, 8q by ż1 dt Bpxq “ 2 0 xt ` 1

(1.1)

This function is analytic, with the series expansion at x “ 0 given by Bpxq “

8 ÿ p´1qn n x . 2n ` 1 n“0

Simple case examination shows that Bpxq satisfies ? $ arctan x ’ ’ ? px ą 0q ’ ’ x ’ & Bpxq “ 1 px “ 0q ’ ? ’ ’ ’ arctanh ´x ’ % ? p´1 ă x ă 0q. ´x 1

(1.2)

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Among various relations which Bpxq satisfies, ˆ ˙ 1 π 1 “ ? ´ Bpxq, B x x 2 x

px ą 0q

and 1 Bp´1 ` xq “ ´ log x ` log 2 ` Opxq as x Ñ 0` 2 are particularly worth to notice. Then the Ahmed integrals App, q, rq are defined by ż1 Bpqppx2 ` 1qq def App, q, rq “ dx, 2 0 pr ` 1qpx ` 1 ˜ and similarly the associated Ahmed integrals App, q, rq are defined by ˆ ˙ pq def ˜ App, q, rq “ App, q, rq ` B prpq ` 1qq B q`1 c b pq 1 “ App, q, rq ` ? arctan rpq ` 1q ¨ arctan . pqr q`1

(1.3)

(1.4)

These integrals reduce to the classical Ahmed integrals for special choices of parameters, as we can see ? ˆ ˙ ż1 1 tan´1 x2 ` α ? dx “ A , α, α ´ 1 , (1.5) α 0 px2 ` 1q x2 ` α which holds for α ą 1. The definition (1.3) and (1.4) seem very bizarre and visually unappealing at first glance. But this definition is designed to capture some symmetries innate in the Ahmed integrals, as we will see. 1.2. Identities involving Ahmed integrals. We first show the following identity. Theorem 1.6. For p, q, r ą 0, we have ˜ ˜ r, pq “ Apr, ˜ p, qq. App, q, rq “ Apq,

(1.7)

Proof. We perform a direct calculation. By partial fraction, we have ˆż 1 ˙ ż1 1 dy App, q, rq “ dx 2 2 2 0 pr ` 1qpx ` 1 0 qppx ` 1qy ` 1 ż1ż1 dxdy “ 2 2 2 0 0 ppr ` 1qpx ` 1q pqppx ` 1qy ` 1q ˙ ˆ ż1ż1 1 qy2 1`r “ ´ dxdy. 2 p1 ` rqpx2 ` 1 pqy2 x2 ` qy2 ` 1 0 0 qry ` r ` 1 Then by definition (1.1) of B, the former fraction becomes ˆ ˙ ż1ż1 1`r qr dxdy “ B pppr ` 1qq B . 2 2 r`1 0 0 pqry ` r ` 1q pp1 ` rqpx ` 1q

COXETER’S INTEGRALS

3

Similarly, the latter becomes ˆ ˙ ż1 qy2 pqy2 qy2 dx “ 2 B 2 2 2 qy ` 1 qy2 ` 1 0 pqy x ` qy ` 1 d c q y pqy2 a “ arctan . 2 p qy ` 1 qy2 ` 1 by (1.2). So we have ˆ App, q, rq “ B pppr ` 1qq B

qr r`1

˙

d c ż1 q pqy2 y a ´ arctan dy, p 0 pqry2 ` r ` 1q qy2 ` 1 qy2 ` 1 and it remains to evaluate the last integral. But it is easy to observe that „  ? b q ry d 2 a arctan rpqy ` 1q “ dy pqry2 ` r ` 1q qy2 ` 1 thus we obtain d c ż1 q y pqy2 a arctan dy p 0 pqry2 ` r ` 1q qy2 ` 1 qy2 ` 1 d « ff1 b 1 pqy2 2 arctan rpqy ` 1q arctan “ ? pqr qy2 ` 1 0 a ż1 2 arctan rpqy ` 1q a ´ dy r pqy2 ` 1q 0 pp1 ` pqqy2 ` 1q ˆ ˙ pq “ B prpq ` 1qq B ´ Apq, r, pq, q`1 from which we deduce (1.7) as desired.



This theorem allows us to transform one Ahmed integral to another one, and in some limiting case we can obtain a meaningful results. For instance, the limiting condition r Ñ 0 applied to the first part of the identity (1.7), together with Apq, 0, pq “ B pq pp ` 1qq, immediately yields ˙ ˆ pq , App, q, 0q “ B pqpp ` 1qq ` Bppq ´ B q`1 and especially for p “ q “ 1, (1.5) gives ? ? ż1 ? tan´1 x2 ` 1 3 2 π ? ? dx “ Ap1, 1, 0q “ arctan 2 ´ p2 2 ´ 1q. 2 2 2 4 0 px ` 1q x ` 1 But in general, (1.7) alone gives no specific information about the actual value. So we need the following identity.

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Theorem 1.8. For a, b, c, d ą 0, we have c ˙ c ˙ ˆ ˆ ad bc a b`d d c b`d b ` A , , A , , bc b`d c b ad b`d a d d d ˜ ¸ c c π ad bc a c “ arctan ` arctan ´ arctan ¨ arctan . 2 b d bpa ` b ` dq dpb ` c ` dq (1.9) Proof. We begin from the expression ? c c ż1ż1 a c abcd dxdy. I “ arctan ¨ arctan “ 2 2 b d 0 0 pax ` bqpcy ` dq By exploiting the partial fraction decomposition 1 A`B 1 1 “ “ ` , AB ABpA ` Bq ApA ` Bq BpA ` Bq we have

?

ż1ż1 I“

abcd dxdy ` ` cy2 ` b ` dq 0 0 ? ż1ż1 abcd ` dxdy 2 ` dqpax2 ` cy2 ` b ` dq pcy 0 0 pax2

bqpax2

def

“ I1 ` I2

By symmetry, it suffices to evaluate I1 . By simple calculation, ? ż1ż1 abcd I1 “ dydx 2 2 2 0 0 pax ` bqpax ` cy ` b ` dq ? ˆ ˙ ż1 abcd c “ B dx 2 2 ax2 ` b ` d 0 pax ` bqpax ` b ` dq ˆ ˆ 2 ˙˙ ? ż1 ? c abcd π ax ` b ` d ? “ ´B dx 2 2 ax2 ` b ` d c 0 cpax ` bq ˜ c ? ˆ 2 ˙¸ ż1 abd ad ax ` b ` d π 1 ? “ ´ B dx 2 pax2 ` bq ax2 ` b ` d bc pa{bqx2 ` 1 c 0 c ? ˆ ˙ ż abd ad b`d d π 1 a ? dx ´ “ A , , . 2 0 pax2 ` bq ax2 ` b ` d bc b`d c b Then the integral in the last line is evaluated as follows: By the substitution x ÞÑ t´1 , ? ? ż ż π 1 π 8 abd abd t ? a dx “ dt 2 0 pax2 ` bq ax2 ` b ` d 2 1 pbt2 ` aq pb ` dqt2 ` a « ff8 c π b ppb ` dqt2 ` aq “ arctan 2 ad 1 d π ad “ arctan . 2 bpa ` b ` dq

COXETER’S INTEGRALS

5

Therefore we have π I1 “ arctan 2

d

π I2 “ arctan 2

d

and likewise

ad ´A bpa ` b ` dq

ˆ

bc ´A dpb ` c ` dq

ˆ

a b ` d ad , , c c bc

˙

c b ` d bc , , a a ad

˙

,

.

Combining these results yields (1.9).



An asymmetric form of (1.9) is given by ˆ ˙ ? 1 1 1 1 pqr App, q, rq ` ? A , , pqr q p r d d ˜ ¸ c b 1 pr r`1 π arctan ´ arctan ppr ` 1q ¨ arctan “ ` arctan . 2 p`1 qr rpq ` 1q (1.10) As a special case of (1.10) with pq “ 1 and r “ 1, we have ˜ c ¸2 c ˙ ˆ π 1 1 1 2 arctan , , α, 1 “ arctan ´ A α 2 α`1 2 α hence in conjunction with (1.5), we obtain ? ˆ ˙ ż1 tan´1 x2 ` 2 1 5π2 ? dx “ A , 2, 1 “ , 2 96 0 px2 ` 1q x2 ` 2

(1.11)

which is the original result of Ahmed in [Ahm02]. 2. Calculation of Coxeter’s integrals 2.1. Reduction of Coxeter’s integral to Ahmed integral. A Coxeter’s integral is an integral of the form c ż θ0 cos θ ` 1 arctan dθ. (2.1) a cos θ ` b 0 We impose additional condition a ą |b| and min0ďθďθ0 pa cos θ ` bq ě 0. Our aim in this chapter is to show that any Coxeter’s integral is reduced to an Ahmed integral. We make use of simple change of variable to obtain d c ż θ0 ż θ0 2 cos2 pθ{2q cos θ ` 1 arctan dθ “ arctan dθ a cos θ ` b a ` b ´ 2a sin2 pθ{2q 0 0 ˜ ¸ ż θ0 {2 cos θ “2 arctan a dθ, 0 pa ` bq{2 ´ a sin2 θ where the substitution θ ÞÑ 2θ is used in the last line. This shows that it suffices to consider the integral of the form ˜ ¸ żφ cos θ a Ipα, β, φq “ arctan dθ, (2.2) 0 α β2 ´ sin2 θ

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where α ą 0, φ P r0, π{2s and sin φ ă β ă 1. With this notation we readily identify (2.1) as ˜ ¸ c c ż θ0 ? cos θ ` 1 a ` b θ0 arctan dθ “ 2I a, , . a cos θ ` b 2a 2 0 Then the substitution sin θ ÞÑ β sin θ with φ˜ “ arcsinpsin φ{βq, followed by the substitution tan θ ÞÑ t tan φ˜ gives Ipα, β, φq ¸˜ ¸ ˜a ż φ˜ β cos θ 1 ´ β2 sin2 θ a “ dθ arctan αβ cos θ 0 1 ´ β2 sin2 θ ¸ ˜a ¸˜ ż1 1 ` t2 p1 ´ β2 q tan2 φ˜ tan φ˜ dt β a . “ arctan αβ 1 ` t2 tan2 φ˜ 0 1 ` t2 p1 ´ β2 q tan2 φ˜ Then by comparison with the definition (1.3), it follows that ˆ ˙ 2 tan φ˜ ˜ 1 , β Ipα, β, φq “ A p1 ´ β2 q tan2 φ, , α α2 β2 1 ´ β2 hence after some simple algebraic manipulations, we obtain d c ˙ ˆ ż θ0 cos θ ` 1 1 ´ cos θ0 a ´ b 1 ´ cos θ0 2 a`b . arctan dθ “ 2 A , , a cos θ ` b a cos θ0 ` b 2 a cos θ0 ` b a ` b a ´ b 0 (2.3) 2.2. Calculation of Classical Coxeter’s Integrals. Now let’s consider the following classical Coxeter’s integrals: ˆ ˆ ˙ ˙ żπ żπ 2 3 cos θ cos θ I1 “ arccos arccos dθ, I2 “ dθ, ˆ 1 ` 2 cos θ ˙ ˆ 1 ` 2 cos θ ˙ ż0π ż0π 2 3 1 1 arccos arccos I3 “ dθ, I4 “ dθ, 1 ` 2 cos θ 1 ` 2 cos θ 0 0 ˆ ˙ ˆ ˙ żπ ż arccosp 1 q 3 3 1 ´ cos θ 1 ´ cos θ I5 “ arccos dθ, I6 “ arccos dθ, 2 cos θ c c 2 cos θ ż0π ż0π 2 3 cos θ cos θ I7 “ arccos dθ, I8 “ arccos dθ. 1 ` 2 cos θ 1 ` 2 cos θ 0 0 By the half-angle formula, for |A| ă 1, c arccos A “ 2 arccos Thus we obtain ż θ0

ˆ arccos

0

cos θ 1 ` 2 cos θ

A`1 “ 2 arctan 2 ż θ0

˙ dθ “ 2

c

c

3 cos θ ` 1 dθ 4 cos θ ` 2

c

cos θ ` 1 dθ, 3 cos θ ` 1

arccos 0

ż θ0 “2

arctan 0

1´A . 1`A

COXETER’S INTEGRALS

and likewise ż θ0

ˆ arccos

0

1 1 ` 2 cos θ

ż θ0

˙ dθ “ 2

c arccos

0

ż θ0

7

cos θ ` 1 dθ 2 cos θ ` 1

c

cos θ dθ arctan cos θ`1 0 ¸ ˜ c ż θ0 cos θ π ´ arctan “2 dθ 2 cos θ ` 1 0 c ż θ0 cos θ ` 1 dθ, “ πθ0 ´ 2 arctan cos θ 0 c ˙ ˆ ż θ0 ż θ0 cos θ ` 1 1 ´ cos θ dθ “ 2 arccos arccos dθ 2 cos θ 4 cos θ 0 0 ¸ ˜ c ż θ0 3 cos θ ´ 1 π ´ arccos dθ “2 2 4 cos θ 0 c ż θ0 cos θ ` 1 dθ, “ πθ0 ´ 2 arctan 3 cos θ ´ 1 0 c c ż θ0 ż θ0 cos θ cos θ ` 1 arccos dθ “ arctan dθ, 2 cos θ ` 1 2 cos θ ` 1 0 0 “2

So that ˆ ˙ ˙ 1 1 4 1 I2 “ ? A , ,2 , I1 “ 4 A 1, , 2 , 2 5ˆ 2 5 ˙ 2 2 ? π π 1 I3 “ ´ 4 28 A p8, 2, 1q , I4 “ ´ 4A , 2, 1 , 2 3 ˆ ˙ ˆ2 ˙ 2 ? 1 1 π 1 I5 “ 2π arctan ? ´ 2 28 A 8, 1, , I6 “ ´ 4 A 2, 1, , 2 3ˆ 2˙ ˙ 2 ˆ 1 2 1 2 , ,3 , I8 “ A , ,3 . I7 “ 2 A 2 3 8 3 ? ? Here the notation 8 Ap8, q, rq is understood as the limit case of p App, q, rq, hence by (1.10) with p Ñ 8, d d ˜ ¸ ? ? π 1 r`1 8 Ap8, q, rq “ ? arctan ` arctan r ´ arctan . (2.4) 2 qr qr rpq ` 1q ˆ

Hence I3 and I5 are given by I3 “ I5 “ Also, with aid of (1.7) and (1.11), we obtain ˆ ˙ 1 13π2 A 2, 1, “ , 2 288

π2 . 6

ˆ ˙ 1 5π2 A 1, , 2 “ 2 96

and we obtain I1 “

5π2 , 24

I4 “

π2 , 8

I6 “

11π2 . 72

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References [Ahm02] Zafar Ahmed. Definitely an integral. American Mathematical Monthly, 109:670–671, 2002.