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2 SOS440
Among various relations which Bpxqsatisfies,1
xB1
x 2 ?x Bpxq, px0q
and
Bp1 xq 12
log x log 2 Opxq asx0`
are particularly worth to notice.
Then theAhmed integralsApp, q, rqare defined by
App, q, rq def1
0
Bpqppx2 `1qqpr 1qpx2 `1 dx, (1.3)
and similarly theassociated Ahmed integrals App, q, rqare
defined by
App, q, rq def App, q, rq B prpq 1qqB
pqq 1
App, q, rq 1?pqr
arctan
brpq 1q arctan
c pq
q 1 .(1.4)
These integrals reduce to the classical Ahmed integrals for
special choices of parameters,
as we can see 10
tan1?
x2 `px2 `1q
?x2 ` dxA
1
,, 1
, (1.5)
which holds for 1. The definition (1.3) and (1.4) seem very
bizarre and visuallyunappealing 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, r0, we haveApp, q, rq Apq, r,pq Apr,p,
qq. (1.7)
Proof. We perform a direct calculation. By partial fraction, we
have
App, q, rq 1
0
1
pr 1qpx2 `11
0
dy
qppx2 `1qy2 `1
dx
1
01
0
dxdy
ppr 1qpx2 `1q pqppx2 `1qy2 `1q1
0
10
1
qry2 `r 1
1 rp1 rqpx2 `1
qy2
pqy2x2 `qy2 `1
dxdy.
Then by definition (1.1) ofB, the former fraction becomes10
10
1 rpqry2 `r 1q pp1 rqpx2 `1q dxdyB pppr 1qqB
qr
r 1
.
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COXETERS INTEGRALS 3
Similarly, the latter becomes
10
qy2
pqy2x2 `qy2 `1 dx qy2
qy2 `1B pqy2
qy2 `1c
q
p
yaqy2 `1
arctan
d pqy2
qy2 `1 .
by (1.2). So we have
App, q, rq B pppr 1qqB
qr
r 1
c
q
p
10
y
pqry2 `r 1qa
qy2 `1arctan
d pqy2
qy2 `1 dy,
and it remains to evaluate the last integral. But it is easy to
observe that
d
dy
arctan
brpqy2 `1q
q
?r y
pqry2 `r 1qa
qy2 `1thus we obtain
cq
p
10
y
pqry2 `r 1qa
qy2 `1arctan
d pqy2
qy2 `1 dy
1?pqr
arctan
brpqy2 `1q arctan
d pqy2
qy2 `1
ff10
1
0
arctan arpqy2 `1qpp1 pqqy2 `1q arpqy2 `1q dy B prpq 1qqB
pq
q 1
Apq, r,pq,
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 r0applied to the first part of the identity
(1.7), together with Apq, 0,pq B pq pp 1qq,immediately yields
A
pp, q, 0
q B
pq
pp
1
qqB
pp
q B pq
q 1 ,and especially for pq1, (1.5) gives1
0
tan1?
x2 `1px2 `1q
?x2 `1 dx
Ap1, 1, 0q 3?
2
2 arctan
?2
4p2
?2 1q.
But in general, (1.7) alone gives no specific information about
the actual value. So we need
the following identity.
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4 SOS440
Theorem 1.8. For a, b, c, d0, we have
cadbc A
ab d,
b
d
c ,
d
b` c
bc
ad A c
b d,b
d
a ,
b
d
2
arctan
d ad
bpa b dq`arctand
bc
dpb c dq
arctan
ca
barctan
cc
d.
(1.9)
Proof. We begin from the expression
Iarctanc
a
barctan
cc
d1
0
10
?abcd
pax2 `bqpcy2 `dq dxdy.
By exploiting the partial fraction decomposition
1
AB A B
ABpA Bq 1
ApA Bq` 1
BpA Bq,
we have
I1
0
10
?abcd
pax2 `bqpax2 `cy2 `b dq dxdy
`1
0
10
?abcd
pcy2 `dqpax2 `cy2 `b dq dxdydef I1 I2
By symmetry, it suffices to evaluateI1. By simple
calculation,
I1 1
0 1
0
?abcd
pax2
`b
qpax2
`cy2
`b
d
q
dydx
1
0
?abcdpax2 `bqpax2 `b dqB
c
ax2 `b d
dx
1
0
?abcd
cpax2 `bq
2
?c?
ax2 `b d B
ax2 `b dc
dx
1
0
2
?abd
pax2 `bq?ax2 `b d c
ad
bc
1
pa{bqx2 `1B
ax2 `b dc
dx
2
10
?abd
pax2 `bq?
ax2 `b d dxc
ad
bcA
a
b d,b d
c,
d
b
.
Then the integral in the last line is evaluated as follows: By
the substitution xt1,
2
10
?abdpax2 `bq?ax2 `b d dx 28
1?abd tpbt2 `aqa
pb dqt2 `a dt
2
arctan
c b
adppb dqt2 `aq
ff81
2
arctan
d ad
bpa b dq .
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COXETERS INTEGRALS 5
Therefore we have
I1
2arctan d ad
bpa b dqAa
c,
b
d
c,
ad
bc ,
and likewise
I2 2
arctan
d bc
dpb c dqA
c
a,
b da
,bc
ad
.
Combining these results yields (1.9).
An asymmetric form of (1.9) is given by
?pqrApp, q, rq 1?
pqrA
1
q,
1
p,
1
r
2arctan d
1
rpq 1q`arctan c
pr
p 1arctan bpp
r
1
q arctan d
r 1
qr
.
(1.10)
As a special case of (1.10) with pq1 andr1, we have
A
1
, , 1
2arctan
c 1
11
2
arctan
c2
2,
hence in conjunction with(1.5), we obtain10
tan1?
x2 `2px2 `1q
?x2 `2 dx
A
1
2, 2, 1
5
2
96 , (1.11)
which is the original result of Ahmed in [Ahm02].
2. Calculation ofCoxeters integrals
2.1. Reduction of Coxeters integral to Ahmed integral. A
Coxeters integral is an in-
tegral of the form 00
arctan
c cos 1a cos b d. (2.1)
We impose additional condition a |b|and min00pa cos bq 0. Our
aim in thischapter is to show that any Coxeters integral is reduced
to an Ahmed integral.
We make use of simple change of variable to obtain00
arctan
c cos 1a cos b d
00
arctan
d 2cos2p{2q
a b 2a sin2p{2q d
20{2
0
arctan
cos a
pa bq{2 a sin2
d,
where the substitution2is used in the last line. This shows that
it suffices to considerthe integral of the form
Ip,,q
0
arctan
cos
a2 sin2
d, (2.2)
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6 SOS440
where0, P r0, {2sand sin 1. With this notation we readily
identify (2.1)as
00
arctanc
cos 1a cos b d2I
?a, ca b2a
, 02
.
Then the substitution sin sin witharcsinpsin {q, followed by the
substitutiontan ttangives
Ip,,q
0
arctan
a1 2 sin2 cos
cos a
1 2 sin2 d
1
0
arctan
a1 t2p1 2q tan2
a1 t2p1 2q tan2
tandt
1 t2 tan2 .
Then by comparison with the definition(1.3), it follows that
Ip,,q tan
A
p1 2q tan2 , 1
22,
2
1 2
,
hence after some simple algebraic manipulations, we obtain
00
arctan
c cos 1a cos b d2
d 1 cos 0a cos 0 b A
a b
2
1 cos 0a cos 0 b ,
2
a b ,a ba b
.
(2.3)
2.2. Calculation of Classical Coxeters Integrals. Now lets
consider the following clas-
sical Coxeters integrals:
I1
2
0
arccos
cos
1 2 cos
d, I2
3
0
arccos
cos
1 2 cos
d,
I3
2
0
arccos
1
1 2 cos
d, I4
3
0
arccos
1
1 2 cos
d,
I5 arccosp 13 q
0
arccos
1 cos
2cos
d, I6
3
0
arccos
1 cos
2cos
d,
I7
2
0
arccos
c cos
1 2 cos d, I8
3
0
arccos
c cos
1 2 cos d.
By the half-angle formula, for|A| 1,
arccosA2 arccos cA
1
2 2 arctan c1
A
1 A .Thus we obtain0
0
arccos
cos
1 2 cos
d20
0
arccos
c3cos 14cos 2 d
20
0
arctan
c cos 13cos 1 d,
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COXETERS INTEGRALS 7
and likewise
00
arccos 11 2 cos
d
2 00
arccos c cos 1
2cos 1 d
20
0
arctan
c cos
cos 1 d
20
0
2arctan
c cos
cos 1
d
0 20
0
arctan
ccos 1
cos d,
00
arccos
1 cos
2cos
d2
00
arccos
ccos 1
4cos d
2 0
0
2arccos c3cos
1
4cos
d
0 20
0
arctan
c cos 13cos 1 d,0
0
arccos
c cos
2cos 1 d0
0
arctan
c cos 12cos 1 d,
So that
I1 4 A
1,1
2, 2
, I2 4?
5A
1
5,
1
2, 2
,
I3 2
24
?28 A p8, 2, 1q , I4
2
34 A
1
2, 2, 1
,
I5 2 arctan 1?2 2 ?28 A
8, 1, 12 , I6 2
34 A
2, 1,1
2
,
I7 2 A
1
2,
2
3, 3
, I8 A
1
8,
2
3, 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 8,?8 Ap8, q, rq
2?
qr
arctan
d 1
rpq 1q`arctan?
r arctand
r 1qr
. (2.4)
HenceI3 and I5 are given by
I3I5 2
6 .
Also, with aid of (1.7) and(1.11), we obtain
A
2, 1,
1
2
13
2
288 , A
1,
1
2, 2
5
2
96
and we obtain
I1 52
24, I4
2
8, I6 11
2
72 .
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References
[Ahm02] Zafar Ahmed. Definitely an integral.American
Mathematical Monthly, 109:670671, 2002.
