Calculation of Coxeter' s integrals . Reduction to Ahmed integrals .
Attheendofthedocument
https://acrobat.adobe.com/id/urn:aaid:sc:EU:0009eeb4-7402-432b-97ed-75b8d955663c
8classicalCoxeterintegralsareshown.WeintendtocalculatethemusingMATHEMATICA(version14.0).TodothiswewillfirstreducethemtoAhmedintegrals.Usingthedefinitionsandresultscontainedintheaforementioneddocument,wehavedocument,wehave:■
ACoxeter'sintegralisanintegralofdeformArcTanθwiththeconditiona≥bandb+aCos[θ]>0for0<θ<α.
α
∫
0
1+Cos[θ]
b+aCos[θ]
■
Wedefineforanyp,q,r>0,thefollowingfunction:
A0[p_,q_,r_]:=pqr+(1+(1+))xFullSimplify
1
∫
0
ArcTan+
2
q
2
p
2
q
2
x
2
q
2
p
2
q
2
x
2
p
2
r
2
x
■
IfFurthermorep*q*r=1,wedefineA1[p_,q_,r_]:=+(1+(1+))x->+--
1
∫
0
ArcTan+
2
q
2
p
2
q
2
x
2
q
2
p
2
q
2
x
2
p
2
r
2
x
2
π
8
1
2
2
ArcTan1
1+
r2
q
2
ArcTanp
1+
2
r
2
ArcTan
1+
q2
p
■
Inthespecialcasewherep*q=1andr=1wedefineAhmed[α_]:=Integrate,{x,0,1}ArcTan-
ArcTan
α+
2
x
α+
1+2
x
2
2
x
α
π
2
1
1+α
1
2
2
ArcTan
2
α
■
A1[∞,q_,r_]:=ArcTan+ArcTan[r]-ArcTan+1//FullSimplify
π
2
1
r+1
2
q
2
r
qr
ThefollowingareverifiedAhmed[α]->A0,
1
α
α
,1■
Inaddition,forA<1,youhave
ArcCos[A]=2ArcCos=2ArcTan.
A+1
2
1-A
1+A
■
Sia≥bandb+aCos[θ]>0for0<θ<α,ismet:ArcTanθ=2A0*,,.
α
∫
0
1+Cos[θ]
b+aCos[θ]
a-b
2
1-Cos[α]
b+aCos[α]
2
a+b
a+b
a-b
Finally,wecalculatethe8classicalCoxeterintegrals:
●
ForI1andI2:ArcCosθ=2ArcTanθ=4A0,,
α
∫
0
Cos[θ]
1+2Cos[θ]
α
∫
0
Cos[θ]+1
3Cos[θ]+1
1-Cos[α]
1+3Cos[α]
1
2
2
◆
I1=ArcCosθ->4A11,,
π
2
∫
0
Cos[θ]
1+2Cos[θ]
1
2
2
->5
2
π
24
◆
I2=ArcCosθ->4A0,,(*)
π
3
∫
0
Cos[θ]
1+2Cos[θ]
1
5
1
2
2
=2
2
π
15
●
ForI3andI4:ArcCosθ=2ArcTanθ=απ-A1,
α
∫
0
1
1+2Cos[θ]
α
∫
0
Cos[θ]
1+Cos[θ]
1-Cos[α]
2Cos[α]
2
,1◆
I3=ArcCosθ=2ArcTanθ=-4A1[Infinity,
π
2
∫
0
1
1+2Cos[θ]
π
2
∫
0
Cos[θ]
1+Cos[θ]
2
π
2
2
,1]=2
π
6
◆
I4=ArcCosθ=2ArcTanθ=-4A1,
π
3
∫
0
1
1+2Cos[θ]
π
3
∫
0
Cos[θ]
1+Cos[θ]
2
π
3
1
2
2
,1=2
π
8
●
ForI5eI6:ArcCosθ=2ArcTanθ=πα-4A1,1,
α
∫
0
1-Cos[θ]
2Cos[θ]
α
∫
0
3Cos[θ]-1
Cos[θ]+1
2(1-Cos[α])
3Cos[α]-1
1
2
◆
I5=ArcCosθ=2ArcTanθ=πArcCos-4A1Infinity,1,=
ArcCos
1
3
∫
0
1-Cos[θ]
2Cos[θ]
ArcCos
1
3
∫
0
3Cos[θ]-1
Cos[θ]+1
1
3
1
2
2
π
6
◆
I6=ArcCosθ=2ArcTanθ=-4A1=
π
3
∫
0
1-Cos[θ]
2Cos[θ]
π
3
∫
0
3Cos[θ]-1
Cos[θ]+1
2
π
3
2
,1,1
2
11
2
π
72
●
ForI7eI8:ArcCosθ=ArcaTanθ=2A1,
α
∫
0
Cos[θ]
2Cos[θ]+1
α
∫
0
1+Cos[θ]
Cos[θ]
1-Cos[α]
2Cos[θ]
2
,1◆
I7=ArcCosθ=2A1[Infinity,
π
2
∫
0
Cos[θ]
2Cos[θ]+1
2
,1]=2
π
6
◆
I8=ArcCosθ=ArcaTanθ=2A1,
π
3
∫
0
Cos[θ]
2Cos[θ]+1
π
3
∫
0
1+Cos[θ]
Cos[θ]
1
2
2
,1=5
2
π
48
(*) All except I2 are easy to calculate.
The value of I2 and many more Coxeter integrals are
are given in the document
are given in the document
I hope that in the next version of MATHEMATICA this type of integrals will be incorporated and can be calculated formally .
Thank you .
I hope that in the next version of MATHEMATICA this type of integrals will be incorporated and can be calculated formally .
Thank you .