CalculationofaCoxeterintegral​​​​
MATHEMATICAnocalculaformalmentelasiguienteintegraldeCoxeter:
In[]:=
IntegrateArcCos
Cos[w]
1+2Cos[w]
,{w,0,Pi/3}
Out[]=
π
3
∫
0
ArcCos
Cos[w]
1+2Cos[w]
w
Usandoalgunosresultadoscontenidosenelartículo​​
"Some properties on generalized Ahmed’s integralSangchul LeeDecember 1,2013"​​
setiene:​​
IntegrateArcCos
Cos[w]
1+2Cos[w]
,{w,0,Pi/3}==2IntegrateArcTan
1+Cos[x]
1+3Cos[x]
,{x,0,Pi/3}==
4
5
Integrate
ArcTan
1
2
+
2
x
10

1
2
+
2
x
10
1+
3
2
x
5
,{x,0,1}==
2
2
π
15
SetratadeevaluarlaúltimaintegralusandoMATHEMATICA.Paraellodefinimos:
In[]:=
​​myintegral=
4
5
integrate
ArcTan
1
2
+
2
x
10

1
2
+
2
x
10
1+
3
2
x
5
,{x,0,1};
In[]:=
​​A0[p_,q_,r_]:=pqrintegrate
ArcTan
2
q
+
2
p
2
q
2
x

2
q
+
2
p
2
q
2
x
(1+
2
p
(1+
2
r
)
2
x
)
,{x,0,1}
In[]:=
L[z_]:=Sum
2n+1
z
2
(2n+1)
,{n,0,Infinity}
In[]:=
L[z]
Out[]=
1
4
zLerchPhi
2
z
,2,
1
2

In[]:=
AA[p_,q_,r_]:=A0[p,q,r]+ArcTanr
2
q
+1
*ArcTan
pq
1+
2
q
,
2
π
4
+2L
pqr-1
pqr+1
-integrate
pqrArcTan[t]
2
p
2
q
2
r
+
2
t
,t,0,
1+
2
p
q-integrate
pqrArcTan[t]
2
p
2
q
2
r
+
2
t
,t,0,
1+
2
q
r-integrate
pqrArcTan[t]
2
p
2
q
2
r
+
2
t
,t,0,p
1+
2
r

In[]:=
myintegral
Out[]=
4integrate
ArcTan
1
2
+
2
x
10

1
2
+
2
x
10
1+
3
2
x
5
,{x,0,1}
5
Ennuestrocaso:
In[]:=
AA
1
5
,
1
2
,
2

Out[]=

1
3
πArcTan
1
15
+
integrate
ArcTan
1
2
+
2
x
10

1
2
+
2
x
10
1+
3
2
x
5
,{x,0,1}
5
,
2
π
4
-2integrate
ArcTan[t]
5
1
5
+
2
t
,t,0,
3
5
-integrate
ArcTan[t]
5
1
5
+
2
t
,t,0,
3
+
-1+
1
5
LerchPhi
7
2
-
3
5
2
,2,
1
2

21+
1
5

In[]:=
{myintegral,resultado}=
4
1
∫
0
ArcTan
1
2
+
2
x
10

1
2
+
2
x
10
1+
3
2
x
5
x
5
,
2
π
-8Integrate
ArcTan[t]
5
1
5
+
2
t
,t,0,
3
5
-4Integrate
ArcTan[t]
5
1
5
+
2
t
,t,0,
3
+4
-1+
1
5
LerchPhi
7
2
-
3
5
2
,2,
1
2

21+
1
5
-
4
3
πArcTan
1
15
//FullSimplify
Out[]=

4
1
∫
0
5
10
ArcTan
1
2
+
2
x
10

5+
2
x
5+3
2
x

x
5
,
2
9
π(11π-6ArcCot[
15
]-9ArcTan[
15
])+(-3+
5
)LerchPhi
7
2
-
3
5
2
,2,
1
2
+
2
Log[2]
+ArcCsch[2](-ArcCsch[2]+Log[4])-Log[16]Log[40]+8Log[2]Log[5-
5
]+3
2
Log[-1+
5
]
+6PolyLog2,
1
1-
5
+2PolyLog2,-
1
2
1/3
(-1)
(-1+
5
)+2PolyLog2,
1
2
2/3
(-1)
(-1+
5
)-6PolyLog2,
1
1+
5
+2PolyLog2,
1-
3
1+
5
+2PolyLog2,
1+
3
1+
5
+PolyLog2,
(-1+
5
)
-+
15
-PolyLog2,-
(1+
5
)
-+
15
+PolyLog2,-
(-1+
5
)
+
15
-PolyLog2,
(1+
5
)
+
15

In[]:=
Nmyintegral,resultado,
2
2
π
15
//Chop
Out[]=
{1.31595,1.31595,1.31595}
​​
Miproblemaymipreguntaessisepuedesimplificarmásel"resultado"hastallegaralvalorreal
2
2
π
15
.
​
​