if you only want to see the final result just scroll to the very end and only run the subchapter "Final analytic expression for QNEC"
and look at the three input expressions in red *)
Definitions
◼ Basic defini�ons
In[2]:= (*line element in 1805.08782*)
ds2= -(r^2+G1[r]^2)dt^2+dr^2(r^2+G2[r]^2) +r^2 dphi^2;
G1[r_]:=1-8 GN h;
G2[r_]:=1-8 GN h1-1(r^2+1)^2 h-1;
In[5]:= (* transformation to z=1/r *)
rTOz=r→1/z, dr→-1/z^2 dz;
In[6]:= (* h=0 gives global AdS *)
Collectds2/. h→0/. rTOz,dt, dz, dphi
Out[6]= dt2 -1- 1
z2 + dz2
1+z12z4
+dphi2 z2
In[7]:= (* transformation to z=1/r *)
aux1=ds2/. rTOz//Simplify ; f1[z_]:=1+z^21-8 GN h^2;
f2[z_]:=1+z^21-8 GN h1- (1+1/z^2)^1-2 h^2;
replF=ff1→f1, ff2→f2;
(* check if f1 and f2 are defined consistently*)
1/z^2-dt^2 f1[z] +dphi^2+dz^2f2[z] ⩵aux1//Simplify
Out[11]= True
In[12]:= (* standard assumptions to be used in Simplify routines below *)
assume =Assumptions →
x>0, x<1, l>0, l<Pi, h>1/4, zcut>0, zast>0, Nast>0, Cosl>0, Cotl
22>0;
In[13]:= (* useful later for simplification *)
repldummy =dummy1 →Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 2
2,
dummy2 →Hypergeometric2F1 3 2, 2,5
2+2 h,-Tanl 2
2
, dummy3 →Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl 22 ; subl=l-2+4 h→0, l4 h→0, l2+4 h→0, l4+4 h→0, l6+4 h→0, l8+4 h→0,
1 l
2-8 h
→0, 1 l
-8 h
→0, 1 l
-2-8 h
→0, 1 l
-8-4 h
→0, 1
l
-6-4 h
→0, 1 l
-4-4 h
→0, 1 l
-2-4 h
→0, 1 l
-4 h
→0, 1 l
2-4 h
→0;
◼ Lagrangian and Noether charges
In[15]:= (* the Lagrangian for the area functional *)
Lz_, dz_, dt_:=1/z Sqrt-dt^2 ff1[z] +dz^2ff2[z] +1;
In[16]:= (* Noether charge for spatial homogenity *)
Q1=Lz, dz, dt-dz*DLz, dz, dt, dz-dt*DLz, dz, dt, dt//Simplify
Out[16]= 1
z 1-dt2ff1[z] +ff2[z]dz2
In[17]:= (* Q1 being a Noether charge means that it is independent of z
and can be evaluated at z=zast, where dz z=zast=0 per definition*) Q1zast= Q1/. z->zast/. dz->0/. dt->dtzast
replNast=Nast→1/zast/Q1zast;
Out[17]= 1
zast 1-dtzast2ff1[zast]
In[19]:= (* Noether charge Q2 for spatial homogeneity ddydLddt=0 *)
Q2=DLz, dz, dt, dt
Out[19]= - dt ff1[z]
z 1-dt2ff1[z] + dz2
ff2[z]
In[20]:= (* after definingΛ:=-Q2/Q1 we can express dt in terms of Λ *)
replDt=SolveΛ== -Q2/Q1, dt[[1]]
repldtzast=dtzast→ dt/. replDt/. z→zast
Out[20]= dt→ Λ
ff1[z]
Out[21]= dtzast→ Λ
ff1[zast]
In[22]:= (* using Q2 and Λ in the expression for Q1 gives a differential equation for z;
we take the positive branch corresponding to the interval (y=0,z=0) to (y=y*,z=zast) *) replDz=Solve1/ (zast*Nast)⩵Q1/. replDt, dz[[2]]
Out[22]= dz→ z2Λ2-z2ff1[z] +Nast2zast2ff1[z] ff2[z] z ff1[z]
In[23]:= repleps=eps→h*GN;
replGN=GN→epsh;
Calculations
◼ Integral yielding lambda
In[25]:= (* integral for the shifted time : λ=
∫0λdt=∫0t*dt+∫tλ*dt⩵2∫0zastdz Λ z f1[z]
f1[z] f2[z] z2Λ2-z2f1[z]+Nast2zast2f1[z]
*)
replz=z→zast*x, dz→zast*dx;
dtInt=Simplify2 zast dtdz/. replDz/. replDt/. replF/. replNast/. repldtzast/. replz/. replF, assume ;
int0=SimplifyIntegrateSeriesdtInt/. replGN,{eps, 0, 1}/. eps→0, x, assume ; ExpandNormal SimplifySeries
DSeriesdtInt/. replGN,{eps, 0, 1}, eps/. eps→0,{Λ, 0, 2}, assume ; int1=Integrate-
16(x zast+x^3 zast^3)1+ 1
x2zast2-2 hΛ 1-x2 1+x2zast25/2
, x+
Integrate 48 x3zast3Λ 1-x2 1+x2zast25/2
, x; tInt0=Limitint0+eps int1, x→0, assume ; tInt1=Limitint0+eps int1, x→1, assume ;
tInt=Simplify[tInt1-tInt0, assume ]
Out[31]= 32 eps zast3Λ
1+zast22
+2 ArcTan zastΛ
1+zast2 1+zast2-Λ2
-
8 eps π 1+zast1 2-2 hzastΛGamma [2+2 h] (1+2 h)1+zast2Gamma 32+2 h
In[32]:= (* solving for Λ; we have to pick the positive branch
correspondingt to positive dtdy(zast) for positiveλ *)
solLambda =Solveλ==SimplifySeries[tInt,{Λ, 0, 2}], Assumptions →1+zast^2>0//
Normal ,Λ[[1]];
replLambda =Λ →FullSimplifySeriesΛ/. solLambda ,{eps, 0, 1}, assume //Normal ;
In[34]:= CollectFullSimplify2 zast
FullSimplifySeriesΛ/. solLambda ,{eps, 0, 1}, assume //Normal λ, eps
Out[34]= 1+zast2+eps -16 zast2+
4 π 1+zast1 2-2 h1+zast2Gamma [1+2 h] Gamma 32+2 h
◼ Integral yielding l+lamda
In[35]:= (* integral for the shifted length l+λ=
∫0(l+λ)dy=∫0y*dy+∫yl+λ* dy=∫0z*dz+dydz++∫z0*dz-dydz-=
2∫0z*dz+dydz+=2∫0zastdz z f1[z]
f2[z] z2Λ2-z2f1[z]+Nast2zast2f1[z]
; *)
repldyInt=FullSimplifySolvedz/. replDz ⩵2 dzdy/. replz, dy, zast>0//
Flatten;
dyInt=dy/. repldyInt/. dx→1/. replNast/. repldtzast/. replF;
integrand=CollectFullSimplifyNormal Simplify
SeriesSeriesSimplifyNormal SeriesdyInt/. replGN,{eps, 0, 1}, Assumptions → x>0, x<1, zast>0, h>1/4/.
replLambda ,{eps, 0, 1},{λ, 0, 2},
Assumptions → x>0, x<1, zast>0, h>1/4,{eps,λ}
Out[37]= 2 x zast - 1
-1+x2 1+x2zast2
+ x1+zast2 λ2
4 1-x2 zast1+x2zast23/2 +
eps 1
4 zastx 64 x2zast4 1-x2 1+x2zast25/2
+ 64 x4zast6 1-x2 1+x2zast25/2
-
641+x2zast1 2-2 hzast2 1-x2 1+x2zast2
+
1
4 zastxλ2 - 16 zast2
1-x2 1+x2zast25/2
+ 24 x2zast2 1-x2 1+x2zast25/2
+ 8 x2zast4
1-x2 1+x2zast25/2
- 8 1+ 1 zast2
2 h 1+ 1
zast2 1+ 1 x2zast2
-2 h
1+zast2 1-x2 1+x2zast23/2 + 8 π x2 1+ 1
x2zast2
2 hzast4 1+zast2 1+x2zast2 x2zast4
1-2 h
Gamma [1+2 h] 1-x2 1+x2zast25/2Gamma 3 2+2 h
In[38]:= I2=SimplifyIntegrateintegrand/. eps→0,{x, 0, 1}, Assumptions →{x>0, x<1, zast>0};
I3=eps IntegrateSimplify
x 64 x2zast4
1-x2 1+x2zast252
+ 64 x4zast6
1-x2 1+x2zast252
4 zast ,
{x, 0, 1}, Assumptions →zast>0;
I4=eps Integrate
x -641+
1
x2zast2-2 hzast2 1-x2 1+x2zast2
4 zast ,{x, 0, 1}, Assumptions → zast>0, h>1/4; I5=epsλ2Integrate
1
4 zastx - 16 zast2
1-x2 1+x2zast25/2
+ 24 x2zast2 1-x2 1+x2zast25/2
+ 8 x2zast4 1-x2 1+x2zast25/2
, {x, 0, 1}, Assumptions →zast>0;
I6=epsλ2Integrate 1
4 zastx - 8 1+ 1 zast2
2 h 1+ 1
zast2 1+ 1 x2zast2
-2 h
1+zast2 1-x2 1+x2zast23/2 +
8 π x2 1+ 1 x2zast2
2 hzast4 1+zast2 1+x2zast2 x2zast4
1-2 h
Gamma 1+2 h 1-x2 1+x2zast25/2Gamma 3
2+2 h , {x, 0, 1}, Assumptions → zast>0, h>1/4;
In[43]:= (* total result to linear order in eps and quadratic inλ *)
yInt=CollectI2+I3+I4+I5+I6//FullSimplify ,{eps,λ}
Out[43]= λ2
4 zast+2 ArcTan[zast] + eps -16 zast
1+zast2+16 ArcTan[zast] +
π zast-1+4 h1+zast2-2 hλ2Gamma [1+2 h] Gamma 32+2 h
- 8 π zast1+4 h1+zast2-2 hGamma [1+2 h]
Hypergeometric2F1 1 2, 1,3
2+2 h,-zast2 Gamma 3 2+2 h
◼ Ge�ng zast
In[44]:= (* master equation determining z_* to all orders in lambda *)
master =Collect
SimplifyNormal SeriesSeriesl+λ-yInt/. zast→z00+z01 eps+z10λ+z11λeps+ z20λ^2+z21λ^2 eps,{eps, 0, 1},{λ, 0, 2},{eps,λ};
In[45]:= replz00=FlattenSimplifySolve[(master /. λ →0/. eps→0)⩵0, z00], assume [[1]]; replz01=SimplifyFlatten
SolveSimplifymaster /. λ →0/. replz00 eps, assume ⩵0, z01[[1]];
replz10=FlattenSolve[(D[master ,λ] /.λ →0/. eps→0)⩵0, z10][[1]]; replz11=
FlattenFullSimplifySolveSimplifyD[master ,λ] /.λ →0/. replz10 eps2
1+z0022 ⩵0, z11[[1]];
replz20=SimplifyFlattenSolveD[master ,{λ, 2}] /. eps→0/. replz10/. replz11 ⩵0, z20[[1]];
replz21=SimplifyFlattenSolveSimplifyD[master ,{λ, 2}] /. replz20/. replz10/. replz11/. replz01 eps2 ⩵0, z21[[1]];
In[51]:= z=CollectSimplify
z00+z01 eps+z10λ+z11λeps+z20λ^2+z21λ^2 eps/. replz20/. replz21/.
replz10/. replz11/. replz00/. replz01,{eps,λ}; replzast=zast→z
replzast2=zast0→SeriesCoefficientzast/. replzast,{eps, 0, 0}, zast1→SeriesCoefficientzast/. replzast,{eps, 0, 1};
Out[52]= zast→ 3 Csc[l] 4(1+Cos[l])
+λ Csc[l]2 2(1+Cos[l])
-Cos[2 l]Csc[l]2 2(1+Cos[l])
+λ2 Csc[l] 4(1+Cos[l])
-3 Csc[l]2Sin[2 l] 8(1+Cos[l])
- Csc[l]2Sin[3 l]
4(1+Cos[l])
+eps 6 Csc[l] 1+Cos[l]
-4 l Csc[l]2 1+Cos[l]
+4 l Cos[2 l]Csc[l]2 1+Cos[l]
+
4 π Cscl
2-1-4 hGamma [1+2 h]Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22 Secl
23 Gamma 3
2+2 h-2 Csc[l]2Sin[3 l] 1+Cos[l]
+
λ -8 l Csc[l] 1+Cos[l]
+ 32 π Cscl
2-5-4 hCsc[l]3Gamma [1+2 h]Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22Secl
2 Gamma 3 2+2 h+ 4 l Csc[l]2Sin[2 l]
1+Cos[l] +
4 π Gamma [1+2 h]Secl
221-2 hTanl
24 h Gamma 12+2 h
+
λ2 - 2 Csc[l] 1+Cos[l]
+10 l Cot[l]Csc[l] 1+Cos[l]
-11 l Csc[l]2 1+Cos[l]
-3 l Cos[2 l]Csc[l]2 1+Cos[l]
+ 48 π Cscl
2-7-4 hCsc[l]4Gamma [1+2 h]Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22Secl
2 Gamma 3 2+2 h+ π Cscl
21-4 hGamma [1+2 h]Hypergeometric2F1 1 2, 1,3
2+2 h, -Tanl
22Secl
23 2 Gamma 3
2+2 h +3 Csc[l]2Sin[2 l] 1+Cos[l]
+
-1+16 h2 π Cscl
2Gamma [1+2 h]Hypergeometric2F1 1 2, 1,3
2+2 h, -Tanl
22Secl
2 Secl 22
-2 hTanl
24 h 2 Gamma 3
2+2 h + 4 π Gamma [1+2 h] Secl
22 1
-2 hTanl
21+4 h Gamma 1 2+2 h+
(1+4 h) π Gamma [1+2 h]Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22 Secl
22 1
-2 hTanl
21+4 h 2 Gamma 3
2+2 h + π Gamma [1+2 h]Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl 22 Secl
22 1
-2 hTanl
23+4 h Gamma 3 2+2 h- 1
2(3+4 h) (5+4 h)Gamma 3
2+2 h
π Gamma [1+2 h] Secl
2 Secl 22
-2 hTanl 24 h 15 Cscl
2+32 h Cscl
2+16 h2Cscl
2+ (5+4 h) (7+8 h+ (-3+8 h)Cos[l]) Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl
22Secl
25Tanl 2- 24 Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22Secl
25Tanl 23
In[54]:= (* small l expansion of z_* *)
Collect
Normal ExpandSeriesz,l, 0, 3/. l4 h→0/. 1 l
-4 h
→0/. l1+4 h→0/. l3+4 h→0, {eps,λ}
Out[54]= l 2+l3
24+ 1 2+l2
8 λ+ - 1 4 l+ l
12+53 l3
1440 λ2+eps -2 l3
3 -2 l2λ+ -4 l 3 -53 l3
45 λ2
In[55]:= CollectFullSimplify[z/. eps→0],λ/.1+Cosl →2Cosl2^2/. Sinl →2 Sinl2Cosl2
Out[55]= 1
2λSecl 22+1
8λ2(-3 Cot[l] +Csc[l])Secl
22+Tanl 2
In[56]:= (* this is the exact expression for z_* to leading order in eps *)
LOla=Tanl2+λ 2 Cosl2^2+λ^2/4Tanl2 Cosl2^2-1Sinl
Out[56]= 1
2λSecl
22+Tanl 2+1
4λ2 -Csc[l] +Secl
22Tanl 2
In[57]:= Simplify[% - %%]⩵0
Out[57]= True
◼ Area integral
General defini�ons
In[58]:= (* area integrand*)
R[x]:= x2Λ2+ (Nast^2-x^2)ff1[x zast];
dA=2 Nast Sqrtff1[x zast] x R[x]Sqrtff2[x zast]; dA==
SimplifyLz, dz, dtdy/. replDt/. replDz/. repldyInt/. replz/. dx→1, assume
Out[60]= True
Divergent part
In[61]:= (* the integral diverges for x→0 *)
Simplify
SeriesSeriesCoefficientdA/. replF/. replGN,{eps, 0, 0},{x, 0, 0}, assume
Out[61]= 2
x+O[x]1
In[62]:= (* next we split up the integral∫zcut/zast1 Lⅆx=
∫01Lⅆx-∫0zcut/zastLⅆx and subtract 0=∫012/xⅆx-∫0zcut/zast2/x ⅆx+2 Logzcut
zast *) AdivLO=Integrate[2/x, x];
Adiv1LO=AdivLO/. x→1;
AdivcutLO=AdivLO/. x→zcut/zast;
AcutLO=Adiv1LO-AdivcutLO
Out[65]= -2 Logzcut zast
Convergent part, LO
In[66]:= dALOpert=SimplifyNormal
SeriesNormal SeriesNormal SeriesSeriesdA-2/x/. replNast/. repldtzast/. replF/. replGN/. replLambda ,{λ, 0, 2},{eps, 0, 0}/.
replzast,{eps, 0, 0},{λ, 0, 2}, assume
Out[66]= -2 x+
2 1-x4+(-1+x1+Cos[l]2
)2Cos[l]
x -
xλ22-x2+-1+x2Cos[l] 1+Cos[l]
1-x4+(-1+x2)2Cos[l]
1+x2+Cos[l] -x2Cos[l]2
+
2 xλ 1+Cos[l]
1-x4+(-1+x2)2Cos[l] Tanl2 -1-x2+-1+x2Cos[l]
In[67]:= (* the near boundary part ∫0zcut/zast... of the
integral is of O(zcut^2) and can be neglected *) CollectIntegrateSimplify
SeriesSeriesCoefficientdA-2/x/. replF/. replGN,{eps, 0, 0},{x, 0, 2}, assume ,{x, 0, zcut/zast}, zcut
Out[67]= -zcut2-1+Nast2zast2+Λ2 2 Nast2zast2
In[68]:= (* Mathematica needs some help to manage
the O(eps^0) integral for the regularized area *)
In[69]:= I7=IntegratedALOpert, x;
I7dn=SimplifyNormal Series[I7,{x, 0, 0}], assume ;
I7up=SimplifyNormal Series[I7,{x, 1, 0}], assume /. x→1;
In[72]:= (* Evaluating the limit of AppellF1 requires
advanced educated guessing and hindsight, see below *)
In[73]:= N -I Cotl
2AppellF11,1 2,1
2, 2, 1 x2,-
Cotl
22
x2 x^2-
2 Log[x] -2 I ArcTanCotl
2-Log4 Cosl
22 /. x→10^(-5) /. l→Pi10
Out[73]= 4.8761×10-11+2.66454×10-15ⅈ
In[74]:= (* graphical checks that these two
expressions are equivalent up to O(x^2) terms *) PlotRe -I Cotl
2AppellF11,1 2,1
2, 2, 1 x2,-
Cotl
22
x2 x^2-
2 Log[x] -2 I ArcTanCotl
2-Log4 Cosl
22 /. l→Pi10,{x, 0, 0.001}
Out[74]=
0.0002 0.0004 0.0006 0.0008 0.0010
1.×10-7 2.×10-7 3.×10-7 4.×10-7 5.×10-7
In[75]:= PlotIm -I Cotl
2AppellF11,1 2,1
2, 2, 1 x2,-
Cotl
22
x2 x^2-
2 Log[x] -2 I ArcTanCotl
2-Log4 Cosl
22 /. l→Pi10,{x, 0, 0.001}
Out[75]=
0.0002 0.0004 0.0006 0.0008 0.0010
-2.×10-15 -1.×10-15 1.×10-15 2.×10-15
In[76]:= AregPertLO=FullSimplify
Normal SeriesI7up-I7dn/.
ⅈAppellF11,1
2,1
2, 2, 1
x2,-Cot
l 22
x2 Cotl
2
x2 →-2 Log[x] + 2 I ArcTanCotl
2+Log4 Cosl
22/. replzast/. eps→0,{λ, 0, 2}
Out[76]= - λ2
2+2 Cos[l]
+Log[2(1+Cos[l])] -λTanl 2
In[77]:= (* contribution from the log to LO *)
AcutPertLO=CollectFullSimplify
Normal SeriesSimplifySeriesAcutLO/. replzast,{eps, 0, 0},{λ, 0, 2}, λ
Out[77]= 2λCsc[l] +λ2 -3
2Cot[l]Csc[l] -Csc[l]2
2 -2 Log[zcut(Cot[l] +Csc[l])]
In[78]:= (* trivial sanity check that adding and taking limits commutes *)
AcheckLO=FullSimplifyNormal SeriesAcutLO+I7up-I7dn/. 1
x2ⅈAppellF11,1 2,1
2, 2, 1 x2,-
Cotl
22
x2 Cotl 2 → -2 Log[x] +2 I ArcTanCotl
2+Log4 Cosl 22/. replzast/. eps→0,{λ, 0, 2};
FullSimplifyAcheckLO-AcutPertLO-AregPertLO
Out[79]= 0
HEE, LO
In[80]:= FullSimplifyCotl+Cscl-Cotl2
Out[80]= 0
In[81]:= (* holographic entanglement entropy O(eps^0) *)
HEEcLO=1/ (4 GN) (AcutPertLO+AregPertLO) //Normal ; HEELO=
FullSimplifyFullSimplifyFullSimplifyHEEcLO/.λ →0, Assumptions →{zcut>0}/.
1+Cosl →2Cosl2^2/.Cotl+Cscl →Cotl2/. Logzcut Cotl
2->Log[zcut] +LogCotl 2/. Log1+Cosl-2 LogCotl
2 →Log2Cosl2^2 Cotl2^2/.
Log2-2 Cosl →Log4Sinl2^2 replGbyc= {GN→3/ (2 c)};
SimplifySeriesHEELO,l, 0, 2, Assumptions →l>0/. replGbyc,l>0, zcut>0
Out[82]=
-2 Log[zcut] +Log4 Sinl22 4 GN
Out[84]= 1
3c Log l
zcut-c l2
72 +O[l]3
QNEC, LO
In[85]:= (* QNEC O(eps^0) *)
QNEC=
Simplify1 2 Pi(D[HEEcLO,{λ, 2}] +6/c D[HEEcLO,{λ, 1}]^2) /.λ →0/. replGbyc
Out[85]= - c 12π
In[86]:= (* Daniels definitionof the lightlike deformation : k=(λ/2,λ/2) O(eps^0) *)
HEElahalf=HEEcLO/.λ → λ/2;
QNEClahalf=
1 2 Pi DHEElahalf,{λ, 2}+6/c DHEElahalf,{λ, 1}^2/.λ →0/. replGbyc//
Simplify
Out[87]= - c 48π
In[88]:= (* lightlike projection of the EMT O(eps^0) *)
Tkk=1 2 Pi(-c/24); QNEClahalf⩵Tkk
Out[89]= True
NLO
In[90]:= (* O(eps^1) integral for the regularized area; this is length,
so we split it into O(1) int8, O(λ) int9 and O(λ^2) int10 parts;
thos parts have to be massaged individuallyfor Mathematica to succeed *) preparedANLOpert=
SeriesSeriesdA/. replNast/. repldtzast/. replF/. replGN/. replLambda , {λ, 0, 2},{eps, 0, 1}//Normal ;
dANLOpert=eps SeriesSeriesCoefficientpreparedANLOpert/. zast→(zast0+eps zast1), {eps, 0, 1}/. replzast2,{λ, 0, 2}//Normal ;
In[92]:= (* let us start with O(1) *)
int8=FunctionExpandSimplifyExpanddummy3 dANLOpert/.λ →0, assume /. 16 π x -1+x2 -1-x2-Cosl+x2Cosl
1+Cosl Cscl 2
-4 h
Gamma 1+2 hHypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22
-1+x2 -1-x2-Cosl+x2Cosl2Gamma 3
2+2 h - 16 π x Cosl
-1+x2 -1-x2-Cosl+x2Cosl
1+Cosl
Cscl 2-4 h
Gamma 1+2 hHypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22
-1+x2 -1-x2-Cosl+x2Cosl2Gamma 3
2+2 h → dummy1 dummy3 /. 1+
Cotl
22 x2
-2 h
→dummy2 dummy3 ;
int81=IntegrateSimplifyint8/. dummy1 →0/. dummy2 →0/. dummy3 →1, assume , x;
int82=IntegrateSimplifyint8/. dummy1 →0/. dummy2 → 1+ Cotl
22 x2
-2 h
/. dummy3 →0, assume , x;
int83=IntegrateSimplifyint8/. dummy1 →
16 π x√-1+x2 -1-x2-Cosl+x2Cosl 1+CoslCscl 2-4 h Gamma 1+2 hHypergeometric2F1 1
2, 1,3
2+2 h,-Tanl 22
-1+x2 -1-x2-Cosl+x2Cosl2Gamma 3
2+2 h -
16 π x Cosl√-1+x2 -1-x2-Cosl+x2Cosl 1+Cosl
Cscl
2-4 hGamma 1+2 h Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl 22
-1+x2 -1-x2-Cosl+x2Cosl2Gamma 3
2+2 h /. dummy2 →0/. dummy3 →0, assume , x;
In[96]:= I8up=Simplify
SimplifySimplifyNormal Seriesint81+int82+int83,{x, 1, 0}, assume /. Gamma 2+2 h → 1+2 h2 h Gamma 2 h/.
Hypergeometric2F1 2 h,1
2+2 h,3
2+2 h,-Tanl 22+ 4 h Hypergeometric2F1 1
2+2 h, 1+2 h,3
2+2 h,-Tanl 22 →
4 h+1 Cosl2^4 h, assume ;
I8dn=SimplifyNormal Seriesint81+int82+int83,{x, 0, 0}/. Cotl
22 x2
-2 h
→0;
I8=SimplifyI8up-I8dn, assume
Out[98]= - 8 eps π Cscl
2-2-4 hGamma [1+2 h] Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22Secl
22 Gamma 3
2+2 h - 4 eps(1+4 h) π Gamma [2 h]Sinl
24 h Gamma 3
2+2 h
+8 eps l Tanl 2
In[99]:= Simplify
Dint91+int92+int93+int94+int95, x-ExpandDdANLOpert,λ/.λ →0/. l→Pi2/. h→2/. eps→1/10, assume
Out[99]= - 1
1575-1+x2 1+x26
4 x 1-x4 315π -2+x2 1+x23-2 243+x2 972-80 Hypergeometric2F1 1
2, 1,11
2 ,-1 - 32 Hypergeometric2F1 1
2, 1,11 2 ,-1+ 4 x6 -387+4 Hypergeometric2F11
2, 1,11
2 ,-1 - 6 x4 -243+8 Hypergeometric2F11
2, 1,11
2 ,-1 + 2 x8 279+8 Hypergeometric2F11
2, 1,11 2 ,-1
In[100]:= (* now consider O(λ);
to avoid issues when expanding around x=1 we take instead the directed limit from below otherwise the integrals do not evaluate correctly! *) int9=ExpandDdummy8 dANLOpert,λ/.λ →0/.
Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 2
2
→dummy4 dummy8 /.
Tanl
24 h→dummy5 dummy8 /. 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-1-2 h
→dummy6 dummy8 /.
1+ 161+Cosl2Sinl2 x2-3+CsclSin3 l2
-2 h
→dummy7 dummy8 ;
int91=IntegrateSimplifyint9/. dummy4 →0/. dummy5 →0/. dummy6 →0/.
dummy7 →0/. dummy8 →1, assume , x;
int92=IntegrateSimplifyint9/. dummy4 →Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22/. dummy5 →0/.
dummy6 →0/. dummy7 →0/. dummy8 →0, assume , x; int93=IntegrateSimplifyint9/. dummy4 →0/. dummy5 →Tanl
2
4 h
/. dummy6 →0/. dummy7 →0/. dummy8 →0,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4, x; int94=IntegrateSimplifyint9/. dummy4 →0/. dummy5 →0/.
dummy6 → 1+ 161+Cosl2Sinl2 x2-3+CsclSin3 l2
-1-2 h
/. dummy7 →0/. dummy8 →0, assume , x;
int95=IntegrateSimplifyint9/. dummy4 →0/. dummy5 →0/. dummy6 →0/. dummy7 → 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-2 h
/. dummy8 →0, assume , x; I9up1=SimplifyLimitint91, x→1, Direction→1, assume ,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4; I9up2=SimplifyLimitint92, x→1, Direction→1, assume ,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4; I9up3=SimplifyLimitint93, x→1, Direction→1, assume ,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4; I9up4=SimplifyLimitint94, x→1, Direction→1, assume ,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4; I9up5=SimplifyLimitint95, x→1, Direction→1, assume ,
Assumptions → x>0, x<1, l>0, l<Pi, h>1/4; I9up=I9up1+I9up2+I9up3+I9up4+I9up5;
I9dn=
SimplifyExpandNormal Seriesint91+int92+int93+int94+int95,{x, 0, 0}/. Cotl
22 x2
-2 h
→0, assume ; I9=SimplifyI9up-I9dn, assume λ
Out[113]= 1
2epsλCscl 2-4 h 1
Gamma [3+2 h]
π Cscl
2Gamma [1+2 h] -32(1+h) (1+h+h Cos[l])Gamma [2+2 h] Gamma 3
2+2 h
- Gamma [3+2 h] (5+4 h) (7+8 h+ (-2+8 h)Cos[l])Gamma 5
2+2 h- 2(13+16 h+4(1+4 h)Cos[l])Gamma 7
2+2 h (1+Cos[l])Gamma 5
2+2 hGamma 7
2+2 h Secl 2+ 32 Cosl
23Sinl
2 -2 π Gamma [1+2 h] (1+Cos[l])Gamma 3 2+2 h+ Gamma 1
2+2 hHypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22 + Cscl
22+4 hGamma 1
2+2 hGamma 3
2+2 hSin[l]l+Sin[l] (1+Cos[l])3Gamma 1
2+2 hGamma 3
2+2 h + 8 π Gamma [2+2 h]Tanl2
(1+2 h)Gamma 32+2 h
In[114]:= (* finally consider O(λ^2); same remarks as above;
evaluation of this part takes a while *)
int10=ExpandDdummy8 dANLOpert,{λ, 2}/.λ →0/. Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl
22 →dummy1 dummy8 /. Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22 →dummy2 dummy8 /. Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22 →dummy3 dummy8 /. 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-2 h
→dummy4 dummy8 /.
1+ 161+Cosl2Sinl2 x2-3+CsclSin3 l2
-1-2 h
→dummy5 dummy8 /.
1+ 161+Cosl2Sinl2 x2-3+CsclSin3 l2
-2-2 h
→dummy6 dummy8 /.
1+161+Cosl2Sinl2
-3+CsclSin3 l2
-2 h
→dummy7 dummy8 ; int101=IntegrateSimplify
int10/. dummy1 →0/. dummy2 →0/. dummy3 →0/. dummy4 →0/. dummy5 → 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-1-2 h
/. dummy6 →0/.
dummy7 →0/. dummy8 →0, assume , x;
int102=IntegrateSimplifyint10/. dummy1 →0/. dummy2 →0/. dummy3 →0/. dummy4 → 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-2 h
/. dummy5 →0/. dummy6 →0/. dummy7 →0/. dummy8 →0, assume , x;
int103=IntegrateSimplifyint10/. dummy1 →0/. dummy2 →0/.
dummy3 →Hypergeometric2F1 5 2, 3,7
2+2 h,-Tanl 22/. dummy4 →0/. dummy5 →0 /. dummy6 →0/.
dummy7 →0/. dummy8 →0, assume , x; int104=IntegrateSimplify
int10/. dummy1 →0/. dummy2 →Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 2
2
/.
dummy3 →0/. dummy4 →0/. dummy5 →0/. dummy6 →0/. dummy7 →0/. dummy8 →0, assume , x; int105=IntegrateSimplify
int10/. dummy1 →Hypergeometric2F1 3 2, 2,5
2+2 h,-Tanl
22/. dummy2 →0/. dummy3 →0/. dummy4 →0/. dummy5 →0/.
dummy6 →0/. dummy7 →0/. dummy8 →0, assume , x;
int106=IntegrateSimplifyint10/. dummy1 →0/. dummy2 →0/. dummy3 →0/.
dummy4 →0/. dummy5 →0 /. dummy6 → 1+ 161+Cosl2Sinl2
x2-3+CsclSin3 l2
-2-2 h
/. dummy7 →0/. dummy8 →0, assume , x;
int107=IntegrateSimplifyint10/. dummy1 →0/. dummy2 →0/. dummy3 →0/. dummy4 →0/. dummy5 →0 /. dummy6 →0/.
dummy7 → 1+161+Cosl2Sinl2
-3+CsclSin3 l2
-2 h
/. dummy8 →0, assume , x; int10rest=int10/. dummy1 →0/. dummy2 →0/. dummy3 →0/. dummy4 →0/.
dummy5 →0 /. dummy6 →0/. dummy7 →0/. Tanl
24 h→dummy9 dummy8 /. Tanl
21+4 h→dummy10 dummy8 ; int108=IntegrateSimplifyint10rest/. dummy8 →0/. dummy9 →0/.
dummy10 ->Tanl
21+4 h, assume , x;
int109=IntegrateSimplifyint10rest/. dummy8 →0/. dummy9 →Tanl 24 h/. dummy10 →0, assume , x;
int110=IntegrateSimplifyint10rest/. dummy8 →1/. dummy9 →0/. dummy10 →0, assume , x;
I10up=SimplifyLimitint101+int102+int103+int104+int105+int106+
int107+int108+int109+int110, x→1, Direction→1, assume , Assumptions → x>0, x<1, l>0, l<Pi, h>1/4;
I10dn=SimplifyExpandSimplifyNormal Seriesint101+int102+int103+int104+ int105+int106+int107+int108+int109+int110,{x, 0, 0}, assume /. x Tanl
2
4 h
→0, assume ; I10=SimplifyI10up-I10dn, assume λ^2/2
Simplify::time :
Time spent on a transformation exceeded 300.` seconds, and the transformation was aborted. Increasing the value of TimeConstraint option may improve the result of simplification .
Simplify::time :
Time spent on a transformation exceeded 300.` seconds, and the transformation was aborted. Increasing the value of TimeConstraint option may improve the result of simplification .
Out[128]= 1 32epsλ2
1 (1+2 h)Gamma 5
2+2 h
π -128(1+2 h)23+7 h+6 h2+8 h(1+h)Cos[l] +h(1+2 h)Cos[2 l]
Cscl
2-4 hCsc[l]2Gamma [1+2 h] + 1
1+h2132-2 hCosl
23(1+Cos[l])-3-2 h Csc[l]-4 h -(1+h) (-2+Cos[l]) 1+Cos[l] (2+2 h+Cos[l] +2 h Cos[l])
Gamma [2+2 h] + 2 (1+2 h)Cosl
23(-1+Cos[l])Gamma [3+2 h] + 4-h(1+Cos[l])2-2 hCscl
22Secl
26 2(1+2 h) (5+3 Cos[l])
3+7 h+6 h2+8 h(1+h)Cos[l] +h(1+2 h)Cos[2 l]Gamma [1+2 h] + 24 h(3+Cos[l]) (2+2 h+Cos[l] +2 h Cos[l])
Gamma [2+2 h]Sinl
22 Sin[l]4 h - 8 Cscl
2-4(1+h)Secl
22 215+32 h+16 h2 π Cscl
210Gamma [1+2 h] Gamma 3
2+2 hSin[l]6-1
4Gamma 1 2+2 h 815+32 h+16 h2 l+2 Cotl
2 Cscl
28+4 hGamma 3
2+2 hSin[l]5+ π Gamma [1+2 h] 960 Cosl
22Cos[l]Cotl
24-6415+32 h+16 h2
3+h+12 h2+-5+16 h2Cos[l] +h(-1+4 h)Cos[2 l]
Cotl
24Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22+ 128(5+4 h) (7+8 h+ (-3+8 h)Cos[l])Cotl
22 Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl 22- 3072 Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl 22+
15 Cscl
210Sin[l]6+32 h Cscl
210Sin[l]6+ 16 h2Cscl
210Sin[l]6+32 h Cos[l]Cscl
210Sin[l]6+ 16 h2Cos[l]Cscl
210Sin[l]6-15 Cscl
212Sin[l]8- 32 h Cscl
212Sin[l]8-16 h2Cscl
212Sin[l]8 (3+4 h) (5+4 h) (1+Cos[l])3Gamma 1
2+2 hGamma 3 2+2 h
In[129]:= (* after this tour de force we add all results to get the O(eps) contribution
to the area integral apart from the log-term that we split away *) ApertNLO=I8+I9+I10
Out[129]= - 8 eps π Cscl
2-2-4 hGamma [1+2 h] Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22Secl
22 Gamma 3
2+2 h - 4 eps(1+4 h) π Gamma [2 h]Sinl24 h
Gamma 3
2+2 h
+ 1
32epsλ2 1
(1+2 h)Gamma 5
2+2 h
π -128(1+2 h)23+7 h+6 h2+8 h(1+h)Cos[l] +h(1+2 h)Cos[2 l]Cscl 2-4 h Csc[l]2Gamma [1+2 h] + 1
1+h2132-2 hCosl
23(1+Cos[l])-3-2 hCsc[l]-4 h -(1+h) (-2+Cos[l]) 1+Cos[l] (2+2 h+Cos[l] +2 h Cos[l])Gamma [
2+2 h] + 2 (1+2 h)Cosl
23(-1+Cos[l])Gamma [3+2 h] + 4-h(1+Cos[l])2-2 hCscl
22Secl
26 2(1+2 h) (5+3 Cos[l])
3+7 h+6 h2+8 h(1+h)Cos[l] +h(1+2 h)Cos[2 l]Gamma [1+2 h] + 24 h(3+Cos[l]) (2+2 h+Cos[l] +2 h Cos[l])Gamma [2+2 h]Sinl
22 Sin[l]4 h - 8 Cscl
2-4(1+h)Secl 22 215+32 h+16 h2 π Cscl
210Gamma [1+2 h]Gamma 3
2+2 hSin[l]6- 1
4Gamma 1
2+2 h 815+32 h+16 h2 l+2 Cotl
2 Cscl 28+4 h Gamma 3
2+2 hSin[l]5+ π Gamma [1+2 h] 960 Cosl
22Cos[l]Cotl
24-6415+32 h+16 h2
3+h+12 h2+-5+16 h2Cos[l] +h(-1+4 h)Cos[2 l]
Cotl
24Hypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22+ 128(5+4 h) (7+8 h+ (-3+8 h)Cos[l])Cotl
22 Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl
22-3072 Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22+15 Cscl
210Sin[l]6+32 h Cscl
210Sin[l]6+16 h2 Cscl
210Sin[l]6+32 h Cos[l]Cscl
210Sin[l]6+16 h2Cos[l]Cscl
210Sin[l]6-15 Cscl
212Sin[l]8-32 h Cscl
212Sin[l]8-16 h2Cscl
212Sin[l]8 (3+4 h) (5+4 h) (1+Cos[l])3Gamma 1
2+2 hGamma 3
2+2 h +8
eps l Tanl 2+ 1
2 eps λ Csc
l 2-4 h
1
Gamma [3+2 h]
π Cscl
2Gamma [1+2 h] -32(1+h) (1+h+h Cos[l])Gamma [2+2 h]
Gamma 3
2+2 h
-
Gamma [3+2 h] (5+4 h) (7+8 h+ (-2+8 h)Cos[l])Gamma 5 2+2 h- 2(13+16 h+4(1+4 h)Cos[l])Gamma 7
2+2 h (1+Cos[l])Gamma 5
2+2 hGamma 7
2+2 h Secl 2+
32 Cosl
23Sinl
2 -2 π Gamma [1+2 h] (1+Cos[l])Gamma 3 2+2 h+
Gamma 1
2+2 hHypergeometric2F1 1 2, 1,3
2+2 h,-Tanl 22 + Cscl
22+4 hGamma 1
2+2 hGamma 3
2+2 hSin[l]l+Sin[l] (1+Cos[l])3Gamma 1
2+2 hGamma 3
2+2 h + 8 π Gamma [2+2 h]Tanl2
(1+2 h)Gamma 32+2 h
In[130]:= (* now let us add again the log part and expand
to first order in eps and second order in lambda *)
Acuttotal=SimplifyNormal SeriesSeriesAcutLO/. replzast,{eps, 0, 1},{λ, 0, 2}
Out[130]= 2λCsc[l] -1
2λ2(1+3 Cos[l])Csc[l]2-2 Logzcut Cotl
2+16 eps(1+Cos[l])Csc[l]2 -1
32Cscl
2Secl
234 l+4λ+9 lλ2+4 lλ2Cos[l] +-4 l-4λ+3 lλ2Cos[2 l] - 6 Sin[l] -4λ2Sin[l] -4 lλSin[2 l] -6λ2Sin[2 l] +2 Sin[3 l]+ π Cscl
2-4(2+h)Gamma [1+2 h] Secl 22
-2 hSin[l] -415+32 h+16 h2 λ
(-1+λCot[l])Cscl
29+4 hGamma 3
2+2 hSecl
23Sin[l]Tanl 24 h+ Gamma 1
2+2 h -λ2Cscl
25+4 hSecl
2 15 Cscl
24+32 h Cscl 24+16 h2Cscl
24+ (5+4 h) (7+8 h+ (-3+8 h)Cos[l])Cscl 22 Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl
22Secl 26- 24 Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22Secl 28 Tanl
24 h+1
415+32 h+16 h2Cscl 29 Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl
22Secl 25 4 Secl
22 2 h+9λ2 Secl
22 2 h-4λ Secl 22 2 h Sin[2 l] +2λ2Cscl
24 hTanl
24 h+2 hλ2Cscl 24 h Tanl
24 h+24 h2λ2Cscl
24 hTanl
24 h+2λ2Cos[l] 2 Secl
22 2 h+-3+16 h2Cscl
24 hTanl 24 h + Cos[2 l] -4+3λ2 Secl
22 2 h+ 2 h(-1+4 h)λ2Cscl
24 hTanl
24 h 16(3+4 h) (5+4 h)Gamma 1
2+2 hGamma 3 2+2 h
In[131]:= (* simple check that limits of small eps and small lambda commute *)
Acutcheck=
SimplifyNormal SeriesSeriesAcutLO/. replzast,{λ, 0, 2},{eps, 0, 1}; SimplifyAcutcheck-Acuttotal
Out[132]= 0
Total area
In[133]:= (* area consist of three parts: the regular O(1) piece,
the regular O(eps) piece and the log-contribution,
both to O(1) and O(eps); we treat different orders in λ separately, called area0, area1 and area2 the latter is split further into
four pieces to reduce the time for Mathematica to Simplify *) arearaw=CollectSimplifyAregPertLO+ApertNLO+Acuttotal, assume ,{eps,λ}; area0=Simplifyarearaw/. Hypergeometric2F1 1
2, 1,3
2+2 h,-Tanl 2
2
→dummy1 /. Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl 2
2
→dummy2 /. Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl 2
2
→dummy3 /.
Sqrt1+Cosl →Sqrt[2]Cosl2/.λ →0, assume //FullSimplify ; area1=SimplifyDarearaw/. Hypergeometric2F11
2, 1,3
2+2 h,-Tanl
22 →dummy1 /. Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl
22 →dummy2 /. Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl 2
2
→dummy3 /.
Sqrt1+Cosl →Sqrt[2]Cosl2,λ/.λ →0, assume //FullSimplify ; area21=SimplifyDdummy4 arearaw/. Hypergeometric2F11
2, 1,3
2+2 h,-Tanl 22 → dummy1 dummy4 /. Hypergeometric2F1 3
2, 2,5 2+2 h, -Tanl
22 →dummy2 dummy4 /. Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22 →dummy3 dummy4 /.
Sqrt1+Cosl →Sqrt[2]Cosl2,{λ, 2} 2/. dummy1 →0/. dummy2 →0/. dummy3 →0/. dummy4 →1, assume ;
area22=SimplifyCollectDdummy4 arearaw/. Hypergeometric2F1 1 2, 1,3
2+ 2 h,-Tanl
22 →dummy1 dummy4 /. Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl 22 → dummy2 dummy4 /. Hypergeometric2F1 5
2, 3,7
2+2 h,-Tanl
22 →dummy3 dummy4 /.
Sqrt1+Cosl →Sqrt[2]Cosl2,{λ, 2} 2, dummy4 /. dummy1 →0/. dummy2 →0/. dummy4 →0, assume ;
area23=SimplifyCollectDdummy4 arearaw/. Hypergeometric2F1 1 2, 1,3
2+ 2 h,-Tanl
22 →dummy1 dummy4 /. Hypergeometric2F1 3
2, 2,5
2+2 h,-Tanl 22 →