Im=Sin-
1
1+t
(1+t)
tArg[1+t]
1+
2
t
1
21+
2
t
(1+)
2
t
Arg[1+t]
1+
2
t
tLog[1+]
2
t
2(1+)
2
t
But the MRB constant (CMRB)=Csch[πt]Imn.
∞
∫
0
1
1+t
(1+t)
CMRB=NSum[Cos[Pin](n^(1/n)-1),{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision500]
Out[]=
0.187859642462067120248517934054273230055903094900138786172004684089477231564660213703296654433107496903842345856258019061231370094759226630438929348896184120837336626081613602738126379373435283212552763962171489321702076282062171516715408412680448363541671998519768025275989389939144579835055613509648521071207844423095868129497688526949564204255586483670441042527952471060666092633974834103115781678641668915460034222258838002545539689294711421221891050983287122773080200364452153905363950553321735
NIntegrateCsch[πt]Im,{t,0,Infinity},WorkingPrecision20
1
1+t
(1+t)
Out[]=
0.18785964246206712051
NIntegrateCsch[πt]Sin-,{t,0,Infinity},WorkingPrecision20
tArg[1+t]
1+
2
t
1
21+
2
t
(1+)
2
t
Arg[1+t]
1+
2
t
tLog[1+]
2
t
2(1+)
2
t
Out[]=
0.18785964246206712353
In[]:=
TimingCMRB-NIntegrateCsch[πt]Im,{t,0,Infinity},WorkingPrecision500,Method"Trapezoidal"
1
1+t
(1+t)
Out[]=
{2.34375,-3.00×}
-496
10
In[]:=
TimingCMRB-NIntegrateCsch[πt]Sin-,{t,0,Infinity},WorkingPrecision500,Method"Trapezoidal"
tArg[1+t]
1+
2
t
1
21+
2
t
(1+)
2
t
Arg[1+t]
1+
2
t
tLog[1+]
2
t
2(1+)
2
t
Out[]=
{2.39063,-3.00×}
-496
10
Inapreviousreplywehad
In[]:=
Timing[CMRB-NIntegrate[Csc[πt]Im[(1-t^(1/t))],{t,1,InfinityI},WorkingPrecision500,Method"DoubleExponential"]]
Out[]=
{2.25,-3.00×}
-496
10
Nowwehave
CMRB=Re((-1)csc(πt))t.
∞
∫
1
1/t
t
In[]:=
Timing[CMRB-INIntegrate[Re[Csc[πt](t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision500,Method"DoubleExponential"]]
Out[]=
{2.3125,-3.00×}
-496
10
0=∞∫1(1/tt-1)Re(csc(πt))t.
0=(-1)Re(csc(πt))t.
∞
∫
1
1/t
t
Timing[N[INIntegrate[(Re[Csc[πt]])(t^(1/t)-1),{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{1.85938,0}
Q=∞∫1Re(1/tt-1)csc(πt)t=∞∫1Im((1-1/tt)csc(πt))t.
Q=Re(-1)csc(πt)t=Im((1-)csc(πt))t.
∞
∫
1
1/t
t
∞
∫
1
1/t
t
Timing[Q=N[INIntegrate[Csc[πt]Re[(t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{23.0781,-0.10140904834001897354784794206523556559332349957408439415056089296571421188649799373309577398669321526864592292038098918512543392706217249642683673371900540797185401511906146581436666108354602466457454608304564559215252062489969450647133679118113551001450006470728274511882073812791922316447197828371101025304131116541836872476946717657217445214250863386411181234407320824451107398817105828827311462932592569333983845794888828277136232512108579431744492548432182332462941559129986359369813989517742364}
Timing[Q-N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{23.7656,0.×}
-501
10
(-1)csc(πt)t-CMRB-Q=0.
∞
∫
1
1/t
t
INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]-Q-CMRB
Out[]=
9.3471×+0.×
-94
10
-101
10
INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074916-0.1014090483400189735478479420652355655933234995740843941505608929657142118864979937330957739866932152
NIntegrate[Csc[πt]Im[(1-t^(1/t))],{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074916
CMRB=∞∫1(1/tt-1)csc(πt)t-∞∫1Im((1-1/tt)csc(πt))t.
CMRB=(-1)csc(πt)t-Im((1-)csc(πt))t.
∞
∫
1
1/t
t
∞
∫
1
1/t
t
CMRB-(INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]-Quiet[N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision200],100]])
Out[]=
5.3×-0.×
-99
10
-101
10
Interestingly, that is the same as
Interestingly, that is the same as
CMRB=∞∫1Im(1-1/tt)csc(πt)t.
CMRB=Im(1-)csc(πt)t.
∞
∫
1
1/t
t
Thus we have
Thus we have
∞∫1Re(1/tt-1)csc(πt)t=∞∫1Im((1-1/tt)csc(πt))t
Re(-1)csc(πt)t=Im((1-)csc(πt))t
∞
∫
1
1/t
t
∞
∫
1
1/t
t
In[]:=
Quiet[N[INIntegrate[Csc[πt]Re[(t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision1000],500]]-Quiet[N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision200],100]]
Out[]=
0.×
-101
10
Related integral equations include
Related integral equations include
Out[]//TableForm=
It might be hard to prove a pattern here, but here is where that comes from.
In[]:=
CMRB=NSum[Cos[Pin](n^(1/n)-1),{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision100];Table[Rationalize[CMRB-(NIntegrate[Im[(t^(1/t)-t^(n))](-Csc[πt]),{t,1,InfinityI},WorkingPrecision100,Method"Trapezoidal"]),10^-6],{n,1,22}]
Out[]=
,,,,,,,,-,,,,-,,,,-,,,,-,
1
4
1
2
5
8
1
2
1
4
1
2
25
16
1
2
29
4
1
2
695
8
1
2
5459
4
1
2
929585
32
1
2
3202289
4
1
2
221930585
8
1
2
4722116519
4
1
2
Thisstronglyindicatesthat
Whichmakessence,sinceCMRB=1∫∞Im(1-1/tt)csc(πt)t.Thenweareleftwiththefollowingforthesamevaluesofn,
Whichmakessence,sinceCMRB=Im(1-)csc(πt)t.Thenweareleftwiththefollowingforthesamevaluesofn,
1
∫
∞
1/t
t
is faster than the following where you have to have WorkingPrecision2000 for 1000 accurate digits.
is faster than the following where you have to have WorkingPrecision2000 for 1000 accurate digits.
But, this second one is faster after moving Limit[g[x],x->Infinity]=1 to Limit[g[x]-1,x->Infinity]=0
But, this second one is faster after moving Limit[g[x],x->Infinity]=1 to Limit[g[x]-1,x->Infinity]=0
From our first equation we get the following interesting one.
From our first equation we get the following interesting one.