Im
1
1+t
(1+t)
=
tArg[1+t]
1+
2
t

1
21+
2
t

(1+
2
t
)
Sin
Arg[1+t]
1+
2
t
-
tLog[1+
2
t
]
2(1+
2
t
)

But the MRB constant (CMRB)=
∞
∫
0
Csch[πt]Im
1
1+t
(1+t)
n
.
CMRB=NSum[Cos[Pin](n^(1/n)-1),{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision500]
Out[]=
0.187859642462067120248517934054273230055903094900138786172004684089477231564660213703296654433107496903842345856258019061231370094759226630438929348896184120837336626081613602738126379373435283212552763962171489321702076282062171516715408412680448363541671998519768025275989389939144579835055613509648521071207844423095868129497688526949564204255586483670441042527952471060666092633974834103115781678641668915460034222258838002545539689294711421221891050983287122773080200364452153905363950553321735
NIntegrateCsch[πt]Im
1
1+t
(1+t)
,{t,0,Infinity},WorkingPrecision20
Out[]=
0.18785964246206712051
NIntegrateCsch[πt]
tArg[1+t]
1+
2
t

1
21+
2
t

(1+
2
t
)
Sin
Arg[1+t]
1+
2
t
-
tLog[1+
2
t
]
2(1+
2
t
)
,{t,0,Infinity},WorkingPrecision20
Out[]=
0.18785964246206712353
In[]:=
TimingCMRB-NIntegrateCsch[πt]Im
1
1+t
(1+t)
,{t,0,Infinity},WorkingPrecision500,Method"Trapezoidal"
Out[]=
{2.34375,-3.00×
-496
10
}
In[]:=
TimingCMRB-NIntegrateCsch[πt]
tArg[1+t]
1+
2
t

1
21+
2
t

(1+
2
t
)
Sin
Arg[1+t]
1+
2
t
-
tLog[1+
2
t
]
2(1+
2
t
)
,{t,0,Infinity},WorkingPrecision500,Method"Trapezoidal"
Out[]=
{2.39063,-3.00×
-496
10
}
Inapreviousreplywehad
In[]:=
Timing[CMRB-NIntegrate[Csc[πt]Im[(1-t^(1/t))],{t,1,InfinityI},WorkingPrecision500,Method"DoubleExponential"]]
Out[]=
{2.25,-3.00×
-496
10
}
Nowwehave
CMRB=
∞
∫
1
Re((
1/t
t
-1)csc(πt))t.
In[]:=
Timing[CMRB-INIntegrate[Re[Csc[πt](t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision500,Method"DoubleExponential"]]
Out[]=
{2.3125,-3.00×
-496
10
}

0=
∞
∫
1
(
1/t
t
-1)Re(csc(πt))t.

Timing[N[INIntegrate[(Re[Csc[πt]])(t^(1/t)-1),{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{1.85938,0}

Q=
∞
∫
1
Re(
1/t
t
-1)csc(πt)t=
∞
∫
1
Im((1-
1/t
t
)csc(πt))t.

Timing[Q=N[INIntegrate[Csc[πt]Re[(t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{23.0781,-0.10140904834001897354784794206523556559332349957408439415056089296571421188649799373309577398669321526864592292038098918512543392706217249642683673371900540797185401511906146581436666108354602466457454608304564559215252062489969450647133679118113551001450006470728274511882073812791922316447197828371101025304131116541836872476946717657217445214250863386411181234407320824451107398817105828827311462932592569333983845794888828277136232512108579431744492548432182332462941559129986359369813989517742364}
Timing[Q-N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision1000],500]]
Out[]=
{23.7656,0.×
-501
10
}

∞
∫
1
(
1/t
t
-1)csc(πt)t-CMRB-Q=0.
INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]-Q-CMRB
Out[]=
9.3471×
-94
10
+0.×
-101
10

INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074916-0.1014090483400189735478479420652355655933234995740843941505608929657142118864979937330957739866932152
NIntegrate[Csc[πt]Im[(1-t^(1/t))],{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074916

CMRB=
∞
∫
1
(
1/t
t
-1)csc(πt)t-
∞
∫
1
Im((1-
1/t
t
)csc(πt))t.

CMRB-(INIntegrate[Csc[πt](t^(1/t)-1),{t,1,InfinityI},WorkingPrecision100,Method"DoubleExponential"]-Quiet[N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision200],100]])
Out[]=
5.3×
-99
10
-0.×
-101
10


Interestingly, that is the same as

CMRB=
∞
∫
1
Im(1-
1/t
t
)csc(πt)t.

CMRB=
∞
∫
1
Im(1-
1/t
t
)csc(πt)t.
was proved -> here,<- considering
∞
∫
1
g(t)t
=

∞
∫
1
g(t)t
.

Thus we have


∞
∫
1
Re(
1/t
t
-1)csc(πt)t=
∞
∫
1
Im((1-
1/t
t
)csc(πt))t

In[]:=
Quiet[N[INIntegrate[Csc[πt]Re[(t^(1/t)-1)],{t,1,InfinityI},WorkingPrecision1000],500]]-Quiet[N[NIntegrate[Im[Csc[πt](1-t^(1/t))],{t,1,InfinityI},WorkingPrecision200],100]]
Out[]=
0.×
-101
10

CMRB=1/2+
∞
∫
1
Im((1-
2n/t
t
)csc(πt))t/.nϵZ
*
-

Related integral equations include

Out[]//TableForm=
It might be hard to prove a pattern here.
CMRB=NSum[Cos[Pin](n^(1/n)-1),{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision1000];

g(x)=
1/x
x
,CMRB=
1
2

∞
∫
1
(g(t)-g(2-t))csc(πt)t

In[]:=
g[x_]=
1/x
x
;Timing[CMRB-1/2INIntegrate[(Csc[πt])(g[t]-g[2-t]),{t,1,InfinityI},WorkingPrecision1000,Method"Trapezoidal"]]
Out[]=
{8.42188,3.×
-998
10
}

is faster than the following where you have to have WorkingPrecision2000 for 1000 accurate digits.

g(x)=
1/x
x
;CMRB=
∞
∫
1
Re(g(t)csc(πt))t

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.
​
​