Using this
,
we get the following for the MRB constant (CMRB).
f(x_)=-1;g(x_)=exp(πx)f(x);
1/x
x
CMRB=g(n)
∞
∑
n=1
then
g1(x_)=;
1/x
x
CMRB=t
1
2
∞
∫
0
g1(1-t)-g1(t+1)
sin(πt)
then finally,
f1
(
x_
)
=
1-
1/x
x
;
g2(x_)=
f1(1-x)
exp(πx)-exp(-πx)
;
u:=1-t
;
CMRB=2t.
1
∫
0
Img2
t
u
2
u
Using this
we get the following, where m is the MRB constant (CMRB).
f(x_)=-1;g(x_)=exp(πx)f(x);
1/x
x
m=g(n)
∞
∑
n=1
In[]:=
f[x_]=-1+x^(1/x);g[x_]=f[x]Exp[PiIx];m=N[NSum[g[n],{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision107],100]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074969
gs(x_)=;
1/x
x
m=t
1
2
∞
∫
0
gs(1-t)-gs(t+1)
sin(πt)
In[]:=
gs[x_]=x^(1/x);m-I/2NIntegrate[(gs[(1-t)]-gs[(1+t)])/(Sin[πt]),{t,0,InfinityI},WorkingPrecision100]
Out[]=
0.×
-101
10
fp(x_)=1-;gt(x_)=;u:=1-t;
1/x
x
fp(1-x)
exp(πx)-exp(-πx)
m=2t
1
∫
0
Imgt
t
u
2
u
In[]:=
fp[x_]=1-x^(1/x);gt[x_]=(fp[(-xI+1)])/(Exp[Pix]-Exp[-Pix]);u:=1-t;m-2NIntegrate[Im[gt[t/u]]/u^2,{t,0,1},WorkingPrecision100]
Out[]=
0.×
-101
10