We get the following, where m is the MRB constant and m2 is its integrated analog.
f(x_)=
1/x
x
-1;g(x_)=exp(πx)f(x);
​​
m=
∞
∑
n=1
g(n)
fp(x_)=1-
1/x
x
;
​​
u:=1-t;v:=
t
1-t
g
1
(x_)=
fp(1-x)
exp(πx)-exp(-πx)
;
​​
m=2
1
∫
0
Im(
g
1
(v))
2
u
t
f(x_)=
1/x
x
-1;g(x_)=exp(πx)f(x);
​​
m2=
∞
∫
1
g(x)x
f(x_)=
1/x
x
-1;
​​
u=1-t;v:=
t
1-t
;
g
2
(x_)=
f(1+v)
exp(πv)
;
​​
m2=-
1
∫
0
g
2
(t)
2
u
t
​​​​​​
f(x_)=
1/x
x
-1;g(x_)=exp(πx)f(x);
​​
m=
∞
∑
n=1
g(n)
In[]:=
f[x_]=x^(1/x)-1;g[x_]=f[x]Exp[PiIx];m=N[NSum[g[n],{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision107],100]
Out[]=
0.1878596424620671202485179340542732300559030949001387861720046840894772315646602137032966544331074969
fp(x_)=1-
1/x
x
;
gt(x_)=fp(1-x)/(exp(πx)-exp(-πx))
;
u:=1-t
;
v:=(t/(1-t));
​​
m=2
1
∫
0
Im(gt(v))
2
u
t
In[]:=
​​fp[x_]=1-x^(1/x);
g
1
[x_]=(fp[(-xI+1)])/(Exp[Pix]-Exp[-Pix]);u:=1-t;v:=(t/(1-t));m-2NIntegrate[Im[
g
1
[v]]/u^2,{t,0,1},WorkingPrecision100]
Out[]=
0.×
-101
10
f(x_)=
1/x
x
-1;g(x_)=exp(πx)f(x);​​m2=
∞
∫
1
g(x)x
In[]:=
f[x_]=x^(1/x)-1;g[x_]=f[x]Exp[PiIx];NIntegrate[g[x],{x,1,Infinity}]
Out[]=
0.070776-0.0473806
In[]:=
f[x_]=x^(1/x)-1;m2=NIntegrate[(-1)^xf[x],{x,1,InfinityI},WorkingPrecision51]
Out[]=
0.0707760393115288035395280218302820013657546962033630-0.0473806170703507861072094065026036785731528996931736
f(x_)=
1/x
x
-1;​​u=1-t;v:=
t
1-t
;gs(x_)=
f(1+v)
exp(πv)
;​​m2=-
1
∫
0
gs(t)
2
u
t
In[]:=
f[x_]=x^(1/x)-1;​​u=1-t;v:=(t/(1-t));
g
2
[x_]=f[(1+vI)]/Exp[Piv];Timing[m2-(-INIntegrate[
g
2
[t]/u^2,{t,0,1},WorkingPrecision51])]
Out[]=
{0.125,0.×
-52
10
+0.×
-52
10
}