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We get the following, where m is the MRB constant and m2 is its integrated analog.
f(x_)=-1;g(x_)=exp(πx)f(x);
1/x
x
m=g(n)
∞
∑
n=1
fp(x_)=1-;
1/x
x
u:=1-t;v:=(x_)=;
t
1-t
g
1
fp(1-x)
exp(πx)-exp(-πx)
m=2t
1
∫
0
Im((v))
g
1
2
u
f(x_)=-1;g(x_)=exp(πx)f(x);
1/x
x
m2=g(x)x
∞
∫
1
f(x_)=-1;
1/x
x
u=1-t;v:=;(x_)=;
t
1-t
g
2
f(1+v)
exp(πv)
m2=-(t)t
1
∫
0
g
2
2
u
f(x_)=-1;g(x_)=exp(πx)f(x);
1/x
x
m=g(n)
∞
∑
n=1
In[]:=
f[x_]=x^(1/x)-1;g[x_]=f[x]Exp[PiIx];m=N[NSum[g[n],{n,1,Infinity},Method"AlternatingSigns",WorkingPrecision107],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=2Im(gt(v))t
1
∫
0
2
u
In[]:=
fp[x_]=1-x^(1/x);[x_]=(fp[(-xI+1)])/(Exp[Pix]-Exp[-Pix]);u:=1-t;v:=(t/(1-t));m-2NIntegrate[Im[[v]]/u^2,{t,0,1},WorkingPrecision100]
g
1
g
1
Out[]=
0.×
-101
10
f(x_)=-1;g(x_)=exp(πx)f(x);m2=g(x)x
1/x
x
∞
∫
1
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},WorkingPrecision51]
Out[]=
0.0707760393115288035395280218302820013657546962033630-0.0473806170703507861072094065026036785731528996931736
f(x_)=-1;u=1-t;v:=;gs(x_)=;m2=-t
1/x
x
t
1-t
f(1+v)
exp(πv)
1
∫
0
gs(t)
2
u
In[]:=
f[x_]=x^(1/x)-1;u=1-t;v:=(t/(1-t));[x_]=f[(1+vI)]/Exp[Piv];Timing[m2-(-INIntegrate[[t]/u^2,{t,0,1},WorkingPrecision51])]
g
2
g
2
Out[]=
{0.125,0.×+0.×}
-52
10
-52
10