TheLaplacetransformanalogytotheCMRB
Given the following expression.
In[]:=
ReImPlot[
x
(-1)
(
1/x
x
-1),{x,-1,1}]
Out[]=
-1.0
-0.5
0.5
1.0
-6
-4
-2
2
4
6
Consider the term
f[x_]=
x
(-1)
(
1/x
x
-1);
of the MRB constant
In[]:=
CMRB=
∞
∑
x=1
f[x];
In[]:=
N[CMRB]
Out[]=
0.18786
In the above replies we investigated
M1=
∞
∫
1
x
(-1)
(
1/x
x
-1)x
M2=
∞
∫
1
x
(-1)
(
1/x
x
-1)x
Out[]=
∞
∫
1
x
(-1)
-1+
1
x
x
x
In[]:=
Quiet[N[M1]]
Out[]=
0.0707768-0.047384
WhataboutthefollowingLaplacetransformanalogy,M0,totheCMRB,​​​​M0=(
ℒ
x
[
1/x
x
-1](s)/.s-π)​​=

t
-
0
∞
∫
t
x
(-1)
(
1/x
x
-1)x​​=

t
-
0
∞
∫
t
Exp[IPi](
1/x
x
-1)x?
In[]:=
Needs["NumericalCalculus`"];M0=Quiet[NLimit[Integrate[f[x],{x,t,Infinity}],t->0,Direction-1]]
Out[]=
-0.168485-0.484112
Thusly, we are inquiring about
In[]:=
M0=
ℒ
x
[
1/x
x
-1](s)/.s-π;
In[]:=
M0a=N[LaplaceTransform[(
1/x
x
-1),x,s]/.s->-IPi]
Out[]=
-0.168483-0.484114
In[]:=
Mn=Quiet[NLimit[NIntegrate[f[x],{x,t,1}],t->0,Direction-1]]
Out[]=
-0.239261-0.436734
Mn=
lim
t
+
0
1
∫
t
f(x)x
1
∫
t
f(x)x
Then it makes perfect sense that (
M0=
ℒ
x
[
1/x
x
-1](s)/.s-π
)-
Mn=
lim
t
+
0
1
∫
t
f(x)x
-
M2=
∞
∫
1
x
(-1)
(
1/x
x
-1)x
=0.
Rationalize[N[M0-Mn-M2],
-5
10
]
Out[]=
0