So far it has not been proven that
anything in mathematics is difficult.
Norbert a’Campo
Source: Riemann’s Zeta Function. H.M. Edwards 1974.
J(x)
is the function which begins at 0 for
x=0
and increases by a jump of 1 at primes, by a jump of
1
2
at prime squares, by a jump of
1
3
at prime cubes etc.
J(x)
is therefore a rational number.
Riemann denotes the function
J(x)
by
f(x)
in his 1859 paper: “On the number of primes less than a given magnitude”.
In[]:=
JRiemann[x_]:=Sum[With[{fi=FactorInteger[n]},If[Length[fi]==1,1/Last[First[fi]],0]],{n,2,x}]
Example:
In[]:=
JRiemann[10]
Out[]=
16
3
In[]:=
Plot[JRiemann[x],{x,0,20},Filling->Axis,PlotLabel->"JRiemann"]
Out[]=
J(x)
can be expressed by the prime counting function
π(x)
:
J(x)=
∞
∑
n=1
π(
1/n
x
)
n
=
log(2,x)
∑
n=1
π(
1/n
x
)
n
​The sum is finite because
π(
1/n
x
)=0
for
1/n
x
<2
.
In[]:=
QuietSolve
1/n
x
==2,n
Out[]=
n
Log[x]
Log[2]

In[]:=
Jπ[x_]:=SumPrimePi[
1/n
x
]n,{n,1,Log[2,x]}
Example :
In[]:=
Jπ[10]
Out[]=
16
3
In[]:=
Plot[Jπ[x],{x,0,20},Filling->Axis,PlotLabel->"Jπ"]
Out[]=
The Möbius inversion formula then gives:
π(x)=
∞
∑
n=1
μ(n)
n
J(
1/n
x
)=
log(2,x)
∑
n=1
μ(n)
n
J(
1/n
x
)
​The sum is finite because
J(
1/n
x
)=0
for
1/n
x
<2
.
In[]:=
πJ[x_]:=SumMoebiusMu[n]Jπ[
1/n
x
]n,{n,1,Log[2,x]}
Example:
In[]:=
πJ[100]
Out[]=
25
In[]:=
Plot[πJ[x],{x,0,20},Filling->Axis,PlotLabel->"πJ"]
Out[]=
Riemann gave an explicit formula for
J(x)
:
J(x)=Li(x)-
∑
ρ
Li(
ρ
x
)+
∞
∫
x
1
t(
2
t
-1)logt
t-log(2)
for
x>1
Li(x)
is the LogIntegral given by:
Out[]=
x
∫
0
1
log(t)
tli(x)
Ei(x)
is the ExpIntegralEi given by:
Out[]=
-
∞
∫
-x
-t

t
tEi(x)
This is an usefull identity:
Out[]=
Ei(log(x))li(x)
The main result of this notebook is the following identity:(*)
∞
∫
x
1
t(
2
t
-1)logt
t=-
∞
∑
n=1
Ei(-2nlogx)
for
x>1
First we calculate a series expansion around ∞ of the integrand:
In[]:=
Series
1
t(
2
t
-1)Log[t]
,{t,∞,10}
Out[]=
1
Log[t]
3
t
+
1
Log[t]
5
t
+
1
Log[t]
7
t
+
1
Log[t]
9
t
+
11
O
1
t

Now we integrate termwise:
In[]:=
Integrate
1
Log[t]
3
t
,{t,x,∞},Assumptions->x>1
Out[]=
-ExpIntegralEi[-2Log[x]]
In[]:=
Integrate
1
Log[t]
5
t
,{t,x,∞},Assumptions->x>1
Out[]=
-ExpIntegralEi[-4Log[x]]
So we conclude (
x>1
):
Out[]=
∞
∫
x
1
t(
2
t
-1)log(t)
t-
∞
∑
n=1
Ei(-2nlog(x))
Now you can see, that
-2n
,
nϵ
are the trivial zeros of the Zeta function:
In[]:=
{Zeta[-2],Zeta[-4]}
Out[]=
{0,0}
The non trivial zeros are called
ρ
with
Imρ≠0
:
In[]:=
{N[ZetaZero[1]],Zeta[ZetaZero[1]]}
Out[]=
{0.5+14.1347,0}
Here you can see some zeros and the simple pole in the complex plane:
Out[]=
Some zeros and the pole of the Zeta function
The first term is caused by the simple pole of Zeta at 1:
The second term is caused by the non trivial zeros and the third term is caused by the trivial zeros of the Zeta function. The forth term is simply:
or if you prefer the RiemannXi function:
Here is a working example:
Here 20 nontrivial zeros and 2 trivial zeros are used:
Here the Möbius inversion formula is again used:
Here 10 nontrivial zeros and 1 trivial zero are used:

Appendix PrimePi table:

CITE THIS NOTEBOOK

Riemann's explicit formula made more explicit​
by Oliver Seipel​
Wolfram Community, STAFF PICKS, January 3, 2024
​https://community.wolfram.com/groups/-/m/t/3095772