Hidden inside the prime numbers is a Mellin transform, and
hidden inside that transform is the Prime Number Theorem.
This post follows that trail from an elementary Tauberian
lemma through contour integration and the analytic structure of the Riemann
zeta function. By carefully extracting the main term from an averaged
prime-counting function, we arrive at the celebrated asymptotic law
Alfred Tauber (1866–1942) was an Austrian mathematician best
known for his foundational work in mathematical analysis, particularly in
summability theory.
A Tauberian Lemma
Lemma: Let 
be a positive non-decreasing
function, and as 
let
Then
Proof: Take 
a positive number. The
assumption
yields
for 
, with 
depending on .
Take 
then
Since 
is non-decreasing
Thus,
Send 
Send 
Similarly, by considering
we obtain
This
Example: Let 
Then
But
so the lemma applies:
Example: Consider
with 
, where 
.
Let 
. Then
or
which implies that
That is,
Abelian Theorems go from
the behavior of a function to the behavior of its transform.
Tauberian Theorems go in the
reverse direction: the deduce the behavior of the underline function from the
behavior of its average or integral. Because information is lost when you
integrate, reversing the arrow requires an extra constraint known as the Tauberian
condition.
The lemma requires is positive and non- decreasing.
Without the Tauberian condition, the
integral can behave nicely while the function 
oscillates wildly or grows erratically without
settling down.
Let us see how this lemma helps us
prove the Prime Number Theorem.
Main topic
Recall A Mellin Transform Hidden in the
Prime Distribution,
where we derived the Mellin transform:
with
and
analytic for 
.
Also, we established that
for and fixed 
. Hence 
.
This result, 
, originates
from a sequence of nested integrals.
Starting from the base function:
we defined successively
which are connected through the identity:
Provided 
is holomorphic in a
vertical strip 
and has sufficient decay
on vertical lines (see Mellin Transform in The Riemann Hypothesis Revealed),
the Mellin inversion formula gives
Substituting ,
A cleaner form follows from 
:
Since
we obtain
Notice what this means conceptually:
1. In the contour deformation whose right
edge is the Mellin inversion line, the residue theorem picks up residues coming
from the poles of 
occurring at 
, every nontrivial zero 
of 
, and at every trivial zero 
is not single-valued
there. We must first choose branch cuts emanating from the zeros. The contour
has to detour around those cuts. Thus does not contribute
residues at zeros. Instead, it contributes integrals along the branch cuts.
Incidentally, we can avoid this.
To justify the Mellin inversion
rigorously, the key step is proving that
is absolutely integrable on the vertical line 
.
Since we’ve already shown (see A Mellin Transform Hidden in the
Prime Distribution)
that 
is bounded throughout the half-plane , the only serious issue is
controlling
on that line.
Here, 
, so for fixed 
and 
. Thus, the factor has constant modulus along the vertical line.
If we take 
, then since (as shown, Elementary Asymptotic Analysis of the
Riemann Zeta Function and its Derivative)
controlling 
becomes immediate.
Indeed, these estimates
imply for large 
,
since 
is bounded and 
term is 
. Consequently,
Now,
On the remaining finite
interval 
(the choice of 
is arbitrary), the
integrand 
is continuous and bounded,
so its integral is finite.
Therefore
remains bounded for 
, away from the singularity at 
.
■
So, 
is available and the
formula
is valid without invoking any
information near the critical strip.
Recovering the Mellin inversion
formula
Typically, the inversion formula is
recovered by integrating over a rectangular contour with vertices 
and taking 
. Nevertheless, special care must be
taken to investigate how the integrand behaves near .
In the neighborhood of :
For 
(see: The Riemann Hypothesis Revealed, Zeta’svalues near one)
hence
and since 
near 
(here 
), we have
Furthermore, as shown in A Small Tast from My New Book: Season 2
Episode 7,
where 
, is a root of zeta in the
critical strip, 
and 
. Thus, near 
1. The sum of zeros is empty
2. The error
term, when 
, 
.
Thus, we can write
or
where
as 
and remains bounded for 
, 
. In particular, 
for 
.
■
Putting everything together, for
To recover 
, we proceed in two steps.
Step 1: Evaluate 
.
Notice that
So on the left of we have a first order pole at and a second order pole at 
, hence
and
Thus,
Step 2: Evaluate
.
Normally we consider the rectangular
contour with vertices
The left edge approaches the line 
, where 
has the logarithmic infinity at . To avoid this point, we indent the
contour by a semicircle of radius centered at . Cauchy’s theorem may then be
applied to the resulting contour. Finally, letting and 
, we recover the above integral as
part of an identity.
Let’s see the details.
Over the segments 
and 
Notice that
Thus,
which means: no contribution from the
horizontal segments.
The vertical segment 
with the indentation by the semicircle at .
As
Thus,
■
Consequently,
The estimate.
Set
First we need to check integrability near 
and as 
Integrability
near 
. The local expansion gives
and
hence we have
The denominator 
is smooth and nonzero at .
Thus,
near zero.
Since
(
) the function 
is integrable near zero.
Integrability
as 
.
The assumption that 
for 
gives
which is integrable at infinity.
So 
.
By the Riemann-Lebesgue lemma,
Consequently,
and hence
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