1. SOME THOUGHTS ON DIVERGENT AND ASYMPTOTIC
SERIES
Jacob Barnett
It is my intention of writting this paper, to be able to give a convenient method in
order to apply asymptotic and divergent series to solve integral equations like
∞
this g ( s ) = s ∫ dtf (t ) K ( st ) or in order to extend the prime number theorem to be
0
able to evaluate asymptotic sums in the form ∑p
p≤ x
k
, also i would like to give a
∞
∫x
m
finite meaning to divergent integrals of the form dx for integer m.
a
In order to do that, first i would need to define a generalized Borel transform of a
divergent series in the form
∞ ∞
 ∞ a  ∞
∑ an = ∫ dx  ∑ n x n  f ( x) with ∫ dx f ( x) x
n
= Ψn (1)
n =0 0  n =0 Ψ n  0
In case f(x) is just the exp(-x) one recovers the normal Borel transform, my next
steep then is to consider the following asumptions.
1) the Borel generalized transform of a convergent series will be equal to
the sum of this series
2) i will use a function g(s) that admits an expansion of the form
∞
an
g ( s ) = ∑ n , this is very convenient for our purposes
n =0 s
∞
then for our integral equation of the form g ( s ) = s ∫ dtf (t ) K ( st ) , using our
0
assumptions 1) and 2) we have the result
∞
an ∞  ∞ a  ∞
g (s) = ∑ = ∫ dt  ∑ n n t n K (t ) and K (n) = ∫ dtK (t )t n
ˆ (2)
n=0 s
n
0
ˆ
 n = 0 s K ( n)  0
Then from the definition (2) , we may consider that the solution to the integral
∞ ∞
a
equation g ( s ) = s ∫ duf (u ) K ( su ) will be given by ∑ n t n = f (t ) (after a
0
ˆ
n = 0 K ( n)
t
change of variable = u ) . As an example, let be the integral equation for the
s

2. ∞
π (e t )
prime counting function evaluated at epx(-t) ln ζ ( s ) = s ∫ dt , the left side
0
e st − 1
is just the natural logarithm of the Riemann Zeta function, in this case
∞
K (n) = Γ(n + 1)ζ (n + 1) = ∫ dt (et − 1) −1 t n , following this idea the solution to the
ˆ
0
∞
ln k ( x)
prime number counting function should be of the form π ( x) ≈ ∑ ,
k =1 k !ζ ( k + 1) k
1
from the prime number theorem for big n we have that an ≈ , if one sets
n
t = e x inside the solution one gets the Taylor series for the exponential integral
(except a constant) . Of course the inverse can also be made, given any power
series, we can obtain the integral equation it fullfills, for example
∞
( − x) n ζ (2n) ∞
 1
 
f ( x) = −∑ , using the definition = − ∫ dtfrac   t n −1 , Here
n =1 ( n − 1)!ζ (2 n) 2n 0  t
 
the fractional part of a number is given by the x − [ x] , then using the fact that
for a convergent power series its Borel transform must be equal than its sum
defined in the normal sense we find
∞
(− x)n
∞
dt  ∞ ( −t ) n   x
e − 1 = −∑
−x
= ∫ ∑  frac 
 t 

(3)
n =1 n! 0
t  n =1 ζ (2n)(n − 1)!   
To end this section , i would like to show an strange property of sums and Mellin
transforms, for example in order to solve the integral equation
∞
g ( s ) = s ∫ dtf (t ) K ( st ) , we can turn this into an Algebraic equation by using the
0
∞
∫x ˆ
s −1
Mellin transform of a function defined by, dxf ( x) = F ( s ) for the case of our
0
∞
integral equation g ( s ) = s ∫ dtf (t ) K ( st ) , this is equivalent to the Algebraic
0
ˆ ˆ ˆ
equation G ( s ) = K ( s ) F (1 − s ) , if we have a function that can be expanded as an
∞
alternating power series g ( x) = ∑ (−1) n a (n) x n , using the example (2)
n=0
∞ ∞
∞
dt  ∞

g ( x) = ∑ (−1) a (n) x = ∫  ∑ (−1) n t n x n K (t )
n n
with a (n) = ∫ dtK (t )t n −1 (4)
n =0 0
t  n =0  0
1
The series inside the first equation in (4) is just equal to and its Mellin
1 + xt
∞
t s −1 π
transform is just ∫ dt
0
=
1 + t sin(π s )
, so aplying the Mellin transform inside (4)

3. ∞
 ∞  π
we get dx  ∑ (−1)n a (n) x n  x n −1 =
∫  n =0 a (− s ) , i think this can be very
0  sin(π s )
interesting to evaluate integrals from the Taylor series of the intengrand.
Prime number theorem and beyond:
For me, one of the most interesting parts of mathematics are primes, it is
wonderful how although we can not give the number of primes less than a given
quantity x in a computable way, we still can know their asymptotic behaviour in
x
dt
the form π ( x) ≈ ∫ = Li ( x ) , from this equation we can get the probability of
2
ln t
π ( x) 1
that a randomly chosen integer will be a prime ≈ , in this sectio i will
x ln x
try generalizing the Prime number theorem to include approximates value for
the sums ∑ p k with k a real and positive number, first of all i need the
p≤ x
identities
x m dx ∞
(m + 1) k ln k x dx
∫ ln x = ln ln( x) + ∑ ∫ x ln x = ln ln( x) (5)
k =1 k !.k
Then , in order to obtain an asymptotics of the sums ∑p
p≤ x
k
, considering that
x m dx
powers of x are smooth , i shall replace the sum by an integral ∫ ≈ ∑ pm ,
ln x p ≤ x
here the idea is to weight the sum by means of an integral involving the
π ( x) 1
probability that a number is prime, which is just ≈ i this sense for
x ln x
positive m and for the product of all primes less than x we find
x m dx Li ( x m +1 )  d  Li ( x m +1 )  
∫ ln x ≈ ∑x p ≈ m + 1 ∏ p ≈ exp  dm  m + 1  
m
(6)
  
p≤ p≤x
   m=0
The last sum in (6) is just to take the formal derivative with respect to m at m=0
to get ∑ p m → ∑ ln p , and then exponentitate, for m=-1 using the expression
p≤x p≤ x
(5) ∑p
p≤x
−1
≈ ln ln x , the sum is still divergent , but it diverges even slower than
∞
1
the Harmonic sum ∑n
n =1
. For a function that can be expanded into Taylor
series near x=0 the following asymptotic formula holds .
∞
Li ( x n +1 ) d n f (0)
∑x
p≤
f ( p) ≈ ∑
n = 0 ( n + 1).n ! dx n
, if the sum over all primes is convergent , it can be
calculated by resummation of Taylor series taking a big big x.

4. ∞ ∞
Sum of a divergent series ∑ nk and a divergent integral
n =1
∫ x dx
k
:
a
I know that , perhaps many of the readers will think i am a lunatic if i say that i
have managed to sum divergent series like (for k >0) the following
− Bn +1
1 + 2 k + 3k + ................ = = ζ (− n) for integer n (7)
n +1
However there is a method to obtain (7) with no much difficulty, first we need
the following two identities involving the exponential of the derivative operator
 d  ∞
d n f ( x) a n ∞
 d  1
exp  a  f ( x) = f ( x + a) = ∑ ∑ exp  n dx  =
  (8)
 dx  n =0 dx n n ! n =0 d 
1 − exp  
 dx 
∞
x B
Together with the expansion for the Bernoulli numbers = ∑ n xn ,
exp( x) − 1 n =0 n !
with all this we can obtai the ‘sum’ ( i will call it the sum even the series may be
divergent) of any function in the following form
d
∞ ∞
 d  1
∑
n =0
f ( x + n) = ∑ exp  n  f ( x) =
n =0  dx  d 
dx
d 
f ( x) (9)
  1 − exp  
 dx   dx 
Setting f ( x) = x m , and evaluating the sum at n=0 we obtain
−B
1 + 2 k + 3k + ................ = n +1 valid for every integer n , for n=-1 , for the
n +1
Harmonic series i can use the definition of Euler constant
∞
1
∑ n − ln ∞ = γ = 0.5772... then in order to obtain a finite number for the divergent
n =1
∞
1
series i throw away the logarithm and keep only the finite part so ∑ = 0.5772...
n =1 n
∞
∫x
m
This method can be extended to divergent integrals in the form dx , if we
a
m
insert f ( x) = x inside the Euler-Maclaurin summation formula
∞ ∞ a
m − s m −1− s
∫x dx + ζ ( s − m) − ∑ i m − s + a m − s
2 ∫
m−s
dx = x
a a i =1
∞
a Natural number (10)
∞
B2 r Γ(m − s + 1)
−∑ (m − 2r + 1 − s ) ∫ x m − 2 r − s dx
r =1 (2r )!Γ ( m − 2r + 2 − s ) a

5. ∞
∫x
m
This expression (10) is a recurrence formula for the divergent integrals dx ,
a
here s is a regulator so for big s all the integrals are convergent , is we take the
∞
− Bn +1
limit as s approaches to 0 and use the definition obtained before ∑ i n . =
i =1 n +1
valid for n=0 or positive we can manage to regularize any divergent integral of
∞
∫x
m
the form dx for every m except m=-1.
a
For m=-1 if we use the Stirling series for Gamma function
 1 1 ∞
B2 r x1− 2 n
log Γ( x) =  x −  log x − x + log ( 2π ) + ∑ (11)
 2 2 r =1 2 r ( 2 r − 1)
Taking the derivative with respect to x and inserting it inside the EulerMaclaurin
∞
dx
summation formula , we find for the case m=-1 ∫ = − ln a , in fact if we take
0
x+a
this strange formula for granted we recover the following result for the Harmonic
∞
1 Γ '(a )
series (usng the EulerMaclaurin summation formula again) ∑ =− ,
n =0 n + a Γ( a )
∞
1
for a=1 i get ∑ = 0.5772... as i got previously for the Harmonic series.
n =1 n
For more complicate functions, let us suppose we can expand
k
f (u ) = ∑c
m =−∞
m (u + a )m by means of a Laurent convergent series , then the
∞
∫ duf (u ) ( u + a )
−s
expression for positive a can be regularized using formulae
0
k ∞
(10) and (11) , with the change of variable u + a = x , it becomes ∑
m =−∞
cm ∫ x m dx ,
a
for m smaller than -1 there is no problem in defining the integrals , for m bigger
than -1 or equal to m=-1 i use formulae (10) and (11) to obtain finite results.
For the case of multiple integrals in several variables
−s
∞ ∞ ∞
 n 
∫ 0
dx1 ∫ dx2 ..........∫ dxn G ( x1 , x2 ,......, xn )  a +
 ∑  xi2  a positive integer (12)
0 0  i =1 
The idea is to define a big s, so the integral will be convergent and by the rules
of Calculus we can make a change of variables to n-dimensional hyperspherical
−s
coordinates G (r , θ1 ,......, θ n −1 )r n −1 ( a + r ) if we are lucky and the integral is

6. ∞
∫ drG(r )r ( a + r )
n −1 −s
invariant under rotation , then (12) becomes , this is just a
0
1-dimensional integral that can be again regularized by formulae (10) and (11),
unfortunately in many cases the integrand will not be invariant under rotation, so
we will have to approximate the integral over the angles by a sum dΩ ≈ ∑ Ωi in
i
order to get an integral that will depend only on the modulus of the position
∞
∑ ∫ drG
−s
vector ii ,i2 ,......,in−1 (r )r n −1 ( a + r ) , the problem with this approach is that
i1 ,i2 ,......,in−1 0
we replace a multiple integral on R n by a sum of 1-dimensional integral, for big s
the inetgral will be convergent, the idea here is to analytically continue to the
case s=0 in order to regularize and give a finite meaning to divergent integrals
(in one or several variables) , the change of variable to hyperspherical
coordinates is allowed by the rules of calculus, i have not made any illegitimate
manipulations of the integral so i am very confident that my method will work
and that i have respected all the mathematical rigour of analytic continuation
and calculus of one and several variables.
Riemann Hypothesis and Quantum mechanics:
I have always found interesting the connection between math and physics,
perhaps the most amazing one is the fact that the Riemann Hypothesis, can be
solved by finding a dynamical system whose quantization yields the Riemann
zeros (the imaginary part), i will give a brief sketch of my simple idea.
The idea to prove the Riemann Hypothesis by means of Quantum mechanics, is
d2
just to find a Hamiltonian in one dimension H = − 2 + f ( x) whose eigenvalues
dx
1 
are the roots of the equation ξ  + i E  = 0 , this fact is motivated by the
2 
expansion in a form of an infinite product of the Riemann Xi-function
1  1 ∞  E 
ξ  + i E  = ξ   ∏ 1 −  everytime that E = En this product is 0 , also if
2   2  n =0  En 
we take the logarithmic derivative of this product, and use the representation of
1  1 
the dirac delta function − ℑm   = δ ( x) , we find for the Xi-function
π  x + iε 
1 ξ ' 1  ∞ s ( s − 1)  s 
− ℑm   + i E + iε   = ∑ δ ( E − En ) ξ ( s ) = Γ   ζ ( s )π − s / 2 (13)
π  ξ 2   n =0 2 2
Formula (13) is the density of states of an hypothetical Quantum Hamiltonian in
onde dimension whose energies are the square of the imaginary part of the
2
Riemann zeros En = γ n , if i insert the last result (13) into the Riemann-Weyl
formula that relates primes and zeros of the Riemann zeta function i find

7. ζ 1  1 ζ ' 1  1
π ∑ δ ( E − γ 2 ) = reg  +i E  +  −i E  +
γ ζ 2 2 E ζ 2 2 E
(14)
Γ' 1 E 1 Γ' 1 E 1 log π
 +i
4 
 4 s + Γ  4 − i 2  4 s − 2 s = g(E)
 
Γ 2   
2
So in this case, if we could find a system whose energies were En = γ n , then the
Riemann Hypothesis would follow form the fact that the Hamiltonian operator is
Hermitian (only have real eigenvalues ) and from the expression
1  1 ∞  E 
ξ  + i E  = ξ   ∏ 1 −  , so the Riemann Xi-function evaluated on the
2   2  n =0  En 
1
critical line s = + i z has only Real zeros . In order to get the function f ( x)
2
d2
inside the Hamiltonian H = − 2 + f ( x) , we will use the semiclassical Wilson-
dx
 1
Sommerfeld quantization conditions namely ∫ pdq =  n +  2π n = n( E ) , since
C  2
h
we are working in reduced units where h = = 2m = 1 for simplicity , in the
2π
case of the one dimensional Hamiltonian the Wilson-Sommerfeld quantization
condition yields to a integral equation not for the function f but for its inverse
d −3/ 2  df −1 ( x) 
E
 1  1 df −1
2π  n +  = ∫ pdq =  n +  2π = 2∫ E − x = π −3/ 2   (15)
 2 C  2 0
dx dx  dx 
From (15) and using the group property (assuming it is always valid)
D a +b = D a .D b ,we can give the inverse of the function (potential) in the form
d 1/ 2  1  2 d  
1/ 2
1
f −1 ( x) = 2 π  + N ( x)  + 1/ 2 
arg ζ  + i x +ε  (16)
dx1/ 2  2  π dx  2 
ε → 0 , the first part is the smooth part of the potential , and it is equal to
1 1 i 
arg Γ  + x + ε  − x ln π , (16) can be then defined (in a more compact
π 4 2 
2 d 1/ 2  1  1
form) as f −1 ( x) = 1/ 2 
arg ξ  + i x  +  as the argument of the Riemann
π dx  2  2
Xi-function evaluated on the critical line Re(s)=1/2 , but the first expression is
easier for computational purposes , from the Stirling’s formula the smooth part
of the inverse function is given by the asymptotic equality
x x x 1 1  1   1 +e 
N ( x) ≈ ln  
− − + + O  3/ 2  = O  x 2  e>0 (17)
2π  2π  2π 8 48π x
 x   

8. Of course , i will need to define a half-derivative , this can be made in the form
of a limit, for example the q-th order derivative can be thought as
q
∆ q f ( x) (e ε d / dx
− 1) 1  ∞ q 
q ∑
lim = lim f ( x) = lim (−1) n   f ( x + (q − n)ε )  (18)
ε →∞ εq ε →∞ εq ε →∞ ε
 n =0 n 
∞
q
I have simply used the binomial expansion ( x − 1) = ∑ (−1) n   x q − n with
q
n =0 n
 d   d 
x = exp  ε  and the property exp  a  f ( x) = f ( x + a) to get (18) without
 dx   dx 
any problem, this formula (18) allows us to compute the half derivative
numerically simply by setting q = 0.5 . For x ≤ 0 N(E)=0 , hence f −1 ( x) = 0 for
x ≤ 0 , the inverse of this funciton is just the barrier x=0 since x = f (0) = 0 , then
from our method the inverse of the potential inside the Hamiltonian
d2
H = − 2 + f ( x) is given by the formula
dx
 2 d 1/ 2  1  1
−1  1/ 2 
arg ξ  + i x  +  x>0
f ( x) =  π dx  2  2 (19)
 0 x≤0

I want to thank all the people that have helped me with my ideas, i have tried
submitting it to a math journal, however i have only received negative critics, in
fact a referee told me that the appropiate place for my ideas was the trash and
even insulted me as a retarded and abnormal, it is a pity that sometimes the
ideas are good or bad simply depending on the person who tells them , anyway
i would also like to thank all my future readers and the teachers who gave me
the opportunity to express myself on internet.