an integral expression for the sum of divisors function

The sum of divisors (sigma) function is difficult to work with in
analytic number theory because it is not readily expressible as an
analytic function.

Therefore I thought others might find this interesting: an integral
expression for the sum of divisors function that I discovered.

Recall that
sigma(p^k*q^j*...) = (p^(k+1)-1 / p-1) * (q^(j+1)-1 / q-1) *...,
where p,q,... are distinct primes.

1. Consider first a prime power p^k alone.

Let f(p^k) = kp^(k+1) - (k+1)p^k + 1

Then
the integral of f(p^k) / (1-p)^2 =
p^(k+1) - 1 / p-1 =
1 + p + p^2 +...+p^k =
sigma(p^k)

Note that 1/(1-p)^2 = 1 + 2p + 3p^2 + 4p^3 +...
(Of course the right hand side is not convergent for p>=1, so it's a
good thing we have the closed form expression on the left hand side.)

So we have
the integral of f(p^k) (1 + 2p + 3p^2 + 4p^3 +...) =
1 + p + p^2 +...+p^k =
sigma(p^k)

Also of interest is the derivative of f(p^k):

f'(p^k) = k(k+1)p^k(1-1/p)

Note in particular the term (1-1/p), the inverse of the Euler product
term 1 / (1-1/p).

2. Now consider a positive integer with two or more distinct prime
factors. For brevity of notation, I write the expressions below for
only two prime factors p^k*q^j, but they may be extended to express
any number of prime factors.

Here we have a multiple (iterated) integral expression for sigma
(p^k*q^j):

integral [ f(p^k) / (1-p)^2 (integral [f(q^j) / (1-q)^2] dq) ] dp =
integral [ f(p^k) / (1-p)^2 (sigma(q^j)) dp =
sigma(p^k)*sigma(q^j) =
sigma(p^k*q^j)

Geoffrey Caveney
.

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