# Explaining an Integral Involving the Divisor Function

In a 1973 paper by Martinet, Deshouilliers and Cohen, $A(x)$ is defined as $$A(x)=\lim_{N\to\infty}\frac{\#\{n\leq N\mid \frac{\sigma(n)}{n}≥x \}}{N}$$ where $\sigma(n)$ is the "sum-of-divisors" function $\sigma(n)=\sigma_1(n)=\sum_{d \mid n}d$. It is then stated that the following is "easily provable": $$\int_{0}^\infty x^s A(x)\, \mathrm{d}x=\frac{1}{s+1}\lim_{N\to\infty}\frac{1}{N}\sum_{n≤N}\Big(\frac{\sigma(n)}{n}\Big)^{s+1}$$ However, I cannot see the so-called "easy" proof of this statement. How is this link established?

• This is a good question. The short answer is "this is a Mellin transform." I'm writing a longer answer now. – davidlowryduda Jun 28 '16 at 19:15

The relation $$\phi(s)=\int_0^\infty x^s f(x) \frac{dx}{x} \tag{1}$$ is an example of a Mellin transform. This is something like the Fourier transform, except in different corrdinates. Just as there is an inverse Fourier transform, there is also an inverse Mellin transform. For $f$ and $\phi$ "nice" and satisfying $(1)$, they also satisfy $$f(x)=\frac{1}{2\pi i} \int_{c - i\infty}^{c + i \infty} x^{-s} \phi(s) ds. \tag{2}$$ Similarly, nice $f$ and $\phi$ satisfying $(2)$ also satisfy $(1)$.

So your question is claiming that the Mellin transform of $xA(x)$ is $\frac{1}{s+1} \lim_{N\to \infty} \frac{1}{N} \sum (\sigma(n)/n)^{s+1}$.

I think it's actually easier to show that the inverse Mellin transform of $\frac{1}{s+1} \lim_{N\to \infty} \frac{1}{N} \sum (\sigma(n)/n)^{s+1}$ is $x A(x)$, which is what I'll do below. But to do this, I need to mention one particularly well-known inverse Mellin transform.

Generally, $$\frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} y^{-s} \frac{ds}{s} = \begin{cases} 1 & \text{ if } 0 \leq y < 1 \\ \frac{1}{2} & \text{ if } y = 1 \\ 0 & \text{ if } y > 1. \end{cases} \tag{3}$$ I say this is particularly well-known because this is the key idea behind Perron's Formula, which has played a pivotal role in analytic number theory. It is somewhat common to ignore the $y = 1$ case by simply not considering the case when $y = 1$.

We are now ready to prove their claim. We want to show that $$x A(x) = \frac{1}{2\pi i} \int_{c - i \infty}^{c + i \infty} x^{-s} \lim_{N\to \infty} \frac{1}{N} \sum_{n \leq N} \left( \frac{\sigma(n)}{n}\right)^{s+1} \frac{ds}{s+1},$$ or equivalently (with some convergence conditions that I ignore in order to get to what's really going on), $$A(x) = \lim_{N \to \infty} \frac{1}{N} \sum_{n \leq N} \frac{1}{2\pi i} \int_{(c)} \left( \frac{x}{\sigma(n)/n}\right)^{-(s+1)} \frac{ds}{s+1}.$$ The integral here is essentially $(3)$ but with the change of variables $s \mapsto s + 1$ and with $x (\sigma(n)/n)^{-1} = y$. So we have $$\frac{1}{2\pi i} \int_{(c)} \left( \frac{x}{\sigma(n)/n}\right)^{-s-1} \frac{ds}{s+1} = \begin{cases} 1 &\text{ if } \sigma(n)/n > x \\ \frac{1}{2} &\text{ if } \sigma(n)/n = x \\ 0 &\text{ if } \sigma(n)/n < x.\end{cases}$$ Supposing for ease that we don't choose an $x$ which happens to be equal to $\sigma(n)/n$ for any $n$, then the integral is exactly the indicator function for $\sigma(n)/n > x$. But then the right hand side becomes $$\lim_{N \to \infty} \frac{1}{N} \sum_{n \leq N} [1 \text{ if } \sigma(n)/n > x],$$ which is exactly the definition of $A(x)$.

Since $\frac{1}{s+1} \lim_{N\to \infty} \frac{1}{N} \sum (\sigma(n)/n)^{s+1}$ is the inverse Mellin transform of $xA(x)$, we also get that $xA(x)$ is the Mellin transform of $\frac{1}{s+1} \lim_{N\to \infty} \frac{1}{N} \sum (\sigma(n)/n)^{s+1}$, as we wanted to show. $\spadesuit$

Let $B_N(x) = \# \{ n \le N, \sigma(n)/n < x\}$. Integrating by parts $$\int_0^\infty B_N(x) x^{-s-1}dx = B_N(x)\frac{x^{-s}}{-s} |_0^\infty +\int_0^\infty \frac{x^{-s}}{s}\underbrace{dB_N(x)}_{=B_N'(x)dx} = \frac{1}{s}\sum_{n \le N} (\sigma(n)/n)^{-s}$$

and hence for $\Re(s)$ large enough $$\int_0^\infty B(x) x^{-s-1}dx=\lim_{N \to \infty} \int_0^\infty \frac{B_N(x)}{N} x^{-s-1}dx = \frac{1}{s}\lim_{N \to \infty}\frac{1}{N}\sum_{n \le N} (\sigma(n)/n)^{-s}$$ and for $-\Re(s)$ large enough, assuming it converges $$\int_0^\infty A(x) x^{-s-1}dx=\lim_{N \to \infty} \int_0^\infty \frac{N-B_N(x)}{N} x^{-s-1}dx = \frac{1}{-s}\lim_{N \to \infty}\frac{1}{N}\sum_{n \le N} (\sigma(n)/n)^{-s}$$

• +1. Mellin inversion is overkill. You can think of this as how partial summation allows one to write $\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = s \int_{1}^{\infty} \sum_{n \leq x} a_n x^{-s - 1} \, dx$. Mellin inversion (or more precisely the Perron inversion formula) is the "harder" bit, which is inverting this relation. – Peter Humphries Jul 3 '17 at 2:56