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?
 A: 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
$$\int_0^\infty A(x) x^{-s-1}dx=\lim_{N \to \infty} \int_0^\infty \frac{N-B_N(x)}{N} x^{-s-1}dx = \frac{1}{-s}\lim_{N \to \infty}\frac{1}{N}\sum_{n \le N} (\sigma(n)/n)^{-s}$$
A: 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. $\spadesuit$
