The average fractional part of x/p

For a real number $y$ , write $\displaystyle \Big\{y\Big\}:=y-\lfloor y\rfloor$ for its fractional part. Assuming the prime number theorem, we shall prove that

$\displaystyle \frac{1}{\pi(x)}\sum_{p\le x}\Big\{\frac{x}{p}\Big\}\longrightarrow 1-\gamma\approx 0.4227\cdots,$

where $\gamma$ is Euler’s constant. The quantities $\displaystyle \Big\{x/p\Big\}$ fluctuate in a seemingly irregular way as $p$ runs through the primes. The useful observation is that, after dividing by $x$ , the primes up to $x$ become uniformly distributed through $\displaystyle \Big(0,1\Big]$ with respect to normalized prime counting measure. Since

$\displaystyle \Big\{\frac{x}{p}\Big\}=\Big\{\frac{1}{p/x}\Big\},$

the problem becomes one of averaging the fixed function $\displaystyle f(t):=\Big\{\frac1t\Big\}$ over these scaled primes. The only issue is that $f$ has infinitely many jumps accumulating at $0$ . We first avoid that endpoint, and then show that the discarded interval makes a negligible contribution.

Indeed, for fixed $\displaystyle 0<a<b\le1$ , the prime number theorem gives

$\displaystyle \frac{\pi(bx)-\pi(ax)}{\pi(x)}\longrightarrow b-a.$

Thus the proportion of primes $p\le x$ for which $\displaystyle p/x\in (a,b]$ tends to the length of that interval. By first proving the assertion for step functions and then approximating bounded Riemann-integrable functions by step functions, one obtains, for every fixed $\displaystyle 0<\delta<1$ and every bounded Riemann-integrable function $\displaystyle h:\Big[\delta,1\Big]\to\mathbb R,$

$\displaystyle \frac{1}{\pi(x)}\sum_{\delta x<p\le x}h\Big(\frac{p}{x}\Big)\longrightarrow\int_\delta^1h(t) dt.$

Apply this with $\displaystyle h(t)=\Big\{\frac1t\Big\}.$ On $\displaystyle [\delta,1]$ , this function has only finitely many discontinuities, at the reciprocal integers lying in that interval, and so it is Riemann integrable. Therefore

$\displaystyle \frac{1}{\pi(x)}\sum_{\delta x<p\le x}\Big\{\frac{x}{p}\Big\}\longrightarrow\int_\delta^1\Big\{\frac1t\Big\}dt.$

The primes below $\delta x$ do not affect the limit. Since $\displaystyle 0\le\Big\{\frac1t\Big\}<1,$ we have

$\displaystyle 0\le\frac{1}{\pi(x)}\sum_{p\le\delta x}\Big\{\frac{x}{p}\Big\}\le\frac{\pi(\delta x)}{\pi(x)}\longrightarrow\delta,$

and similarly

$\displaystyle 0\le\int_0^\delta\Big\{\frac1t\Big\}dt\le\delta.$

Letting $\delta\downarrow0$ , we obtain

$\displaystyle \frac{1}{\pi(x)}\sum_{p\le x}\Big\{\frac{x}{p}\Big\}\longrightarrow\int_0^1\Big\{\frac1t\Big\} dt.$

It remains to evaluate the integral. For each integer $m\ge1$ , on the interval $\frac{1}{m+1}<t\le\frac{1}{m},$ one has $\lfloor1/t\rfloor=m$ , and hence $\Big\{\frac{1}{t}\Big\}=1/t-m.$ Thus

$\displaystyle \begin{aligned} \int_0^1\Big\{\frac1t\Big\} dt &=\sum_{m\ge1}\int_{1/(m+1)}^{1/m}\Big(\frac1t-m\Big)dt \\ &=\sum_{m\ge1}\Big(\log\frac{m+1}{m}-\frac1{m+1}\Big). \end{aligned}$

The $M$ -th partial sum is $\displaystyle \log(M+1)-\Big(H_{M+1}-1\Big)=1+\log(M+1)-H_{M+1},$ where $\displaystyle H_N:=\sum_{m\le N}\frac1m$ is the harmonic sum. Since $\displaystyle H_N-\log N\longrightarrow\gamma,$ it follows that

$\displaystyle \int_0^1\Big\{\frac1t\Big\} dt=1-\gamma.$

Consequently,

$\displaystyle \frac{1}{\pi(x)}\sum_{p\le x}\Big\{\frac{x}{p}\Big\}\longrightarrow1-\gamma.$

The limiting average is not $\displaystyle 1/2,$ because the fractional parts $\displaystyle \Big\{\frac{x}{p}\Big\}$ do not themselves become uniformly distributed. Rather, it is the scaled position $\displaystyle p/x\in(0,1]$ of the prime that becomes uniform. The nonlinear transformation $\displaystyle t\mapsto\Big\{\frac{1}{t}\Big\}$ then turns this uniform distribution into one with mean $\displaystyle 1-\gamma.$