$\log$ approximation for $\pi(x)$

It seems that a reasonable $\log$ approximation for $\pi(x)$ can be given, where

$f(y, x) := \log\left(\dfrac{\log(x)}{\log\left(e(y - \lfloor y\rfloor) + x^{1/x}(1 - y + \lfloor y\rfloor)\right)}\right)$

is an interpolating function between $\log(x)\text{ and }\log(\log(x)).$

Comparing $\pi(x)/\dfrac{x}{\log(x)},\pi(x)/\dfrac{x}{f(1+\frac{\log(x)}{x},x)},\pi(x)/\text{li}(x)$: f[y_, x_] := Log[Log[x]/Log[(1 - (y - Floor[y])) x^(1/x) + (y - Floor[y]) E]]
t = 8; r = 10^t; LogLogPlot[{PrimePi[n]/(n/f[1 + Log[n]/n, n]),


Am I correct in assuming that this is an improvement on the $\pi(x)\sim\dfrac{x}{\log(x)}$ approximation?

Re-posted owing to errors in original question

With $y=1+\frac{\log x}{x}$ and consequently $\lfloor y\rfloor=1$ for $x\ge1,$

$$f(y, x) = \log\left(\frac{\log x}{\log\left(\ \frac{e\log x}{x} + x^{1/x}(1 - \frac{\log x}{x})\ \right)}\right)$$

which can more easily be written

$$f(y, x) = \log\log x-\log\log\left(\ \frac{e\log x}{x} + x^{1/x}(1 - \frac{\log x}{x})\ \right)$$

Note that $x^{1/x}$ is approximately $1+\frac{\log x}{x}$ and so the latter argument is about

$$\frac{e\log x}{x} + (1+\frac{\log x}{x})(1 - \frac{\log x}{x})=\frac{e\log x}{x}+1-\frac{\log^2x}{x^2}=1+\frac{e\log x}{x}+o\left(\frac1x\right)$$

That is, the $x^{1/x}(1 - \frac{\log x}{x})$ term is asymptotically negligible. Approximating $$\log\log\left(\frac{e\log x}{x} + x^{1/x}(1 - \frac{\log x}{x})\right)\approx \log\log\left(\frac{e\log x}{x} + 1\right)\approx \log\left(\frac{e\log x}{x}\right)\approx1+\log\log x-\log x$$

and so $f(y,x)\approx\log x-1.$

In fact $\frac{x}{\log x-1}$ is a better approximation of $\pi(x)$ than $\frac{x}{\log x}$. (Sometimes the 1 in the former expression goes by the name "Legendre's constant"; Legendre had guessed its value to be around 1.08.)

A better approximation:

$$\frac{e\log x}{x} + x^{1/x}(1 - \frac{\log x}{x})\approx1+\frac{e\log x}{x}-\frac{\log^2x}{2x^2}-\frac{\log^3x}{3x^3}$$

$$\frac{e\log x}{x}-(1+e^2)\frac{\log^2x}{2x^2}$$

and so

$$f(y,x)\approx\log\log x-\log\left(\frac{e\log x}{x}-(1+e^2)\frac{\log^2x}{2x^2}\right)\\ =\log\log x-\log\left(\frac{e\log x}{x}\right)-\log\left(1-(1+e^2)\frac{x}{e\log x}\cdot\frac{\log^2x}{2x^2}\right)\\ =\log x-1-\log\left(1-(1+e^2)\frac{\log x}{2ex}\right)\\ \approx\log x-1-\frac{1+e^2}{2e}\cdot\frac{\log x}{x}$$

But the right approximation at this order is $\log x-1-1/\log x$ and so $f(y,x)$ is too large by about $1/\log x$.

In summary ("TL;DR"): This is a better approximation than $x/\log x$ but not as good as $\frac{1}{\log x-1-1/\log x}$ which is, in turn, not as good as $\operatorname{li}(x).$

• Thank you for the detailed explanation :) – martin Jun 18 '14 at 19:18

When $y=1+(\log x)/x$, your function $f(y,x)$ equals $$\log\bigg( \frac{\log x}{\log\big(x^{1/x}(1-(\log x)/x)+e(\log x)/x\big)} \bigg).$$ Since $x^{1/x} = 1+(\log x)/x+O((\log x)^2/x^2)$, this is \begin{multline*} \log\bigg( \frac{\log x}{\log\big( 1 + e(\log x)/x + O((\log x)^2/x^2)\big)} \bigg) \\ = \log\bigg( \frac{\log x}{e(\log x)/x + O((\log x)^2/x^2)} \bigg) \\ = \log\bigg( \frac{\log x}{e(\log x)/x} \bigg( 1 + O\bigg( \frac{\log x}x \bigg) \bigg) \bigg) \\ = \log x - 1 + O\bigg( \frac{\log x}x \bigg). \end{multline*} So essentially, your function is just $\log x-1$. And it is well known that $x/(\log x-1)$ is a better approximation to $\pi(x)$ than $x/\log x$, since the former expression matches the first two terms of the asymptotic series $$\mathop{\rm li}(x) = \frac x{\log x} + \frac x{\log^2x} + \frac{2x}{\log^3x} + \cdots.$$

• Yes, this makes sense! – martin Jun 18 '14 at 19:16
• I wrote essentially the same answer (independently) and then carried the approximation one term further, where it disagrees with the series for li. – Charles Jun 18 '14 at 19:18
• (+1, by the way -- good succinct explanation.) – Charles Jun 18 '14 at 19:46