Graph of the Function $y = \dfrac{x}{\ln x}$

$y = \dfrac{x}{\ln x}$ is known as the expression that captures how the primes thin out. The prime number theorem states that $\pi(x)$, the number of primes not exceeding $x$, grows at the same rate as this function.

Domain and sign

The logarithm must be defined and non-zero, so the domain is $x > 0$ with $x \neq 1$. On $0 < x < 1$ we have $\ln x < 0$ and hence $y < 0$, while $y > 0$ for $x > 1$.

Asymptote and end behavior

The line $x = 1$ is a vertical asymptote, approached toward $-\infty$ from the left and toward $+\infty$ from the right. As $x \to 0^{+}$ the numerator tends to $0$ and the denominator to $-\infty$, so $y \to 0^{-}$ and the curve enters the origin from below. As $x \to \infty$ the factor $x$ overwhelms $\ln x$ and the function diverges, so there is no asymptote on the right.

Monotonicity and extrema

The quotient rule gives the derivative.

The denominator is positive, so the sign is that of $\ln x - 1$: negative for $x < e$ and positive for $x > e$. There is therefore a local minimum at $x = e$, of value $e$, so the minimum point $(e, e)$ has equal coordinates.

Range

The branch on $0 < x < 1$ falls steadily from $0^{-}$ near the origin down to $-\infty$, giving $y < 0$, while the branch on $x > 1$ has minimum $e$, giving $y \geq e$. No value in $0 \leq y < e$ is attained. This is exactly the range of $y = \dfrac{e^{x}}{x}$, an amusing coincidence: exchanging logarithm and exponential leaves the same structure.

Concavity

The second derivative is $y'' = \dfrac{2 - \ln x}{x(\ln x)^3}$.

There is thus a single inflection point, at $x = e^{2} \approx 7.389$, where the value is $\dfrac{e^{2}}{2} \approx 3.695$.

The prime number theorem

Writing $\pi(x)$ for the number of primes up to $x$, the prime number theorem says that the ratio of the two tends to $1$.

Intuitively the density of primes near $x$ is about $\dfrac{1}{\ln x}$, so consecutive primes are separated by roughly $\ln x$ on average. The convergence is slow, however, and the values close in only gradually.

A better approximation

The error shrinks dramatically with the logarithmic integral.

Here $\operatorname{li}(10^{6}) \approx 78628$, only about $130$ away from the true value. Substituting $t = e^{u}$ turns that integral into $\int \dfrac{e^{u}}{u}\,du$, the exponential integral. The function $\dfrac{x}{\ln x}$ is the leading term of the asymptotic expansion of $\operatorname{li}(x)$, and the Riemann hypothesis is a statement about how small the difference between $\operatorname{li}(x)$ and $\pi(x)$ can be.