The Prime Number Theorem is usually quoted as \pi(x) \sim x/\log x — the count of primes below x grows like x/\log x. Open almost any analytic number theory paper, though, and you will find the theorem proved in a completely different-looking costume: \psi(x) \sim x, a clean statement about a sum of logarithms with no 1/\log x in sight. A third form, \theta(x) \sim x, sits in between.

This page is about a small miracle of bookkeeping: these three statements are logically equivalent. Prove any one and the other two fall out for free, purely by Abel summation — no new analysis required. Understanding why they are interchangeable, and which one you should actually reach for when doing the hard work, is the single most useful piece of orientation in the whole subject. Spoiler: the analyst's favourite is \psi, and by the end you will see exactly why the awkward-looking one is secretly the nicest.

Meet the three counting functions

All three measure "how much prime is there below x" — they differ only in how each prime is weighted.

So \pi counts, \theta log-weights the primes, and \psi log-weights the primes and their higher powers. Written out, \psi is

because for each prime p the powers p, p^2, \dots, p^k \le x each throw in another \log p, and the largest such k is \lfloor \log x / \log p \rfloor.

A concrete example: all three at x = 10

The primes up to 10 are 2, 3, 5, 7, so \pi(10) = 4. For \theta we log-weight them:

For \psi we also pick up the prime powers \le 10 — namely 4 = 2^2, 8 = 2^3 and 9 = 3^2 — each contributing the log of its base prime:

Notice how close \theta(10) \approx 5.35 and \psi(10) \approx 7.14 already are to x = 10, while \pi(10) = 4 is nowhere near — it is x/\log x \approx 4.34 that \pi chases, not x. That single observation is the whole story of the log-weighting: it rescales the count so the target becomes the clean line y = x.

Seeing it: three staircases, one line

Here are all three, rescaled so they should track y = x: the log-weighted count \theta(x), the prime-power version \psi(x), and — to bring \pi onto the same axes — the product \pi(x)\log x. Watch how all three staircases press up against the diagonal:

The two Chebyshev functions \theta and \psi run almost on top of each other (\psi sits a whisker higher, from the prime powers), and \pi(x)\log x weaves around them. All three hug y = x — a visual statement that the three asymptotic laws are really one law seen through three lenses.

Step 1 — \psi and \theta are barely different

Group the terms of \psi by the power k. All the k = 1 terms are exactly \theta(x). The k = 2 terms are \sum_{p^2 \le x}\log p = \theta(x^{1/2}), the k = 3 terms give \theta(x^{1/3}), and so on. The sum is finite because p^k \le x forces k \le \log_2 x:

Now bound the tail. The trivial estimate \theta(t) \le t \log t (each of the at most t primes contributes at most \log t) gives \theta(x^{1/2}) \le x^{1/2}\log x, and there are only about \log_2 x terms, each no larger than the second. Hence

Since \sqrt{x}\,\log^2 x is dwarfed by x, dividing by x gives (\psi(x)-\theta(x))/x \to 0. Therefore \psi(x) \sim x and \theta(x) \sim x are the same statement — the prime powers are asymptotically negligible. That is the easy half of the equivalence, and it cost us nothing but a crude bound.

Step 2 — passing between \theta and \pi by parts

The jump from a plain count \pi to a log-weighted count \theta is a job for Abel summation — partial summation, the discrete cousin of integration by parts. Writing \theta(x) = \sum_{p \le x} \log p as a Stieltjes integral against the jump-measure d\pi and integrating by parts gives the two conversion formulas that drive everything:

Each is a clean identity — no approximation, valid for all x \ge 2. The first weighs each unit step of \pi by \log p; the second unwinds the weighting back out. To read off asymptotics we just need to know that the integral terms are of lower order than the leading terms — which is exactly what the next worked example shows.

Worked example — turning \theta(x) \sim x into \pi(x) \sim x/\log x

Suppose we have already proved \theta(x) \sim x (the analytically convenient form). Let us extract the classical PNT from it, using the second identity above. The leading term is immediate:

So we only need the integral \int_2^x \theta(t)/(t\log^2 t)\,dt to be small compared with x/\log x. Using \theta(t) = O(t) (Chebyshev's bound — we do not even need the full asymptotic here), the integrand is O(1/\log^2 t), so

The last step is the standard estimate \int_2^x dt/\log^2 t \sim x/\log^2 x (split the range at \sqrt{x}: below it the integrand is bounded, above it \log t > \tfrac12\log x). Putting the pieces together,

because the error term x/\log^2 x is exactly a factor 1/\log x smaller than the main term x/\log x. Dividing through, \pi(x)\big/(x/\log x) \to 1 — precisely \pi(x) \sim x/\log x. The conversion is that mechanical: one integration by parts and one lower-order bound.

The payoff: three theorems that are one theorem

Chaining Step 1 (\psi \leftrightarrow \theta) with Step 2 (\theta \leftrightarrow \pi), and running the arrows both ways, collapses the three statements into one.

Prove any single one and the entire package is yours. The same equipment upgrades error terms too: a bound like \psi(x) = x + O(x\,e^{-c\sqrt{\log x}}) transfers, via these identities, into the corresponding statement about \pi(x) versus the logarithmic integral \operatorname{Li}(x) — the equivalence is not just for leading-order asymptotics but carries the fine structure along with it.

Why \psi is the analyst's darling

If all three are equivalent, why does every serious proof target \psi? Two reasons, both about algebraic cleanliness.

First, its Dirichlet series is beautiful. The von Mangoldt weights are engineered so that \log n = \sum_{d \mid n}\Lambda(d), and taking Dirichlet series turns this identity into

That right-hand side — the logarithmic derivative of the zeta function — is exactly the object whose poles sit at the zeros of \zeta. Because \psi(x) = \sum_{n \le x}\Lambda(n) is the partial-sum (summatory) function of these coefficients, a Perron / contour-integral argument reads \psi(x) \sim x straight off the pole of -\zeta'/\zeta at s = 1. Neither \pi nor \theta has a Dirichlet series anywhere near this tidy.

Second, its arithmetic is smooth. Summing \Lambda over all integers n \le x — not just primes — lets the full weight of the divisor identity \log n = \sum_{d\mid n}\Lambda(d) come to bear, which is what makes elementary manipulations (and the elementary Selberg–Erdős proof) work. Restricting to squarefree primes, as \theta and \pi do, throws that structure away. So the pattern is universal: prove it for \psi, then hand it to \theta and \pi by partial summation.