Chebyshev function

Relationships between ψ(x)/x and ϑ(x)/x

The following theorem relates the two quotients ${\displaystyle \frac{\psi (x)}{x}}$ and ${\displaystyle \frac{\vartheta (x)}{x}}$ .

Theorem: For ${\displaystyle x>0}$, we have

${\displaystyle 0\le \frac{\psi (x)}{x}-\frac{\vartheta (x)}{x}\le \frac{(\log x{)}^2}{2\sqrt{x}\log 2}.}$

This inequality implies that

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\left(\frac{\psi (x)}{x}-\frac{\vartheta (x)}{x}\right)=0.}$

In other words, if one of the ${\displaystyle \psi (x)/x}$ or ${\displaystyle \vartheta (x)/x}$ tends to a limit then so does the other, and the two limits are equal.

Proof: Since ${\displaystyle \psi (x)=\sum _{n\le {\log }_{2}x}\vartheta (x^{1/n})}$, we find that

${\displaystyle 0\le \psi (x)-\vartheta (x)=\sum _{2\le n\le {\log }_{2}x}\vartheta (x^{1/n}).}$

But from the definition of ${\displaystyle \vartheta (x)}$ we have the trivial inequality

${\displaystyle \vartheta (x)\le \sum _{p\le x}\log x\le x\log x}$

so

${\displaystyle \begin{array}{cc}0\le \psi (x)-\vartheta (x)& \le \sum _{2\le n\le {\log }_{2}x}{x}^{1/n}\log (x^{1/n})\\ & \le ({\log }_{2}x)\sqrt{x}\log \sqrt{x}\\ & =\frac{\log x}{\log 2}\frac{\sqrt{x}}{2}\log x\\ & =\frac{\sqrt{x}(\log x{)}^2}{2\log 2}.\end{array}}$

Lastly, divide by ${\displaystyle x}$ to obtain the inequality in the theorem.

Asymptotics and bounds

The following bounds are known for the Chebyshev functions: (in these formulas p_k is the kth prime number; p₁ = 2, p₂ = 3, etc.)

${\displaystyle \begin{array}{cccc}\vartheta (p_k)& \ge k\left(\log k+\log \log k-1+\frac{\log \log k-2.050735}{\log k}\right)& & \text{for }k\ge {10}^{11},\\ \vartheta (p_k)& \le k\left(\log k+\log \log k-1+\frac{\log \log k-2}{\log k}\right)& & \text{for }k\ge 198,\\ |\vartheta (x)-x|& \le 0.006788\frac{x}{\log x}& & \text{for }x\ge 10544111,\\ |\psi (x)-x|& \le 0.006409\frac{x}{\log x}& & \text{for }x\ge e^{22},\\ 0.9999\sqrt{x}& <\psi (x)-\vartheta (x)<1.00007\sqrt{x}+1.78\sqrt[3]{x}& & \text{for }x\ge 121.\end{array}}$

Furthermore, under the Riemann hypothesis,

${\displaystyle \begin{array}{cc}|\vartheta (x)-x|& =O(x^{\frac{1}{2}+\epsilon })\\ |\psi (x)-x|& =O(x^{\frac{1}{2}+\epsilon })\end{array}}$

for any ε > 0.

Upper bounds exist for both ϑ(x) and ψ(x) such that

${\displaystyle \begin{array}{cc}\vartheta (x)& <1.000028x\\ \psi (x)& <1.03883x\end{array}}$

for any x > 0.

An explanation of the constant 1.03883 is given at OEIS: A206431.

The exact formula

In 1895, Hans Carl Friedrich von Mangoldt proved an explicit expression for ψ(x) as a sum over the nontrivial zeros of the Riemann zeta function:

${\displaystyle {\psi }_0(x)=x-\sum _{\rho }\frac{x^{\rho }}{\rho }-\frac{{\zeta }^{\prime }(0)}{\zeta (0)}-\frac{1}{2}\log (1-x^{-2}).}$

(The numerical value of ⁠ζ′(0)/ζ(0)⁠ is log(2π).) Here ρ runs over the nontrivial zeros of the zeta function, and ψ₀ is the same as ψ, except that at its jump discontinuities (the prime powers) it takes the value halfway between the values to the left and the right:

${\displaystyle {\psi }_0(x)=\frac{1}{2}\left(\sum _{n\le x}\mathrm{\Lambda }(n)+\sum _{n<x}\mathrm{\Lambda }(n)\right)=\{\begin{array}{ll}\psi (x)-\frac{1}{2}\mathrm{\Lambda }(x)& x=2,3,4,5,7,8,9,11,13,16,\dots \\ \psi (x)& \text{otherwise.}\end{array}}$

From the Taylor series for the logarithm, the last term in the explicit formula can be understood as a summation of ⁠x^ω/ω⁠ over the trivial zeros of the zeta function, ω = −2, −4, −6, ..., i.e.

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{x^{-2k}}{-2k}=\frac{1}{2}\log \left(1-x^{-2}\right).}$

Similarly, the first term, x = ⁠x¹/1⁠, corresponds to the simple pole of the zeta function at 1. It being a pole rather than a zero accounts for the opposite sign of the term.

Properties

A theorem due to Erhard Schmidt states that, for some explicit positive constant K, there are infinitely many natural numbers x such that

${\displaystyle \psi (x)-x<-K\sqrt{x}}$

and infinitely many natural numbers x such that

${\displaystyle \psi (x)-x>K\sqrt{x}.}$

In little-o notation, one may write the above as

${\displaystyle \psi (x)-x\ne o\left(\sqrt{x}\right).}$

Hardy and Littlewood prove the stronger result, that

${\displaystyle \psi (x)-x\ne o\left(\sqrt{x}\log \log \log x\right).}$

Relation to primorials

The first Chebyshev function is the logarithm of the primorial of x, denoted x#, as we have

${\displaystyle \vartheta (x)=\sum _{p\le x}\log p=\log \prod _{p\le x}p=\log \left(x\mathrm{\#}\right).}$

The prime number theorem is equivalent to ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\vartheta (x)/x)=1}$. From the equivalent form ${\displaystyle \vartheta (x)=x(1+o(1))}$ as ${\displaystyle x\to \mathrm{\infty }}$, where ${\displaystyle o}$ is the little-${\displaystyle o}$ notation, it thus immediately follows that we have

${\displaystyle x\mathrm{\#}=e^{(1+o(1))x}.}$

Relation to the prime-counting function

The Chebyshev function can be related to the prime-counting function as follows. Define

${\displaystyle \mathrm{\Pi }(x)=\sum _{n\le x}\frac{\mathrm{\Lambda }(n)}{\log n}.}$

Then

${\displaystyle \mathrm{\Pi }(x)=\sum _{n\le x}\mathrm{\Lambda }(n){\int }_n^x\frac{dt}{t{\log }^{2}t}+\frac{1}{\log x}\sum _{n\le x}\mathrm{\Lambda }(n)={\int }_2^x\frac{\psi (t)dt}{t{\log }^{2}t}+\frac{\psi (x)}{\log x}.}$

The transition from Π to the prime-counting function, π, is made through the equation

${\displaystyle \mathrm{\Pi }(x)=\pi (x)+\frac{1}{2}\pi \left(\sqrt{x}\right)+\frac{1}{3}\pi \left(\sqrt[3]{x}\right)+\cdots }$

Certainly π(x) ≤ x, so for the sake of approximation, this last relation can be recast in the form

${\displaystyle \pi (x)=\mathrm{\Pi }(x)+O\left(\sqrt{x}\right).}$

The Riemann hypothesis

The Riemann hypothesis states that all nontrivial zeros of the zeta function have real part ⁠1/2⁠. In this case, |x^ρ| = √x, and it can be shown that

${\displaystyle \sum _{\rho }\frac{x^{\rho }}{\rho }=O\left(\sqrt{x}{\log }^{2}x\right).}$

By the above, this implies

${\displaystyle \pi (x)=\mathrm{li}(x)+O\left(\sqrt{x}\log x\right).}$

Smoothing function

The smoothing function is defined as

${\displaystyle {\psi }_1(x)={\int }_0^x\psi (t)dt.}$

It can trivially be shown that ${\displaystyle {\psi }_1(x)\sim \frac{x^2}{2}.}$