Chebyshev function

The second Chebyshev function can be seen to be related to the first by writing it as

${\displaystyle \psi (x)=\sum _{p\leq x}k\log p}$

where k is the unique integer such that p^k ≤ x and x < p^k + 1. The values of k are given in OEIS: A206722. A more direct relationship is given by

${\displaystyle \psi (x)=\sum _{n=1}^{\infty }\vartheta {\big (}x^{\frac {1}{n}}{\big )}.}$

This last sum has only a finite number of non-vanishing terms, as

${\displaystyle \vartheta {\big (}x^{\frac {1}{n}}{\big )}=0\quad {\text{for}}\quad n>\log _{2}x={\frac {\log x}{\log 2}}.}$

The second Chebyshev function is the logarithm of the least common multiple of the integers from 1 to n.

${\displaystyle \operatorname {lcm} (1,2,\dots ,n)=e^{\psi (n)}.}$

Values of lcm(1, 2, ..., n) for the integer variable n are given at OEIS: A003418.

The following theorem relates the two quotients ${\displaystyle {\frac {\psi (x)}{x}}}$ and ${\displaystyle {\frac {\vartheta (x)}{x}}}$ .^[3]

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

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

This inequality implies that

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

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

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

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

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

so

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

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

The following bounds are known for the Chebyshev functions:^[1][2] (in these formulas p_k is the kth prime number; p₁ = 2, p₂ = 3, etc.)

${\displaystyle {\begin{aligned}\vartheta (p_{k})&\geq k\left(\log k+\log \log k-1+{\frac {\log \log k-2.050735}{\log k}}\right)&&{\text{for }}k\geq 10^{11},\\[8px]\vartheta (p_{k})&\leq k\left(\log k+\log \log k-1+{\frac {\log \log k-2}{\log k}}\right)&&{\text{for }}k\geq 198,\\[8px]|\vartheta (x)-x|&\leq 0.006788\,{\frac {x}{\log x}}&&{\text{for }}x\geq 10\,544\,111,\\[8px]|\psi (x)-x|&\leq 0.006409\,{\frac {x}{\log x}}&&{\text{for }}x\geq e^{22},\\[8px]0.9999{\sqrt {x}}&<\psi (x)-\vartheta (x)<1.00007{\sqrt {x}}+1.78{\sqrt[{3}]{x}}&&{\text{for }}x\geq 121.\end{aligned}}}$

Furthermore, under the Riemann hypothesis,

${\displaystyle {\begin{aligned}|\vartheta (x)-x|&=O{\Big (}x^{{\frac {1}{2}}+\varepsilon }{\Big )}\\|\psi (x)-x|&=O{\Big (}x^{{\frac {1}{2}}+\varepsilon }{\Big )}\end{aligned}}}$

for any ε > 0.

Upper bounds exist for both ϑ  (x) and ψ (x) such that^{[4] [3]}

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

for any x > 0.

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

In 1895, Hans Carl Friedrich von Mangoldt proved^[4] an explicit expression for ψ (x) as a sum over the nontrivial zeros of the Riemann zeta function:

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

(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:

${\displaystyle \psi _{0}(x)={\frac {1}{2}}\!\left(\sum _{n\leq x}\Lambda (n)+\sum _{n<x}\Lambda (n)\right)={\begin{cases}\psi (x)-{\tfrac {1}{2}}\Lambda (x)&x=2,3,4,5,7,8,9,11,13,16,\dots \\\,\psi (x)&{\mbox{otherwise.}}\end{cases}}}$

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.

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

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.

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

${\displaystyle \psi (x)-x>K{\sqrt {x}}.}$^[5][6]

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

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

Hardy and Littlewood^[7] prove the stronger result, that

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

The first Chebyshev function is the logarithm of the primorial of x, denoted x #:

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

This proves that the primorial x # is asymptotically equal to e^{(1  + o(1))x}, where "o" is the little-o notation (see big O notation) and together with the prime number theorem establishes the asymptotic behavior of p_n #.

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

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

Then

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

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

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

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

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

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

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

The smoothing function is defined as

${\displaystyle \psi _{1}(x)=\int _{0}^{x}\psi (t)\,dt.}$

Obviously ${\displaystyle \psi _{1}(x)\sim {\frac {x^{2}}{2}}.}$