关于第二切比雪夫函数的等式证明及不等式求证咨询

Proof 1: Equivalent Representations of ψ(n) #

First, let's recap the core definitions we'll work with:

• Chebyshev's ψ Function: $\psi(n) = \sum_{p^m \le n} \log p$, where the sum runs over all prime powers $p^m$ (prime $p$, integer $m \ge 1$) that are ≤ $n$.
• Chebyshev's θ Function: $\vartheta(n) = \sum_{p \le n} \log p$, summing $\log p$ over all primes $p$ ≤ $n$.

Part 1: $\psi(n) = \sum_{p \le n} \left\lfloor \frac{\log n}{\log p} \right\rfloor \log p$ #

For any fixed prime $p$, we need to count how many of its powers ($p^1, p^2, ...$) are ≤ $n$. Solving for the largest integer $m$ such that $p^m ≤ n$:
Take logarithms of both sides: $m \log p ≤ \log n \implies m ≤ \frac{\log n}{\log p}$. The number of valid exponents is exactly $\left\lfloor \frac{\log n}{\log p} \right\rfloor$.

Each prime power $p^m$ contributes $\log p$ to $\psi(n)$, so the total contribution from prime $p$ is $\left\lfloor \frac{\log n}{\log p} \right\rfloor \cdot \log p$. Summing this over all primes $p ≤ n$ accounts for every prime power ≤ $n$, which matches the definition of $\psi(n)$.

Part 2: $\psi(n) = \vartheta(n) + \vartheta(n^{1/2}) + \vartheta(n^{1/3}) + \dots$ #

Consider the term $\vartheta(n^{1/k})$: this sums $\log p$ over all primes $p ≤ n^{1/k}$, which is equivalent to primes where $p^k ≤ n$.

Each prime power $p^m ≤ n$ contributes $\log p$ exactly $m$ times to the infinite sum:

• For each $k$ from 1 to $m$, $p ≤ n^{1/k}$ (since $p^m ≤ n \implies p ≤ n^{1/m} ≤ n^{1/k}$ when $k ≤ m$), so $\log p$ is included in $\vartheta(n^{1/k})$.
• For $k > m$, $p^k > n \implies p > n^{1/k}$, so $\log p$ is not included in $\vartheta(n^{1/k})$.

The total number of times $\log p$ appears is exactly the number of prime powers of $p$ ≤ $n$, which aligns with the definition of $\psi(n)$. Thus, summing $\vartheta(n^{1/k})$ over all $k ≥ 1$ gives $\psi(n)$.

Proof 2: Upper Bound $\psi(n) \le 2 \log 2 \cdot n + 2n^{1/2} \log n$ #

We split $\psi(n)$ into two manageable parts to simplify the bound:

1. Primes $p > \sqrt{n}$: For these primes, $p^2 > n$, so only the first power $p^1$ contributes to $\psi(n)$. This sum is $\vartheta(n) - \vartheta(\sqrt{n})$.
2. Primes $p ≤ \sqrt{n}$: For these primes, the number of prime powers $p^m ≤ n$ is $\left\lfloor \frac{\log n}{\log p} \right\rfloor$, so their total contribution is $\sum_{p \le \sqrt{n}} \left\lfloor \frac{\log n}{\log p} \right\rfloor \log p$.

Step 1: Bound the first part ($p > \sqrt{n}$) #

Using Chebyshev's classic result, we know $\vartheta(x) ≤ 2x \log 2$ for all $x ≥ 1$. This comes from induction and the fact that $\binom{2n}{n} ≤ 2^{2n}$, which implies $\vartheta(2n) - \vartheta(n) ≤ 2n \log 2$. Thus:
$$\vartheta(n) - \vartheta(\sqrt{n}) ≤ \vartheta(n) ≤ 2n \log 2$$

Step 2: Bound the second part ($p ≤ \sqrt{n}$) #

Since $\log p ≥ \log 2$ for all primes $p ≥ 2$, we have $\frac{\log n}{\log p} ≤ \frac{\log n}{\log 2}$. Substitute this into the sum:
$$\sum_{p \le \sqrt{n}} \left\lfloor \frac{\log n}{\log p} \right\rfloor \log p ≤ \frac{\log n}{\log 2} \sum_{p \le \sqrt{n}} \log p = \frac{\log n}{\log 2} \vartheta(\sqrt{n})$$

Again using $\vartheta(\sqrt{n}) ≤ 2\sqrt{n} \log 2$, substitute this in:
$$\frac{\log n}{\log 2} \cdot 2\sqrt{n} \log 2 = 2\sqrt{n} \log n$$

Step 3: Combine the bounds #

Adding the two parts together gives the desired upper bound:
$$\psi(n) \le 2 \log 2 \cdot n + 2n^{1/2} \log n$$

内容的提问来源于stack exchange，提问作者Fats