Second Chebyshev Function

The second Chebyshev function is a function which plays an important role in the prime number theorem. It counts prime powers with a particular weighting, which is generally easier to work with than the true prime counting function.

Definition

The second Chebyshev function is defined as

\[ \psi(x) = \sum_{n \leq x} \Lambda(n)\]

where \(\Lambda\) is the von Mangoldt function.

While we take this as the definition of the function, there are two commonly used alternative expressions for it (sometimes taken as the definition).

Theorem
\[ \psi(x) = \sum_{k = 1}^{\lfloor \log_2(x)\rfloor} \sum_{\substack{p \ \text{prime} \\ p^k \leq x}} \ln(p).\]
Proof

Noting that \(p^k \leq x \implies k \leq \log_p(x) \leq \log_2(x)\), we have

\[\begin{align*} \sum_{n \leq x} \Lambda(n) &= \sum_{n \leq x} \left\{\begin{matrix} \ln(p) & n = p^k \\ 0 & \text{otherwise} \end{matrix}\right\} \\ &= \sum_{k = 1}^\infty \sum_{p^k \leq x} \ln(p) \\ &= \sum_{k = 1}^{\lfloor \log_2(x)\rfloor} \sum_{p^k \leq x} \ln(p). \\ \end{align*}\]

Theorem
\[ \psi(x) = \sum_{p \leq x} \log p \left\lfloor\frac{\log x}{\log p}\right\rfloor\]
Proof
\[\begin{align*} \sum_{p^n \leq x} \log p &= \sum_{p \leq x} (\log p) \cdot \#\{ n \in \mathbb{Z}_{> 0} : p^n \leq x\} \\ &= \sum_{p \leq x} (\log p) \cdot \#\{ n \in \mathbb{Z}_{> 0} : \log_p (x) \leq n\} \\ &= \sum_{p \leq x} (\log p) \cdot \#\left\{ n \in \mathbb{Z}_{> 0} : \frac{\log x}{\log p} \leq n\right\} \\ &= \sum_{p \leq x} \log p \left\lfloor\frac{\log x}{\log p}\right\rfloor \end{align*}\]

We can also relate the definition of this function to the first Chebyshev function.

Theorem
\[ \psi(x) = \sum_{n = 1}^\infty \theta(x^{\frac{1}{n}})\]

where \(\theta\) is the first Chebyshev function.

Proof
\[\begin{align*} \sum_{n = 1}^\infty \theta(x^{\frac{1}{n}}) &= \sum_{n = 1}^\infty \sum_{p \leq x^{\frac{1}{n}}} p \\ &= \sum_{n = 1}^\infty \sum_{p^n \leq x} p \\ &= \sum_{n \leq x} \left\{\begin{matrix} \ln(p) & n = p^k \\ 0 & \text{otherwise} \end{matrix}\right\} \\ &= \sum_{n \leq x} \Lambda(n). \\ \end{align*}\]