📂Number Theory

Selberg Identity Proof

Theorem ¹

$$\Lambda (n) \log n + \sum_{d \mid n } \Lambda (d) \Lambda \left( {{ n } \over { d }} \right) = \sum_{d \mid n} \mu (d) \log^{2} {{ n } \over { d }}$$

Proof

Strategy: Not as hard as it looks. With just the differentiation of arithmetic functions, it can be derived very simply.

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Mangoldt function: $$\sum_{d \mid n} \Lambda ( d ) = \log n$$

According to the definition of the differentiation of arithmetic functions, the Mangoldt function can be expressed using convolution as follows. $$\Lambda \ast\ u = 1 \cdot \log n = u \log n = u '$$ Differentiating both sides, following the product rule, $$\Lambda’ \ast\ u + \Lambda \ast\ u ' = u ''$$ Since $\Lambda \ast\ u = u '$, $$\Lambda’ \ast\ u + \Lambda \ast\ (\Lambda \ast\ u) = u ''$$ Since the Möbius function $\mu$ is the inverse of the unit function $u$, multiplying both sides by $\mu$ gives $$\Lambda ’ + \Lambda^{2} = u '' \ast\ \mu$$

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