Derivative of the Riemann Zeta Function
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Theorem
The derivative of the Riemann zeta function is:
- $\displaystyle \frac{d\zeta}{dz} = -\sum_{n=2}^\infty \frac{\ln(n)}{n^z}$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac{d\zeta}{dz}\) | \(=\) | \(\displaystyle \frac{d}{dz} \left({\sum_{n=1}^\infty n^{-z} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \((1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n=1}^\infty \frac{d}{dz} \left({n^{-z} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n=1}^\infty \left(-{ \ln(n) n^{-z} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Derivative of Exponential Function | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle -\sum_{n=1}^\infty \frac{\ln(n)}{n^z}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle -\sum_{n=2}^\infty \frac{\ln(n)}{n^z}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $\ln 1 = 0$ |
$\blacksquare$
In particular: The step at $(1)$ requires justification. Need to show that the sum converges uniformly and also that the sum of the derivatives converge uniformly. Note that this result may already exist somewhere in the database and so would just need hunting down..
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