d-dn-n-1-H-n-

Question Number 146442 by qaz last updated on 13/Jul/21

\frac{d}{dn} ∣_{n = 1} H_{n} = ?

Answered by mnjuly1970 last updated on 13/Jul/21

= \frac{d}{d n} (\int_{0}^{1} \frac{1 - x^{n}}{1 - x}) = \int_{0}^{1} \frac{- x^{n} l n (x)}{1 - x} d x

= \int_{0}^{1} \frac{- x l n (x)}{1 - x} d x = - \int_{0}^{1} \frac{(1 - x) l n (1 - x)}{x} d x

= {l i}_{2} (1) + \int_{0}^{1} l n (1 - x) d x

= \frac{π^{2}}{6} + \int_{0}^{1} l n (x) d x = - 1 + \frac{π^{2}}{6} \dots .

Answered by mnjuly1970 last updated on 13/Jul/21

ψ (n + 1) := - γ + H_{n}

\frac{1}{n} + ψ (n) = - γ + H_{n}

\frac{d}{d n} (H n) \underset{}{∣} = - 1 + ψ^{'} (1)

= - 1 + \frac{π^{2}}{6} = - 1 + ζ (2) \dots

Commented by qaz last updated on 13/Jul/21

thank you sir .

I think it is possible to use differential chain rule,

same like \frac{\partial}{\partial x} \int_{h (x)}^{g (x)} f (x, t) dt = \int_{h (x)}^{g (x)} \frac{\partial}{\partial x} f (x, t) dt + g {(x)}^{'} f (x, g (x)) - h {(x)}^{'} f (x, h (x))

, to differential to a seiris summation \dots . .

Commented by mnjuly1970 last updated on 13/Jul/21

t h a n k y o u s i r q a z