advanced-calculus-k-1-k-k-n-1-a-n-b-a-b-adapted-from-brilliant-k-0-1-

Question Number 137177 by mnjuly1970 last updated on 30/Mar/21

\dots \dots a d v a n c e d \dots . c a l c u l u s \dots .

Φ = \underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}}

a, b = ? ? (a d a p t e d f r o m b r i l l i a n t)

\dots \dots \dots \dots \dots .

ψ (k) \overset{? ?}{=} - γ + \int_{0}^{1} (\frac{1 - t^{k - 1}}{1 - t}) d t

∴ ψ^{'} (k) = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{1 - t} d t

\underset{k = 1}{\sum^{\infty}} \frac{ψ^{'} (k)}{k} = \int_{0}^{1} \frac{- t^{k - 1} l n (t)}{(1 - t) k} d t

= \int_{0}^{1} \frac{l n (t)}{t (1 - t)} (- \frac{t^{k}}{k}) d t = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ

ϕ = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t (1 - t)} d t = ϕ_{1} + ϕ_{2} w h e r e \dots

= {\int_{0}^{1} \frac{l n (t) . l n (1 - t)}{1 - t} d t = ϕ_{1}} + {\int_{0}^{1} \frac{l n (1 - t) . l n (t)}{t} d t = ϕ_{2}}

ϕ_{1} = {[- \frac{1}{2} l n (t) {l n}^{2} (1 - t)]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t

∴ ϕ_{1} = \frac{1}{2} (2 ζ (3)) = ζ (3) \overset{e a s y}{=} ϕ_{2}

n o t e : \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} d t = 2 ζ (3) (d e r i v e d e a r l i e r)

ϕ = ϕ_{1} + ϕ_{2} = 2 ζ (3) = \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{3}} \dots .

Φ = \underset{n = 1}{\sum^{\infty}} \frac{a}{n^{b}} \dots \dots \Rightarrow a = 2, b = 3