find-1-t-t-t-p-dt-interms-of-p-with-p-gt-0-

Question Number 54367 by maxmathsup by imad last updated on 02/Feb/19

f i n d \int_{1}^{+ \infty} \frac{[t]}{t} t^{- p} d t i n t e r m s o f ξ (p) w i t h p > 0 .

Commented by maxmathsup by imad last updated on 04/Feb/19

l e t A_{p} = \int_{1}^{+ \infty} \frac{[t]}{t} t^{- p} d t \Rightarrow A_{p} = \sum_{n = 1}^{+ \infty} \int_{n}^{n + 1} \frac{n}{t} t^{- p} d t

= \sum_{n = 1}^{\infty} n \int_{n}^{n + 1} t^{- p - 1} d t = \sum_{n = 1}^{\infty} n {[\frac{1}{- p} t^{- p}]}_{n}^{n + 1}

= \sum_{n = 1}^{\infty} (- \frac{n}{p}) {\frac{1}{{(n + 1)}^{p}} - \frac{1}{n^{p}}} = \frac{1}{p} {\sum_{n = 1}^{\infty} \frac{1}{n^{p - 1}} - \sum_{n = 1}^{\infty} \frac{n}{{(n + 1)}^{p}}}

= \frac{1}{p} {\sum_{n = 1}^{\infty} \frac{1}{n^{p - 1}} - \sum_{n = 2}^{\infty} \frac{n - 1}{n^{p}}} = \frac{1}{p} {1 + \sum_{n = 2}^{\infty} \frac{1}{n^{p}}} = \frac{ξ (p)}{p} \Rightarrow

A_{p} = \frac{ξ (p)}{p} .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

\int_{1}^{\infty} \frac{[t]}{t} t^{- p} d t

\int_{1}^{2} 1 \times t^{- p - 1} d t + \int_{2}^{3} 2 \times t^{- p - 1} d t + \int_{3}^{4} 3 \times t^{- p - 1} d t \dots

1 \times {(\frac{t^{- p}}{- p})}_{1}^{2} + 2 \times {(\frac{t^{- p}}{- p})}_{2}^{3} + 3 \times {(\frac{t^{- p}}{- p})}_{3}^{4} + \dots

= (\frac{1}{- p}) [(\frac{1}{2^{p}} - \frac{1}{1^{p}}) + 2 \times (\frac{1}{3^{p}} - \frac{1}{2^{p}}) + 3 \times (\frac{1}{4^{p}} - \frac{1}{3^{p}}) + \dots]

= (\frac{1}{p}) (\frac{1}{1^{p}} + \frac{1}{2^{p}} + \frac{1}{3^{p}} + \frac{1}{4^{p}} + \dots \infty)