let-give-I-x-1-t-E-t-t-x-1-dt-prove-that-x-x-x-1-xI-x-then-chow-that-x-1-x-1-ew-x-gt-1-we-remind-x-n-1-1-n-x-and-x-gt-1-

Question Number 26571 by abdo imad last updated on 26/Dec/17

l e t g i v e I (x) = \int_{1}^{\propto} \frac{t - E (t)}{t^{x + 1}} d t p r o v e t h a t

ξ (x) = \frac{x}{x - 1} - x I (x) t h e n c h o w t h a t {(x - 1)}_{x - 1^{+ e w}} ξ (x) - - > 1

w e r e m i n d ξ (x) = \sum_{n ⩾ 1} \frac{1}{n^{x}} a n d x > 1

Commented by abdo imad last updated on 02/Jan/18

w e h a v e I (x) = \int_{1}^{\propto} \frac{d t}{t^{x}} - \int_{1}^{\propto} \frac{E (t)}{t^{x + 1}} d t b u t

\int_{1}^{\propto} \frac{d t}{t^{x}} = \int_{1}^{\propto} t^{- x} d t = {[\frac{1}{1 - x} t^{1 - x}]}_{t = 1}^{t - >\propto} = \frac{1}{x - 1}

\int_{1}^{\propto} \frac{E (t)}{t^{x + 1}} d t = {l i m}_{n - >\propto} A_{n} w i t h =

A_{n} = \sum_{k = 1}^{n - 1} \int_{k}^{k + 1} \frac{k}{t^{x + 1}} d t = \sum_{k = 1}^{n - 1} k \int_{k}^{k + 1} t^{- x - 1} d t

= \sum_{k = 1}^{n - 1} k {[- \frac{1}{x} t^{- x}]}_{k}^{k + 1} = \sum_{k = 1}^{k = n - 1} \frac{k}{x} (\frac{1}{k^{x}} - \frac{1}{{(k + 1)}^{x}})

x A_{n} = \sum_{k = 1}^{k = n - 1} \frac{1}{k^{x - 1}} - \sum_{k = 1}^{n - 1} \frac{k + 1 - 1}{{(k + 1)}^{x}}

= \sum_{k = 1}^{k = n - 1} \frac{1}{k^{x - 1}} - \sum_{k = 1}^{n - 1} \frac{1}{{(k + 1)}^{x - 1}} + \sum_{k = 1}^{n - 1} \frac{1}{{(k + 1)}^{x}}

= \sum_{k = 1}^{n - 1} \frac{1}{k^{x - 1}} - \sum_{k = 2}^{n} \frac{1}{k^{x - 1}} + \sum_{k = 2}^{n} \frac{1}{k^{x}}

= 1 - \frac{1}{n^{x - 1}} + \sum_{k = 2}^{n} \frac{1}{k^{x}} \Rightarrow {l i m}_{n - >\propto} = \frac{ξ (x)}{x}

I (x) = \frac{1}{x - 1} - \frac{ξ (x)}{x} \Rightarrow x I (x) = \frac{x}{x - 1} - ξ (x)

\Rightarrow ξ (x) = \frac{x}{x - 1} - x I (x)

\Rightarrow (x - 1) ξ (x) = x - x (x - 1) I (x)

\Rightarrow {l i m}_{x - > 1^{+}} (x - 1) ξ (x) = 1 .