1-calculate-1-n-1-1-n-1-t-1-t-dt-2-prove-that-0-1-1-t-1-t-dt-1-is-constant-number-of-euler-

Question Number 40890 by abdo.msup.com last updated on 28/Jul/18

1) c a l c u l a t e \int_{\frac{1}{n + 1}}^{\frac{1}{n}} [\frac{1}{t} - [\frac{1}{t}]] d t

2) p r o v e t h a t \int_{0}^{1} [\frac{1}{t} - [\frac{1}{t}]] d t = 1 - γ

γ i s c o n s t a n t n u m b e r o f e u l e r

Commented by maxmathsup by imad last updated on 01/Aug/18

1) l e t A_{n} = \int_{\frac{1}{n + 1}}^{\frac{1}{n}} {\frac{1}{t} - [\frac{1}{t}]} d t c h a n g e m e n t \frac{1}{t} = x g i v e

A_{n} = - \int_{n}^{n + 1} {x - [x]} (- \frac{d x}{x^{2}}) = \int_{n}^{n + 1} {\frac{1}{x} - \frac{[x]}{x^{2}}} d x

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

= l n (n + 1) - l n (n) + n {\frac{1}{n + 1} - \frac{1}{n}} = l n (n + 1) - l n (n) + \frac{n}{n + 1} - 1

Commented by maxmathsup by imad last updated on 01/Aug/18

2) l e t I = \int_{0}^{1} {\frac{1}{t} - [\frac{1}{t}]} d t c h a n g e m e n t \frac{1}{t} = x g i v e

I = - \int_{1}^{+ \infty} {x - [x]} (- \frac{d x}{x^{2}}) = \int_{1}^{+ \infty} \frac{x - [x]}{x^{2}} d x

= \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{x - [x]}{x^{2}} d x = \sum_{n = 1}^{\infty} A_{n} b u t

\sum_{n = 1}^{\infty} A_{n} = {l i m}_{n \to + \infty} \sum_{k = 1}^{n} A_{k}

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

= l n (n + 1) - \sum_{k = 2}^{n + 1} \frac{1}{k} = l n (n + 1) - (H_{n + 1} - 1)

= 1 - (H_{n + 1} - l n (n + 1)) b u t

H_{n + 1} = l n (n + 1) + γ + o (\frac{1}{n}) \Rightarrow H_{n + 1} - l n (n + 1) = γ + o (\frac{1}{n}) \Rightarrow

\sum_{k = 1}^{n} A_{k} = 1 - γ + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to + \infty} \sum_{k = 1}^{n} A_{k} = 1 - γ = I .