1)calculate ∫_(1/(n+1)) ^(1/n) [(1/t)−[(1/t)]]dt 2)prove that ∫_0 ^1 [(1/t)−[(1/t)]]dt=1−γ γ is constant number of euler


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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
$1) let A_n = ∫_(1/(n+1)) ^(1/n) {(1/t)−[(1/t)]}dt changement (1/t)=x give A_n = −∫_n ^(n+1) {x−[x]}(−(dx/x^2 )) = ∫_n ^(n+1) {(1/x)−(([x])/x^2 )}dx =∫_n ^(n+1) (dx/x) −∫_n ^(n+1) (n/x^2 )dx =[ln(x)]_n ^(n+1) +n [ (1/x)]_n ^(n+1) =ln(n+1)−ln(n) +n{(1/(n+1)) −(1/n)} =ln(n+1)−ln(n) +(n/(n+1)) −1$
$1) l e t A_{n} = \int_{\frac{1}{n + 1}}^{\frac{1}{n}} {\frac{1}{t} - [\frac{1}{t}]} d t c h a n g e m e n t \frac{1}{t} = x g i v e$ $A_{n} = - \int_{n}^{n + 1} {x - [x]} (- \frac{d x}{x^{2}}) = \int_{n}^{n + 1} {\frac{1}{x} - \frac{[x]}{x^{2}}} d x$ $= \int_{n}^{n + 1} \frac{d x}{x} - \int_{n}^{n + 1} \frac{n}{x^{2}} d x = {[l n (x)]}_{n}^{n + 1} + n {[\frac{1}{x}]}_{n}^{n + 1}$ $= l n (n + 1) - l n (n) + n {\frac{1}{n + 1} - \frac{1}{n}} = l n (n + 1) - l n (n) + \frac{n}{n + 1} - 1$
Commented by maxmathsup by imad last updated on 01/Aug/18
$2) let I = ∫_0 ^1 { (1/t)−[(1/t)]}dt changement (1/t)=x give I = −∫_1 ^(+∞) {x−[x]}(−(dx/x^2 )) = ∫_1 ^(+∞) ((x−[x])/x^2 )dx =Σ_(n=1) ^∞ ∫_n ^(n+1) ((x−[x])/x^2 )dx =Σ_(n=1) ^∞ A_n but Σ_(n=1) ^∞ A_n =lim_(n→+∞) Σ_(k=1) ^n A_k Σ_(k=1) ^n A_k =Σ_(k=1) ^n {ln(k+1)−ln(k)} −Σ_(k=1) ^n (1/(k+1)) =ln(n+1)−Σ_(k=2) ^(n+1) (1/k) =ln(n+1)−(H_(n+1) −1) =1−(H_(n+1) −ln(n+1)) but H_(n+1) =ln(n+1) +γ +o((1/n))⇒H_(n+1) −ln(n+1)=γ +o((1/n))⇒ Σ_(k=1) ^n A_k =1−γ +o((1/n)) ⇒lim_(n→+∞) Σ_(k=1) ^n A_k =1−γ =I .$
$2) l e t I = \int_{0}^{1} {\frac{1}{t} - [\frac{1}{t}]} d t c h a n g e m e n t \frac{1}{t} = x g i v e$ $I = - \int_{1}^{+ \infty} {x - [x]} (- \frac{d x}{x^{2}}) = \int_{1}^{+ \infty} \frac{x - [x]}{x^{2}} d x$ $= \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{x - [x]}{x^{2}} d x = \sum_{n = 1}^{\infty} A_{n} b u t$ $\sum_{n = 1}^{\infty} A_{n} = {l i m}_{n \to + \infty} \sum_{k = 1}^{n} A_{k}$ $\sum_{k = 1}^{n} A_{k} = \sum_{k = 1}^{n} {l n (k + 1) - l n (k)} - \sum_{k = 1}^{n} \frac{1}{k + 1}$ $= l n (n + 1) - \sum_{k = 2}^{n + 1} \frac{1}{k} = l n (n + 1) - (H_{n + 1} - 1)$ $= 1 - (H_{n + 1} - l n (n + 1)) b u t$ $H_{n + 1} = l n (n + 1) + γ + o (\frac{1}{n}) \Rightarrow H_{n + 1} - l n (n + 1) = γ + o (\frac{1}{n}) \Rightarrow$ $\sum_{k = 1}^{n} A_{k} = 1 - γ + o (\frac{1}{n}) \Rightarrow {l i m}_{n \to + \infty} \sum_{k = 1}^{n} A_{k} = 1 - γ = I .$
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