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Question Number 88069 by Power last updated on 08/Apr/20

Commented by Power last updated on 08/Apr/20

$⌊ x ⌋ - greatest integer$ $x = ⌊ x ⌋ + {x} 0 ⩽ {x} < 1$
Commented by mathmax by abdo last updated on 08/Apr/20
$A =∫_0 ^1 x{(1/x)}[(1/x)]dx we have (1/x)=[(1/x)]+{(1/x)} ⇒ A =∫_0 ^1 x((1/x)−[(1/x)])[(1/x)]dx =∫_0 ^1 (1−x[(1/x)])[(1/x)]dx we do the changement (1/x) =t ⇒ A =−∫_1 ^(+∞) (1−(1/t)[t])[t] (−(dt/t^2 )) =∫_1 ^(+∞) ((1/t^2 )−(([t]^2 )/t^3 ))dt =∫_1 ^(+∞ ) (dt/t^2 )−∫_1 ^(+∞) (([t]^2 )/t^3 )dt but we have ∫_1 ^(+∞) (dt/t^2 ) =[−(1/t)]_1 ^(+∞) =1 ∫_1 ^(+∞) (([t]^2 )/t^3 )dt =Σ_(n=1) ^∞ ∫_n ^(n+1) (n^2 /t^3 )dt =Σ_(n=1) ^∞ n^2 ∫_n ^(n+1) t^(−3) dt =Σ_(n=1) ^∞ n^2 [−(1/2)t^(−2) ]_n ^(n+1) =−(1/2)Σ_(n=1) ^∞ n^2 {(1/((n+1)^2 ))−(1/n^2 )} =−(1/2)Σ_(n=1) ^∞ ((n^2 /((n+1)^2 ))−1) =−(1/2)Σ_(n=1) ^∞ (((n^2 −(n^2 +2n+1))/((n+1)^2 ))) =−(1/2) Σ_(n=1) ^∞ ((−2n−1)/((n+1)^2 )) =Σ_(n=1) ^∞ ((2n+1)/(2(n+1)^2 )) and this serie is divergente because ((2n+1)/(2(n+1)^2 )) ∼(1/n)( n∼+∞)$
$A = \int_{0}^{1} x {\frac{1}{x}} [\frac{1}{x}] d x w e h a v e \frac{1}{x} = [\frac{1}{x}] + {\frac{1}{x}} \Rightarrow$ $A = \int_{0}^{1} x (\frac{1}{x} - [\frac{1}{x}]) [\frac{1}{x}] d x = \int_{0}^{1} (1 - x [\frac{1}{x}]) [\frac{1}{x}] d x w e d o t h e c h a n g e m e n t$ $\frac{1}{x} = t \Rightarrow A = - \int_{1}^{+ \infty} (1 - \frac{1}{t} [t]) [t] (- \frac{d t}{t^{2}})$ $= \int_{1}^{+ \infty} (\frac{1}{t^{2}} - \frac{{[t]}^{2}}{t^{3}}) d t = \int_{1}^{+ \infty} \frac{d t}{t^{2}} - \int_{1}^{+ \infty} \frac{{[t]}^{2}}{t^{3}} d t b u t w e h a v e$ $\int_{1}^{+ \infty} \frac{d t}{t^{2}} = {[- \frac{1}{t}]}_{1}^{+ \infty} = 1$ $\int_{1}^{+ \infty} \frac{{[t]}^{2}}{t^{3}} d t = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{n^{2}}{t^{3}} d t = \sum_{n = 1}^{\infty} n^{2} \int_{n}^{n + 1} t^{- 3} d t$ $= \sum_{n = 1}^{\infty} n^{2} {[- \frac{1}{2} t^{- 2}]}_{n}^{n + 1} = - \frac{1}{2} \sum_{n = 1}^{\infty} n^{2} {\frac{1}{{(n + 1)}^{2}} - \frac{1}{n^{2}}}$ $= - \frac{1}{2} \sum_{n = 1}^{\infty} (\frac{n^{2}}{{(n + 1)}^{2}} - 1) = - \frac{1}{2} \sum_{n = 1}^{\infty} (\frac{n^{2} - (n^{2} + 2 n + 1)}{{(n + 1)}^{2}})$ $= - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{- 2 n - 1}{{(n + 1)}^{2}} = \sum_{n = 1}^{\infty} \frac{2 n + 1}{2 {(n + 1)}^{2}} a n d t h i s s e r i e i s d i v e r g e n t e$ $b e c a u s e \frac{2 n + 1}{2 {(n + 1)}^{2}} \sim \frac{1}{n} (n \sim + \infty)$
	Answered by mahdi last updated on 08/Apr/20
	$(1/x)=[(1/x)]+{(1/x)}⇒x{(1/x)}[(1/x)]=x((1/x)−[(1/x)])[(1/x)] =[(1/x)]−x[(1/x)]^2 ∫_0 ^1 x{(1/x)}[(1/x)]dx=∫_0 ^1 ([(1/x)]−x[(1/x)]^2 )dx= Σ_(n=1) ^∞ ∫_(1/(n+1)) ^(1/n) ([(1/x)]−x[(1/x)]^2 )dx= Σ_(n=1) ^∞ ∫_(1/(n+1)) ^(1/n) (n−xn^2 )dx=Σ_(n=1) ^∞ [nx−((x^2 n^2 )/2)]_(1/(n+1)) ^(1/n) Σ_(n=1) ^∞ ((1/(n+1))−(1/(2(n+1)^2 )))=Σ_(n=1) ^∞ (((2n+1)/(2(n+1)^2 )))$
	$\frac{1}{x} = [\frac{1}{x}] + {\frac{1}{x}} \Rightarrow x {\frac{1}{x}} [\frac{1}{x}] = x (\frac{1}{x} - [\frac{1}{x}]) [\frac{1}{x}]$ $= [\frac{1}{x}] - x {[\frac{1}{x}]}^{2}$ $\int_{0}^{1} x {\frac{1}{x}} [\frac{1}{x}] dx = \int_{0}^{1} ([\frac{1}{x}] - x {[\frac{1}{x}]}^{2}) dx =$ $\underset{n = 1}{\sum^{\infty}} \int_{\frac{1}{n + 1}}^{\frac{1}{n}} ([\frac{1}{x}] - x {[\frac{1}{x}]}^{2}) dx =$ $\underset{n = 1}{\sum^{\infty}} \int_{\frac{1}{n + 1}}^{\frac{1}{n}} (n - {xn}^{2}) dx = \underset{n = 1}{\sum^{\infty}} {[nx - \frac{x^{2} n^{2}}{2}]}_{\frac{1}{n + 1}}^{\frac{1}{n}}$ $\underset{n = 1}{\sum^{\infty}} (\frac{1}{n + 1} - \frac{1}{2 {(n + 1)}^{2}}) = \underset{n = 1}{\sum^{\infty}} (\frac{2 n + 1}{2 {(n + 1)}^{2}})$
	Commented by mahdi last updated on 08/Apr/20

	$i think : \underset{n = 1}{\sum^{\infty}} (\frac{2 n + 1}{2 {(n + 1)}^{2}}) = + \infty$
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