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Question Number 89244 by 174 last updated on 16/Apr/20

Commented by mathmax by abdo last updated on 16/Apr/20
$A =∫_0 ^1 {(1/x)}^2 dx cha7gement (1/x)=t give A =−∫_1 ^(+∞) {t}^2 (−(dt/t^2 )) =∫_1 ^(+∞) (((t−[t])^2 )/t^2 )dt = ∫_1 ^(+∞) ((t^2 −2t[t]+[t]^2 )/t^2 )dt =∫_1 ^(+∞) (1−2(([t])/t) +(([t]^2 )/t^2 ))dt =Σ_(n=1) ^∞ ∫_n ^(n+1) (1−2(n/t) +(n^2 /t^2 ))dt but ∫_n ^(n+1) (1−2(n/t)+(n^2 /t^2 ))dt =[t−2nln(t)−(n^2 /t)]_n ^(n+1) =1 −2nln(((n+1)/n))−n^2 ((1/(n+1))−(1/n)) =1−2nln(1+(1/n))+(n^2 /(n(n+1))) =1−2nln(1+(1/n))+(n/(n+1)) ln(1+(1/n))∼−2n((1/n)−(1/(2n^2 )))=−2+(1/n^2 ) ⇒ 1−2nln(1+(1/n))+(n/(n+1)) ∼(1/n^2 ) and Σ (1/n^2 ) is convergent....$
$A = \int_{0}^{1} {\frac{1}{x}}^{2} d x c h a 7 g e m e n t \frac{1}{x} = t g i v e$ $A = - \int_{1}^{+ \infty} {t}^{2} (- \frac{d t}{t^{2}}) = \int_{1}^{+ \infty} \frac{{(t - [t])}^{2}}{t^{2}} d t$ $= \int_{1}^{+ \infty} \frac{t^{2} - 2 t [t] + {[t]}^{2}}{t^{2}} d t = \int_{1}^{+ \infty} (1 - 2 \frac{[t]}{t} + \frac{{[t]}^{2}}{t^{2}}) d t$ $= \sum_{n = 1}^{\infty} \int_{n}^{n + 1} (1 - 2 \frac{n}{t} + \frac{n^{2}}{t^{2}}) d t$ $b u t \int_{n}^{n + 1} (1 - 2 \frac{n}{t} + \frac{n^{2}}{t^{2}}) d t = {[t - 2 n l n (t) - \frac{n^{2}}{t}]}_{n}^{n + 1}$ $= 1 - 2 n l n (\frac{n + 1}{n}) - n^{2} (\frac{1}{n + 1} - \frac{1}{n})$ $= 1 - 2 n l n (1 + \frac{1}{n}) + \frac{n^{2}}{n (n + 1)} = 1 - 2 n l n (1 + \frac{1}{n}) + \frac{n}{n + 1}$ $l n (1 + \frac{1}{n}) \sim - 2 n (\frac{1}{n} - \frac{1}{2 n^{2}}) = - 2 + \frac{1}{n^{2}} \Rightarrow$ $1 - 2 n l n (1 + \frac{1}{n}) + \frac{n}{n + 1} \sim \frac{1}{n^{2}} a n d Σ \frac{1}{n^{2}} i s c o n v e r g e n t . . . .$
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