∫_1 ^∞ (({x})/x^3 )dx=1−(π^2 /(12))


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Question Number 143728 by Dwaipayan Shikari last updated on 17/Jun/21
$∫_1 ^∞ (({x})/x^3 )dx=1−(π^2 /(12))$
$\int_{1}^{\infty} \frac{{x}}{x^{3}} d x = 1 - \frac{π^{2}}{12}$
	Answered by mathmax by abdo last updated on 17/Jun/21
	$I=∫_1 ^∞ (({x})/x^3 )dx =∫_1 ^∞ ((x−[x])/x^3 )dx=∫_1 ^∞ (dx/x^2 )−∫_1 ^∞ (([x])/x^3 )dx =[−(1/x)]_1 ^∞ −∫_1 ^∞ (([x])/x^3 )dx=1−∫_1 ^∞ (([x])/x^3 )dx we have ∫_1 ^∞ (([x])/x^3 )dx =Σ_(n=1) ^∞ ∫_n ^(n+1) (n/x^3 )dx =Σ_(n=1) ^∞ n[(1/(−3+1))x^(−3+1) ]_n ^(n+1) =Σ_(n=1) ^∞ n[−(1/(2x^2 ))]_n ^(n+1) =−(1/2)Σ_(n=1) ^∞ n{(1/n^2 )−(1/((n+1)^2 ))} =−(1/2)Σ_(n=1) ^∞ n×(((n+1)^2 −n^2 )/(n^2 (n+1)^2 ))=−(1/2)Σ_(n=1) ^∞ ((2n+1)/(n(n+1)^2 )) ((2x+1)/(x(x+1)^2 ))=(a/x)+(b/(x+1))+(c/((x+1)^2 )) a=1 ,c=1 lim_(x→+∞) xf(x)=0 =a+b ⇒b=−1 ⇒ ((2n+1)/(n(n+1)^2 ))=(1/n)−(1/(n+1)) +(1/((n+1)^2 )) ⇒ Σ_(n=1) ^∞ ((2n+1)/(n(n+1)^2 ))=Σ_(n=1) ^∞ ((1/n)−(1/(n+1)))+Σ_(n=1) ^∞ (1/((n+1)^2 )) =lim_(n→+∞) Σ_(k=1) ^n ((1/k)−(1/(k+1)))+Σ_(n=2) ^∞ (1/n^2 ) =1+(π^2 /6)−1 =(π^2 /6) ⇒I =−(1/2)((π^2 /6))=−(π^2 /(12))$
	$I = \int_{1}^{\infty} \frac{{x}}{x^{3}} dx = \int_{1}^{\infty} \frac{x - [x]}{x^{3}} dx = \int_{1}^{\infty} \frac{dx}{x^{2}} - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx$ $= {[- \frac{1}{x}]}_{1}^{\infty} - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx = 1 - \int_{1}^{\infty} \frac{[x]}{x^{3}} dx we have$ $\int_{1}^{\infty} \frac{[x]}{x^{3}} dx = \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{n}{x^{3}} dx = \sum_{n = 1}^{\infty} n {[\frac{1}{- 3 + 1} x^{- 3 + 1}]}_{n}^{n + 1}$ $= \sum_{n = 1}^{\infty} n {[- \frac{1}{{2 x}^{2}}]}_{n}^{n + 1} = - \frac{1}{2} \sum_{n = 1}^{\infty} n {\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}}$ $= - \frac{1}{2} \sum_{n = 1}^{\infty} n \times \frac{{(n + 1)}^{2} - n^{2}}{n^{2} {(n + 1)}^{2}} = - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{2 n + 1}{n {(n + 1)}^{2}}$ $\frac{2 x + 1}{x {(x + 1)}^{2}} = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}$ $a = 1, c = 1 \lim_{x \to + \infty} xf (x) = 0 = a + b \Rightarrow b = - 1 \Rightarrow$ $\frac{2 n + 1}{n {(n + 1)}^{2}} = \frac{1}{n} - \frac{1}{n + 1} + \frac{1}{{(n + 1)}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{2 n + 1}{n {(n + 1)}^{2}} = \sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{1}{n + 1}) + \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}}$ $= \lim_{n \to + \infty} \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1}) + \sum_{n = 2}^{\infty} \frac{1}{n^{2}}$ $= 1 + \frac{π^{2}}{6} - 1 = \frac{π^{2}}{6} \Rightarrow I = - \frac{1}{2} (\frac{π^{2}}{6}) = - \frac{π^{2}}{12}$
	Commented by mathmax by abdo last updated on 17/Jun/21

	$I = 1 - \frac{π^{2}}{12}$
	Commented by Dwaipayan Shikari last updated on 17/Jun/21

	$T h a n k s s i r$
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