Question-227067

Question Number 227067 by Spillover last updated on 29/Dec/25

Answered by peace2 last updated on 29/Dec/25

(1/x)→y =∫_1 ^∞ (1/y).(([y])/y^2 )dy=Σ_(k=1) ^∞ ∫_k ^(k+1) (k/y^3 )dy =Σ_(k≥1) (k/2)[(1/k^2 )−(1/((1+k)^2 ))] =Σ_(k≥1) (1/2)[(1/k)−(k/((1+k)^2 ))]=(1/2)Σ_(k≥1) [(1/k)−(1/(k+1))+(1/((k+1)^2 ))] =(1/2)Σ_(k≥1) (1/((k+1)^2 ))+(1/2)Σ∫_0 ^1 (x^(k−1) −x^k ) =(1/2)(ζ(2)−1)+(1/2)∫_0 ^1 (1/(1−x))−(x/(1−x))dx =((ζ(2)−1)/2)+(1/2)∫_0 ^1 1dx=((ζ(2))/2)=(π^2 /(12))

$$\frac{\mathrm{1}}{{x}}\rightarrow{y} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{y}}.\frac{\left[{y}\right]}{{y}^{\mathrm{2}} }{dy}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \frac{{k}}{{y}^{\mathrm{3}} }{dy} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{k}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} }\right] \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{k}}−\frac{{k}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} }\right]=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{1}} {\sum}\left[\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\Sigma\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{{k}−\mathrm{1}} −{x}^{{k}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\mathrm{2}\right)−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{{x}}{\mathrm{1}−{x}}{dx} \\ $$$$=\frac{\zeta\left(\mathrm{2}\right)−\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}{dx}=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Answered by Spillover last updated on 29/Dec/25

Answered by Spillover last updated on 29/Dec/25

Answered by Spillover last updated on 29/Dec/25

Commented by Spillover last updated on 29/Dec/25

$${page}\:\mathrm{2} \\ $$

Answered by mr W last updated on 29/Dec/25

∫_0 ^1 x⌈(1/x)⌉dx =Σ_(n=1) ^∞ ∫_(1/(n+1)) ^(1/n) x⌈(1/x)⌉dx =Σ_(n=1) ^∞ ∫_(1/(n+1)) ^(1/n) x(n+1)dx =(1/2)Σ_(n=1) ^∞ (n+1)[(1/n^2 )−(1/((n+1)^2 ))] =(1/2)Σ_(n=1) ^∞ [(1/n^2 )+(1/n)−(1/(n+1))] =(1/2)Σ_(n=1) ^∞ (1/n^2 )+(1/2)Σ_(n=1) ^∞ ((1/n)−(1/(n+1))) =(1/2)×(π^2 /6)+(1/2)×1 =(π^2 /(12))+(1/2) ✓

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\lceil\frac{\mathrm{1}}{{x}}\rceil{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\frac{\mathrm{1}}{{n}+\mathrm{1}}} ^{\frac{\mathrm{1}}{{n}}} {x}\lceil\frac{\mathrm{1}}{{x}}\rceil{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\frac{\mathrm{1}}{{n}+\mathrm{1}}} ^{\frac{\mathrm{1}}{{n}}} {x}\left({n}+\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({n}+\mathrm{1}\right)\left[\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$

Commented by Spillover last updated on 29/Dec/25