1. calculate U_n =∫_0 ^∞ (x^3 −2x+1)e^(−n[x]) dx with n integr natural and n≥1 2. find nature of Σ U


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Question Number 56935 by maxmathsup by imad last updated on 27/Mar/19

$1 . c a l c u l a t e U_{n} = \int_{0}^{\infty} (x^{3} - 2 x + 1) e^{- n [x]} d x w i t h n i n t e g r n a t u r a l a n d n ⩾ 1$ $2 . f i n d n a t u r e o f Σ U_{n}$
Commented by maxmathsup by imad last updated on 28/Mar/19
$1) we have U_n =∫_0 ^∞ x^3 e^(−n[x]) dx −2 ∫_0 ^∞ x e^(−n[x]) dx +∫_0 ^∞ e^(−n[x]) dx ∫_0 ^∞ e^(−n[x]) dx =Σ_(k=0) ^∞ ∫_k ^(k+1) e^(−nk) dx =Σ_(k=0) ^∞ (e^(−n) )^k =(1/(1−e^(−n) )) ∫_0 ^∞ x e^(−n[x]) dx =Σ_(k=0) ^∞ ∫_k ^(k+1) x e^(−nk) dx =Σ_(k=0) ^∞ e^(−nk) ∫_k ^(k+1) xdx =(1/2) Σ_(k=0) ^∞ e^(−nk) {(k+1)^2 −k^2 } =(1/2) Σ_(k=0) ^∞ e^(−nk) {2k+1} =Σ_(k=0) ^∞ k e^(−nk) +(1/2) Σ_(k=0) ^∞ e^(−nk) let w(x)=Σ_(k=0) ^∞ x^k with ∣x∣<1 ⇒w(x)=(1/(1−x)) ⇒w^′ (x)=(1/((1−x)^2 )) also w^′ (x)=Σ_(k=1) ^∞ k x^(k−1) =(1/((1−x)^2 )) ⇒Σ_(k=1) ^∞ kx^k =(x/((1−x)^2 )) ⇒ Σ_(k=0) ^∞ k e^(−nk) =(e^(−n) /((1−e^(−n) )^2 )) ⇒∫_0 ^∞ x e^(−n[x]) dx =(e^(−n) /((1−e^(−n) )^2 )) +(1/(2(1−e^(−n) ))) ∫_0 ^∞ x^3 e^(−n[x]) dx =Σ_(k=0) ^∞ ∫_k ^(k+1) x^3 e^(−nk) dx =Σ_(k=0) ^∞ e^(−nk) (1/4){ (k+1)^4 −k^4 } =(1/4) Σ_(k=0) ^∞ e^(−nk) {(k+1)^2 −k^2 }{(k+1)^2 +k^2 } =(1/4) Σ_(k=0) ^∞ e^(−nk) {2k+1}{2k^2 +2k +1} =(1/4) Σ_(k=0) ^∞ e^(−nk) { 4k^(3 ) +4k^2 +2k +2k^2 +2k+1} =(1/4)Σ_(k=0) ^∞ e^(−nk) {4k^3 +6k^2 +4k +1} =Σ_(k=0) ^∞ k^3 e^(−nk) +(3/2) Σ_(k=0) ^∞ k^2 e^(−nk) +Σ_(k=0) ^∞ k e^(−nk) +(1/4) Σ_(k=0) ^∞ e^(−nk) we have Σ_(k=1) ^∞ kx^k =(x/((x−1)^2 )) ⇒Σ_(k=1) ^∞ k^2 x^(k−1) =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) ⇒Σ_(k=1) ^∞ k^2 x^k =((−x^2 −x)/((x−1)^3 )) =((x^2 +x)/((1−x)^3 )) ⇒ Σ_(k=0) ^∞ k^2 e^(−nk) =((e^(−2n) +e^(−n) )/((1−e^(−n) )^3 )) we have Σ_(k=1) ^∞ k^2 x^k =((x^2 +x)/((1−x)^3 )) ⇒Σ_(k=1) ^∞ k^3 x^(k−1) =(((2x+1)(1−x)^3 +3(1−x)^2 (x^2 +x))/((1−x)^6 )) =(((2x+1)(1−x) +3x^2 +3x)/((1−x)^4 )) =((2x−2x^2 +1−x +3x^2 +3x)/((x−1)^4 )) =((x^2 +4x +1)/((x−1)^4 )) ⇒ Σ_(k=1) ^∞ k^3 x^k =((x^3 +4x^2 +x)/((x−1)^4 )) ⇒Σ_(k=0) ^∞ k^3 e^(−nk) =((e^(−3n) +4e^(−2n) +e^(−n) )/((e^(−n) −1)^2 )) ⇒ the value of U_n is determined .$
$1) w e h a v e U_{n} = \int_{0}^{\infty} x^{3} e^{- n [x]} d x - 2 \int_{0}^{\infty} x e^{- n [x]} d x + \int_{0}^{\infty} e^{- n [x]} d x$ $\int_{0}^{\infty} e^{- n [x]} d x = \sum_{k = 0}^{\infty} \int_{k}^{k + 1} e^{- n k} d x = \sum_{k = 0}^{\infty} {(e^{- n})}^{k} = \frac{1}{1 - e^{- n}}$ $\int_{0}^{\infty} x e^{- n [x]} d x = \sum_{k = 0}^{\infty} \int_{k}^{k + 1} x e^{- n k} d x = \sum_{k = 0}^{\infty} e^{- n k} \int_{k}^{k + 1} x d x$ $= \frac{1}{2} \sum_{k = 0}^{\infty} e^{- n k} {{(k + 1)}^{2} - k^{2}} = \frac{1}{2} \sum_{k = 0}^{\infty} e^{- n k} {2 k + 1}$ $= \sum_{k = 0}^{\infty} k e^{- n k} + \frac{1}{2} \sum_{k = 0}^{\infty} e^{- n k}$ $l e t w (x) = \sum_{k = 0}^{\infty} x^{k} w i t h ∣ x ∣< 1 \Rightarrow w (x) = \frac{1}{1 - x} \Rightarrow w^{^{'}} (x) = \frac{1}{{(1 - x)}^{2}}$ $a l s o w^{^{'}} (x) = \sum_{k = 1}^{\infty} k x^{k - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{k = 1}^{\infty} {k x}^{k} = \frac{x}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{k = 0}^{\infty} k e^{- n k} = \frac{e^{- n}}{{(1 - e^{- n})}^{2}} \Rightarrow \int_{0}^{\infty} x e^{- n [x]} d x = \frac{e^{- n}}{{(1 - e^{- n})}^{2}} + \frac{1}{2 (1 - e^{- n})}$ $\int_{0}^{\infty} x^{3} e^{- n [x]} d x = \sum_{k = 0}^{\infty} \int_{k}^{k + 1} x^{3} e^{- n k} d x = \sum_{k = 0}^{\infty} e^{- n k} \frac{1}{4} {{(k + 1)}^{4} - k^{4}}$ $= \frac{1}{4} \sum_{k = 0}^{\infty} e^{- n k} {{(k + 1)}^{2} - k^{2}} {{(k + 1)}^{2} + k^{2}}$ $= \frac{1}{4} \sum_{k = 0}^{\infty} e^{- n k} {2 k + 1} {2 k^{2} + 2 k + 1}$ $= \frac{1}{4} \sum_{k = 0}^{\infty} e^{- n k} {4 k^{3} + 4 k^{2} + 2 k + 2 k^{2} + 2 k + 1}$ $= \frac{1}{4} \sum_{k = 0}^{\infty} e^{- n k} {4 k^{3} + 6 k^{2} + 4 k + 1}$ $= \sum_{k = 0}^{\infty} k^{3} e^{- n k} + \frac{3}{2} \sum_{k = 0}^{\infty} k^{2} e^{- n k} + \sum_{k = 0}^{\infty} k e^{- n k} + \frac{1}{4} \sum_{k = 0}^{\infty} e^{- n k}$ $w e h a v e \sum_{k = 1}^{\infty} {k x}^{k} = \frac{x}{{(x - 1)}^{2}} \Rightarrow \sum_{k = 1}^{\infty} k^{2} x^{k - 1} = \frac{{(x - 1)}^{2} - 2 (x - 1) x}{{(x - 1)}^{4}}$ $= \frac{x - 1 - 2 x}{{(x - 1)}^{3}} = \frac{- x - 1}{{(x - 1)}^{3}} \Rightarrow \sum_{k = 1}^{\infty} k^{2} x^{k} = \frac{- x^{2} - x}{{(x - 1)}^{3}} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow$ $\sum_{k = 0}^{\infty} k^{2} e^{- n k} = \frac{e^{- 2 n} + e^{- n}}{{(1 - e^{- n})}^{3}}$ $w e h a v e \sum_{k = 1}^{\infty} k^{2} x^{k} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow \sum_{k = 1}^{\infty} k^{3} x^{k - 1} = \frac{(2 x + 1) {(1 - x)}^{3} + 3 {(1 - x)}^{2} (x^{2} + x)}{{(1 - x)}^{6}}$ $= \frac{(2 x + 1) (1 - x) + 3 x^{2} + 3 x}{{(1 - x)}^{4}} = \frac{2 x - 2 x^{2} + 1 - x + 3 x^{2} + 3 x}{{(x - 1)}^{4}} = \frac{x^{2} + 4 x + 1}{{(x - 1)}^{4}} \Rightarrow$ $\sum_{k = 1}^{\infty} k^{3} x^{k} = \frac{x^{3} + 4 x^{2} + x}{{(x - 1)}^{4}} \Rightarrow \sum_{k = 0}^{\infty} k^{3} e^{- n k} = \frac{e^{- 3 n} + 4 e^{- 2 n} + e^{- n}}{{(e^{- n} - 1)}^{2}} \Rightarrow t h e v a l u e o f$ $U_{n} i s d e t e r m i n e d .$
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