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1-calculate-U-n-0-x-3-2x-1-e-n-x-dx-with-n-integr-natural-and-n-1-2-find-nature-of-U-n-

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Question Number 56935 by maxmathsup by imad last updated on 27/Mar/19
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_n
1.calculateUn=∫0∞(x3−2x+1)e−n[x]dxwithnintegrnaturalandn⩾12.findnatureofΣUn
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)wehaveUn=∫0∞x3e−n[x]dx−2∫0∞xe−n[x]dx+∫0∞e−n[x]dx∫0∞e−n[x]dx=∑k=0∞∫kk+1e−nkdx=∑k=0∞(e−n)k=11−e−n∫0∞xe−n[x]dx=∑k=0∞∫kk+1xe−nkdx=∑k=0∞e−nk∫kk+1xdx=12∑k=0∞e−nk{(k+1)2−k2}=12∑k=0∞e−nk{2k+1}=∑k=0∞ke−nk+12∑k=0∞e−nkletw(x)=∑k=0∞xkwith∣x∣<1⇒w(x)=11−x⇒w′(x)=1(1−x)2alsow′(x)=∑k=1∞kxk−1=1(1−x)2⇒∑k=1∞kxk=x(1−x)2⇒∑k=0∞ke−nk=e−n(1−e−n)2⇒∫0∞xe−n[x]dx=e−n(1−e−n)2+12(1−e−n)∫0∞x3e−n[x]dx=∑k=0∞∫kk+1x3e−nkdx=∑k=0∞e−nk14{(k+1)4−k4}=14∑k=0∞e−nk{(k+1)2−k2}{(k+1)2+k2}=14∑k=0∞e−nk{2k+1}{2k2+2k+1}=14∑k=0∞e−nk{4k3+4k2+2k+2k2+2k+1}=14∑k=0∞e−nk{4k3+6k2+4k+1}=∑k=0∞k3e−nk+32∑k=0∞k2e−nk+∑k=0∞ke−nk+14∑k=0∞e−nkwehave∑k=1∞kxk=x(x−1)2⇒∑k=1∞k2xk−1=(x−1)2−2(x−1)x(x−1)4=x−1−2x(x−1)3=−x−1(x−1)3⇒∑k=1∞k2xk=−x2−x(x−1)3=x2+x(1−x)3⇒∑k=0∞k2e−nk=e−2n+e−n(1−e−n)3wehave∑k=1∞k2xk=x2+x(1−x)3⇒∑k=1∞k3xk−1=(2x+1)(1−x)3+3(1−x)2(x2+x)(1−x)6=(2x+1)(1−x)+3x2+3x(1−x)4=2x−2x2+1−x+3x2+3x(x−1)4=x2+4x+1(x−1)4⇒∑k=1∞k3xk=x3+4x2+x(x−1)4⇒∑k=0∞k3e−nk=e−3n+4e−2n+e−n(e−n−1)2⇒thevalueofUnisdetermined.

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