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](Q56935.png)
$$\mathrm{1}.\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\left({x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\right){e}^{−{n}\left[{x}\right]} {dx}\:\:{with}\:{n}\:{integr}\:{natural}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\mathrm{2}.\:{find}\:{nature}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
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 .](Q57019.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} \:{e}^{−{n}\left[{x}\right]} {dx}\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−{n}\left[{x}\right]} {dx}\:+\int_{\mathrm{0}} ^{\infty} \:{e}^{−{n}\left[{x}\right]} {dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{n}\left[{x}\right]} {dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{nk}} {dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−{n}} \right)^{{k}} \:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−{n}\left[{x}\right]} {dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{x}\:{e}^{−{nk}} \:{dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\mathrm{2}{k}+\mathrm{1}\right\} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}\:{e}^{−{nk}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \\ $$$${let}\:{w}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{\infty} \:{x}^{{k}} \:{with}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{w}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{w}^{'} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${also}\:{w}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}\:{x}^{{k}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:{kx}^{{k}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}\:{e}^{−{nk}} \:=\frac{{e}^{−{n}} }{\left(\mathrm{1}−{e}^{−{n}} \right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:{x}\:{e}^{−{n}\left[{x}\right]} {dx}\:=\frac{{e}^{−{n}} }{\left(\mathrm{1}−{e}^{−{n}} \right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{e}^{−{n}} \right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}^{\mathrm{3}} \:{e}^{−{n}\left[{x}\right]} {dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{x}^{\mathrm{3}} \:{e}^{−{nk}} \:{dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \:\:\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\left({k}+\mathrm{1}\right)^{\mathrm{4}} −{k}^{\mathrm{4}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right\}\left\{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:+{k}^{\mathrm{2}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\mathrm{2}{k}+\mathrm{1}\right\}\left\{\mathrm{2}{k}^{\mathrm{2}} \:+\mathrm{2}{k}\:+\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\:\mathrm{4}{k}^{\mathrm{3}\:} \:+\mathrm{4}{k}^{\mathrm{2}} \:+\mathrm{2}{k}\:+\mathrm{2}{k}^{\mathrm{2}} \:+\mathrm{2}{k}+\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \left\{\mathrm{4}{k}^{\mathrm{3}} \:+\mathrm{6}{k}^{\mathrm{2}} \:+\mathrm{4}{k}\:+\mathrm{1}\right\} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} {k}^{\mathrm{3}} \:{e}^{−{nk}} \:+\frac{\mathrm{3}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}^{\mathrm{2}} \:{e}^{−{nk}} \:+\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}\:{e}^{−{nk}} \:+\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:{kx}^{{k}} \:=\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} {k}^{\mathrm{2}} \:{x}^{{k}−\mathrm{1}} \:=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}−\mathrm{1}\right){x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}−\mathrm{1}−\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{−{x}−\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}^{\mathrm{2}} \:{x}^{{k}} \:=\frac{−{x}^{\mathrm{2}} −{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}^{\mathrm{2}} \:{e}^{−{nk}} \:=\frac{{e}^{−\mathrm{2}{n}} \:+{e}^{−{n}} }{\left(\mathrm{1}−{e}^{−{n}} \right)^{\mathrm{3}} } \\ $$$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}^{\mathrm{2}} \:{x}^{{k}} \:=\frac{{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}^{\mathrm{3}} \:{x}^{{k}−\mathrm{1}} \:=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \:+\mathrm{3}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} } \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}−{x}\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:{k}^{\mathrm{3}} \:{x}^{{k}} \:=\frac{{x}^{\mathrm{3}} \:+\mathrm{4}{x}^{\mathrm{2}} \:+{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{\infty} \:{k}^{\mathrm{3}} \:{e}^{−{nk}} \:=\frac{{e}^{−\mathrm{3}{n}} \:+\mathrm{4}{e}^{−\mathrm{2}{n}} \:+{e}^{−{n}} }{\left({e}^{−{n}} \:−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{the}\:{value}\:{of} \\ $$$${U}_{{n}} {is}\:{determined}\:. \\ $$