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Question Number 61530 by maxmathsup by imad last updated on 04/Jun/19
let U_n =∫_0 ^∞ (x^(−2n) /(1+x^4 )) dx with n integr natural and n≥1 1) calculate U_n interms of n 2) find lim_(n→+∞) n^2 U_n 3) study the serie Σ U_n
$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{−\mathrm{2}{n}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:\:{with}\:{n}\:{integr}\:{natural}\:{and}\:\:\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{n}^{\mathrm{2}} \:{U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{serie}\:\Sigma\:{U}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
1) changement x^4 =t give x =t^(1/4) ⇒U_n =∫_0 ^∞ (((t^(1/4) )^((−2)/n) )/(1+t)) (1/4)t^((1/4)−1) dt =(1/4)∫_0 ^∞ (t^(−(1/(2n))+(1/4)−1) /(1+t)) dt =(1/4) ∫_0 ^∞ (t^((1/4)−(1/(2n)) −1) /(1+t))dt let a=(1/4)−(1/(2n)) ⇒a =((2n−4)/(8n)) =((n−2)/(4n))>0 for n>2 and a−1 =((n−2)/(4n)) −1 =((n−2−4n)/(4n)) =((−3n−2)/(4n)) <0 ⇒0<a<1 ⇒ U_n = (1/4) (π/(sin(πa))) =(π/(4sin(π((1/4)−(1/(2n)))))) =(π/(4 sin((π/4)−(π/(2n)))))
$$\left.\mathrm{1}\right)\:{changement}\:{x}^{\mathrm{4}} ={t}\:{give}\:{x}\:={t}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left({t}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)^{\frac{−\mathrm{2}}{{n}}} \:\:\:}{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:{let}\:\:{a}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow{a}\:=\frac{\mathrm{2}{n}−\mathrm{4}}{\mathrm{8}{n}}\:=\frac{{n}−\mathrm{2}}{\mathrm{4}{n}}>\mathrm{0} \\ $$$${for}\:{n}>\mathrm{2}\:\:{and}\:{a}−\mathrm{1}\:=\frac{{n}−\mathrm{2}}{\mathrm{4}{n}}\:−\mathrm{1}\:=\frac{{n}−\mathrm{2}−\mathrm{4}{n}}{\mathrm{4}{n}}\:=\frac{−\mathrm{3}{n}−\mathrm{2}}{\mathrm{4}{n}}\:<\mathrm{0}\:\Rightarrow\mathrm{0}<{a}<\mathrm{1}\:\Rightarrow \\ $$$${U}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:=\frac{\pi}{\mathrm{4}{sin}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\right)}\:=\frac{\pi}{\mathrm{4}\:{sin}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
forgive U_n = ∫_0 ^∞ (x^(−(2/n)) /(1+x^4 )) dx with n≥3
$${forgive}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{−\frac{\mathrm{2}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:{with}\:{n}\geqslant\mathrm{3} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
i have used the result ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) with 0<a<1
$${i}\:{have}\:{used}\:\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\:{with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
2) sin((π/4)−(π/(2n))) =sin((π/4))cos((π/(2n)))−cos((π/4))sin((π/(2n)))=((√2)/2)(cos((π/(2n)))−sin((π/(2n)))) ⇒ cos((π/(2n)))∼1−((((π/(2n)))^2 )/2) =1−(π^2 /(8n^2 )) , sin((π/(2n)))∼ (π/(2n)) ⇒ U_n ∼ (π/(2(√2){1−(π^2 /(8n^2 ))−(π/(2n))})) =(1/(2(√2){ (1/π)−(π/(8n^2 )) −(1/(2n))})) ⇒n^2 U_n ∼ (n^2 /(2(√2){(1/π)−(π/(8n^2 ))−(1/(2n))})) lim_(n→+∞) ((1/π) −(π/(8n^2 )) (1/(2n))) =(1/π) ⇒lim_(n→+∞) n^2 U_n =+∞
$$\left.\mathrm{2}\right)\:{sin}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}{n}}\right)\:={sin}\left(\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)−{cos}\left(\frac{\pi}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)−{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\right)\:\Rightarrow \\ $$$${cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)\sim\mathrm{1}−\frac{\left(\frac{\pi}{\mathrm{2}{n}}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}{n}^{\mathrm{2}} }\:\:\:\:,\:\:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\sim\:\frac{\pi}{\mathrm{2}{n}}\:\Rightarrow \\ $$$${U}_{{n}} \sim\:\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\left\{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}{n}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{n}}\right\}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{\pi}−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\right\}}\:\Rightarrow{n}^{\mathrm{2}} \:{U}_{{n}} \sim\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\left\{\frac{\mathrm{1}}{\pi}−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{n}}\right\}} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\left(\frac{\mathrm{1}}{\pi}\:−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\:=\frac{\mathrm{1}}{\pi}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {n}^{\mathrm{2}} \:{U}_{{n}} =+\infty \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
3) we lim_(n→∞) U_n =(π/(2(√2))) ≠0 ⇒ Σ U_n diverges .
$$\left.\mathrm{3}\right)\:{we}\:\:{lim}_{{n}\rightarrow\infty} {U}_{{n}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\neq\mathrm{0}\:\Rightarrow\:\Sigma\:{U}_{{n}} \:{diverges}\:. \\ $$

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