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Question Number 73327 by mathmax by abdo last updated on 10/Nov/19
let w(x)=∫_0 ^∞ ((lnt)/((x^2 +t^2 )^2 ))dt 1) explicit w(x) 2) calculate U_n =∫_0 ^∞ ((lnt)/((n^2 +t^2 )^2 ))dt find lim_(n→+∞) n^4 U_n and determine nature of tbe serie Σ U_n
$${let}\:{w}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{explicit}\:{w}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({n}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {n}^{\mathrm{4}} {U}_{{n}} \:\:{and}\:{determine}\:{nature}\:{of}\:{tbe}\:{serie}\:\Sigma\:{U}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
1) we have w(x)=∫_0 ^∞ ((ln(t))/((x^2 +t^2 )^2 ))dt let f(x)=∫_0 ^∞ ((ln(t))/(x^2 +t^2 ))dt we have f^′ (x)=−∫_0 ^∞ ((2xln(t))/((x^2 +t^2 )))dt =−2x w(x) ⇒w(x)=−(1/(2x))f^′ (x) we have f(x)=_(t=xu) ∫_0 ^∞ ((ln(xu))/(x^2 (1+u^(2)) )) xdu (we suppose x>0 because f is even) ⇒f(x)=(1/x) ∫_0 ^∞ ((ln(x)+ln(u))/(1+u^2 ))du =((lnx)/x) ∫_0 ^∞ (du/(1+u^2 )) +(1/x)∫_0 ^∞ ((lnu)/(1+u^2 ))du ( ∫_0 ^∞ ((lnu)/(1+u^2 ))du =0) =((πlnx)/(2x)) ⇒f^′ (x)=(π/2){ ((1−lnx)/x^2 )} =((π(1−lnx))/(2x^2 )) ⇒ w(x)=−(1/(2x))×((π(1−lnx))/(2x^2 )) =((π(lnx−1))/(4x^3 )) 2) U_n =∫_0 ^∞ ((lnt)/((n^2 +t^2 )^2 ))dt ⇒ U_n =((π(ln(n)−1))/(4n^3 )) ⇒ lim_(n→+∞) n^4 U_n =lim_(n→+∞) ((nπ(ln(n)−1))/4) =+∞ U_n =(π/4)((ln(n))/n^3 ) −(π/4) (1/n^3 ) Σ((ln(n))/n^3 ) conv. and Σ (1/n^3 ) ⇒ Σ U_n converges.
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{w}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt} \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{xln}\left({t}\right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}{dt}\:=−\mathrm{2}{x}\:{w}\left({x}\right)\:\Rightarrow{w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}{x}}{f}^{'} \left({x}\right) \\ $$$${we}\:{have}\:{f}\left({x}\right)=_{{t}={xu}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({xu}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\left.\mathrm{2}\right)} \right.}\:{xdu}\:\:\left({we}\:{suppose}\:{x}>\mathrm{0}\:{because}\:{f}\right. \\ $$$$\left.{is}\:{even}\right)\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)+{ln}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{{lnx}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:\:\:\left(\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\mathrm{0}\right) \\ $$$$=\frac{\pi{lnx}}{\mathrm{2}{x}}\:\Rightarrow{f}^{'} \left({x}\right)=\frac{\pi}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}−{lnx}}{{x}^{\mathrm{2}} }\right\}\:=\frac{\pi\left(\mathrm{1}−{lnx}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}{x}}×\frac{\pi\left(\mathrm{1}−{lnx}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\:=\frac{\pi\left({lnx}−\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({n}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\Rightarrow\:{U}_{{n}} =\frac{\pi\left({ln}\left({n}\right)−\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{3}} }\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{n}^{\mathrm{4}} \:{U}_{{n}} ={lim}_{{n}\rightarrow+\infty} \frac{{n}\pi\left({ln}\left({n}\right)−\mathrm{1}\right)}{\mathrm{4}}\:=+\infty \\ $$$${U}_{{n}} =\frac{\pi}{\mathrm{4}}\frac{{ln}\left({n}\right)}{{n}^{\mathrm{3}} }\:−\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$$\Sigma\frac{{ln}\left({n}\right)}{{n}^{\mathrm{3}} }\:{conv}.\:{and}\:\:\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\Rightarrow\:\Sigma\:{U}_{{n}} \:\:{converges}. \\ $$

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