Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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.

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{w}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }{dt} \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{xln}\left({t}\right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}{dt}\:=−\mathrm{2}{x}\:{w}\left({x}\right)\:\Rightarrow{w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}{x}}{f}^{'} \left({x}\right) \\ $$$${we}\:{have}\:{f}\left({x}\right)=_{{t}={xu}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({xu}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\left.\mathrm{2}\right)} \right.}\:{xdu}\:\:\left({we}\:{suppose}\:{x}>\mathrm{0}\:{because}\:{f}\right. \\ $$$$\left.{is}\:{even}\right)\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)+{ln}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{{lnx}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:\:\:\left(\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\mathrm{0}\right) \\ $$$$=\frac{\pi{lnx}}{\mathrm{2}{x}}\:\Rightarrow{f}^{'} \left({x}\right)=\frac{\pi}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}−{lnx}}{{x}^{\mathrm{2}} }\right\}\:=\frac{\pi\left(\mathrm{1}−{lnx}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}{x}}×\frac{\pi\left(\mathrm{1}−{lnx}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\:=\frac{\pi\left({lnx}−\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left({n}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\Rightarrow\:{U}_{{n}} =\frac{\pi\left({ln}\left({n}\right)−\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{3}} }\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{n}^{\mathrm{4}} \:{U}_{{n}} ={lim}_{{n}\rightarrow+\infty} \frac{{n}\pi\left({ln}\left({n}\right)−\mathrm{1}\right)}{\mathrm{4}}\:=+\infty \\ $$$${U}_{{n}} =\frac{\pi}{\mathrm{4}}\frac{{ln}\left({n}\right)}{{n}^{\mathrm{3}} }\:−\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$$\Sigma\frac{{ln}\left({n}\right)}{{n}^{\mathrm{3}} }\:{conv}.\:{and}\:\:\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\Rightarrow\:\Sigma\:{U}_{{n}} \:\:{converges}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com