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Question Number 66349 by mathmax by abdo last updated on 12/Aug/19
study the convergence of ∫_1 ^(+∞) ((arctan(x−1))/((x^2 −1)^(4/3) ))dx
$${study}\:{the}\:{convergence}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{arctan}\left({x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Aug/19
changement x−1=t give ∫_1 ^(+∞) ((arctan(x−1))/((x^2 −1)^(4/3) )) dx =∫_0 ^(+∞) ((arctan(t))/(((t+1)^2 −1)^(4/3) ))dt =∫_0 ^∞ ((arctan(t))/((t^2 +2t)^(4/3) ))dt for t∈V(0) ((arctan(t))/((t^2 +2t)^(4/3) ))∼(t/(t^(4/3) (t+2)^(4/3) )) ∼(1/(t^(1/3) .2^(4/3) )) ∫_0 ^1 (dt/(2^(4/3) t^(1/3) )) converges because 0<(1/3)<1 at V(+∞) lim_(t→+∞) t^2 ((arctan(t))/((t^2 +2t)^(4/3) )) =lim_(t→+∞) (π/2) (t^2 /(t^(8/3) (1+(2/t))^(4/3) )) =lim_(t→+∞) (π/(2t^(2/3) )) =0 so the integral ∫_1 ^(+∞) ((arctan(t))/((t^2 +2t)^(4/3) ))dt converges finally this integral is convergent.
$${changement}\:{x}−\mathrm{1}={t}\:{give} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{arctan}\left({x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:{dx}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{arctan}\left({t}\right)}{\left(\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dt}\:\:\:\:{for}\:{t}\in{V}\left(\mathrm{0}\right)\:\:\:\frac{{arctan}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\sim\frac{{t}}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:\sim\frac{\mathrm{1}}{{t}^{\frac{\mathrm{1}}{\mathrm{3}}} .\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\:{converges}\:{because}\:\mathrm{0}<\frac{\mathrm{1}}{\mathrm{3}}<\mathrm{1} \\ $$$${at}\:{V}\left(+\infty\right)\:\:{lim}_{{t}\rightarrow+\infty} \:\:\:\:{t}^{\mathrm{2}} \:\frac{{arctan}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:={lim}_{{t}\rightarrow+\infty} \:\:\:\frac{\pi}{\mathrm{2}}\:\frac{{t}^{\mathrm{2}} }{{t}^{\frac{\mathrm{8}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{2}}{{t}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$$$={lim}_{{t}\rightarrow+\infty} \:\:\:\frac{\pi}{\mathrm{2}{t}^{\frac{\mathrm{2}}{\mathrm{3}}} }\:=\mathrm{0}\:\:\:{so}\:{the}\:{integral}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{arctan}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dt}\:{converges} \\ $$$${finally}\:{this}\:{integral}\:{is}\:{convergent}. \\ $$

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