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Question Number 42812 by maxmathsup by imad last updated on 02/Sep/18
study the convervence of ∫_1 ^(+∞) ((arctan(x−1))/((x^2 −1)^(4/3) )) dx
$${study}\:{the}\:{convervence}\:{of}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{arctan}\left({x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 08/Sep/18
changement x−1 =t give I = ∫_0 ^(+∞) ((arctan(t))/(t^(4/3) (t+2)^(4/3) )) dt ⇒ I = ∫_0 ^1 ((arctan(t))/(t^(4/3) (t+2)^(4/3) )) dt +∫_1 ^(+∞) ((arctan(t))/(t^(4/3) (t+2)^(4/3) ))dt at V(0) ((arctant)/(t^(4/3) (t+2)^(4/3) )) ∼ (t/(t^(4/3) 2^(4/3) )) = (1/(t^(1/3) 2^(4/3) )) and the integral ∫_0 ^1 (dt/(2^(4/3) t^(1/3) )) converges from another side lim_(t→+∞) t^2 ((arctan(t))/(t^(4/3) (t+2)^(4/3) )) =lim_(t→+∞) ((arctan(t))/t^((8/3)−2) ) =lim_(t→+∞) ((arctan(t))/t^(2/3) ) =0 so the integral ∫_1 ^(+∞) ((arctan(t))/(t^(4/3) (t+2)^(4/3) ))dt converges the convergence of I is assured.
$${changement}\:{x}−\mathrm{1}\:={t}\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:{dt}\:\:\:\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:{dt}\:+\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dt} \\ $$$${at}\:{V}\left(\mathrm{0}\right)\:\:\:\frac{{arctant}}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }\:\sim\:\frac{{t}}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \:\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} }\:=\:\:\frac{\mathrm{1}}{{t}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} }\:{and}\:{the}\:\:{integral}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:{converges} \\ $$$${from}\:{another}\:{side}\:\:{lim}_{{t}\rightarrow+\infty} \:{t}^{\mathrm{2}} \:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$$$={lim}_{{t}\rightarrow+\infty} \:\:\:\:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{2}} }\:={lim}_{{t}\rightarrow+\infty} \:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{2}}{\mathrm{3}}} }\:=\mathrm{0}\:{so}\:{the}\:{integral} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({t}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dt}\:{converges}\:{the}\:{convergence}\:{of}\:{I}\:{is}\:{assured}. \\ $$

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