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Question Number 65878 by ~ À ® @ 237 ~ last updated on 05/Aug/19
Calculate lim_(a−>∞) ∫_0 ^∞ (dx/(1+x^a ))
$$\:\:{Calculate}\:\:\:{lim}_{{a}−>\infty} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{{a}} }\: \\ $$
Commented by mathmax by abdo last updated on 05/Aug/19
changement x^a =t give x =t^(1/a) ⇒∫_0 ^∞ (dx/(1+x^a )) =∫_0 ^∞ (1/(1+t))(1/a)t^((1/a)−1) dt =(1/a)∫_0 ^∞ (t^((1/a)−1) /(1+t))dt so for a>1 we get 0<(1/a)<1 ⇒ ∫_0 ^∞ (dx/(1+x^a )) =(1/a) (π/(sin((π/a)))) ⇒lim_(a→+∞) ∫_0 ^∞ (dx/(1+x^a )) =lim_(a→+∞) ((π/a)/(sin((π/a)))) =lim_(t→0) (t/(sint)) =1 ( put t=(π/a)) finally lim_(a→+∞) ∫_0 ^∞ (dt/(1+x^a )) =1
$${changement}\:{x}^{{a}} ={t}\:{give}\:{x}\:={t}^{\frac{\mathrm{1}}{{a}}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{{a}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\frac{\mathrm{1}}{{a}}{t}^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:{so}\:{for}\:{a}>\mathrm{1}\:{we}\:{get}\:\mathrm{0}<\frac{\mathrm{1}}{{a}}<\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{{a}} }\:=\frac{\mathrm{1}}{{a}}\:\frac{\pi}{{sin}\left(\frac{\pi}{{a}}\right)}\:\Rightarrow{lim}_{{a}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{{a}} }\:={lim}_{{a}\rightarrow+\infty} \:\:\frac{\frac{\pi}{{a}}}{{sin}\left(\frac{\pi}{{a}}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \:\frac{{t}}{{sint}}\:=\mathrm{1}\:\:\left(\:\:{put}\:{t}=\frac{\pi}{{a}}\right)\:{finally} \\ $$$${lim}_{{a}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{x}^{{a}} }\:=\mathrm{1} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19
thank you sir ! but at the second line , it′s not clear
$${thank}\:{you}\:{sir}\:!\:\:{but}\:\:{at}\:{the}\:{second}\:{line}\:,\:{it}'{s}\:{not}\:{clear} \\ $$
Commented by mathmax by abdo last updated on 05/Aug/19
x =t^(1/a) ⇒dx =(1/a)t^((1/a)−1) dt....
$${x}\:={t}^{\frac{\mathrm{1}}{{a}}} \:\Rightarrow{dx}\:=\frac{\mathrm{1}}{{a}}{t}^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} \:{dt}…. \\ $$
Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19
let f(a)=∫_0 ^∞ (dx/(1+x^a )) . As (1/(1+x^a ))<(1/x^a ) ⇒ f(a) is real change u=(1/(1+x^a )) then x=((1/u)−1)^(1/a) and dx = (1/a).((−1)/u^2 ).((1/u)−1)^((1/a) −1) du Then f(a)=(1/a)∫_0 ^1 u. (1/u^2 ).(((1−u)/u))^((1/a)−1) du = (1/a) ∫_0 ^1 u^((−1)/a) (1−u)^((1/a)−1) du =(1/a) ∫_0 ^1 u^(1−(1/a) −1) (1−u)^((1/a)−1) du= (1/a) B(1−(1/a) ,(1/a))=(1/a).((Γ(1−(1/a))Γ((1/a)))/(Γ(1))) As ∀ z Γ(1−z)Γ(z)=(π/(sin(πz))) we finally get f(a)= ((π/a)/(sin((π/a)))) then easily lim_(a−>∞) f(a)=1 (by changing b=(π/a))
$$\:\:{let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{{a}} }\:\:\:\:\:\:.\:\:\:{As}\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{{a}} }<\frac{\mathrm{1}}{{x}^{{a}} }\:\Rightarrow\:{f}\left({a}\right)\:{is}\:{real} \\ $$$${change}\:{u}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{{a}} }\:\:{then}\:\:{x}=\left(\frac{\mathrm{1}}{{u}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{a}}} \:\:{and}\:\:\:{dx}\:=\:\frac{\mathrm{1}}{{a}}.\frac{−\mathrm{1}}{{u}^{\mathrm{2}} }.\left(\frac{\mathrm{1}}{{u}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{a}}\:−\mathrm{1}} {du} \\ $$$${Then}\:\: \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}.\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} }.\left(\frac{\mathrm{1}−{u}}{{u}}\right)^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} {du} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{\frac{−\mathrm{1}}{{a}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} \:{du} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{a}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{\mathrm{1}−\frac{\mathrm{1}}{{a}}\:−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{{a}}−\mathrm{1}} {du}=\:\frac{\mathrm{1}}{{a}}\:{B}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}}\:,\frac{\mathrm{1}}{{a}}\right)=\frac{\mathrm{1}}{{a}}.\frac{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{a}}\right)\Gamma\left(\frac{\mathrm{1}}{{a}}\right)}{\Gamma\left(\mathrm{1}\right)} \\ $$$${As}\:\:\forall\:{z}\:\:\Gamma\left(\mathrm{1}−{z}\right)\Gamma\left({z}\right)=\frac{\pi}{{sin}\left(\pi{z}\right)}\:\:\:{we}\:{finally}\:{get} \\ $$$${f}\left({a}\right)=\:\frac{\frac{\pi}{{a}}}{{sin}\left(\frac{\pi}{{a}}\right)}\:\:\:{then}\:\:{easily}\:{lim}_{{a}−>\infty} {f}\left({a}\right)=\mathrm{1}\:\left({by}\:{changing}\:\:{b}=\frac{\pi}{{a}}\right) \\ $$
Commented by ~ À ® @ 237 ~ last updated on 05/Aug/19
As (1/(1+x^a ))<(1/(x^a )) ⇒ f(a) is real when a>1
$${As}\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{{a}} }<\frac{\mathrm{1}}{{x}^{{a}} \:}\:\:\Rightarrow\:\:{f}\left({a}\right)\:{is}\:{real}\:{when}\:{a}>\mathrm{1} \\ $$

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