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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 bymathmax 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 bymathmax 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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