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