Question Number 38198 by maxmathsup by imad last updated on 22/Jun/18
$${we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {ln}\left({x}\right){dx}=−\gamma \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ax}} {ln}\left({x}\right){dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}} {ln}\left(\frac{{x}}{{n}}\right){dx}\:\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} \\ $$
Commented by abdo.msup.com last updated on 24/Jun/18
$$\left.\mathrm{1}\right)\:{changement}\:{ax}={t}\:{give} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left(\frac{{t}}{{a}}\right)\frac{{dt}}{{a}} \\ $$$$=\:\frac{\mathrm{1}}{{a}}\left\{\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left({t}\right){dt}\:−{ln}\left({a}\right)\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\right\} \\ $$$$=\frac{\mathrm{1}}{{a}}\left\{\:\gamma\:\:−{ln}\left({a}\right)\:\left[\:−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \right\} \\ $$$$=\:\frac{\gamma\:+{ln}\left({a}\right)}{{a}} \\ $$$${f}\left({a}\right)=\frac{\gamma\:+{ln}\left({a}\right)}{{a}}\:\:{with}\:{a}>\mathrm{0} \\ $$
Commented by abdo.msup.com last updated on 24/Jun/18
$$\left.\mathrm{2}\right)\:{changement}\:{nx}={t}\:{give} \\ $$$${u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} \:{ln}\left(\:\frac{{t}}{{n}^{\mathrm{2}} }\right)\:\frac{{dt}}{{n}} \\ $$$$=\frac{\mathrm{1}}{{n}}\left\{\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left({t}\right){dt}\:\:−\mathrm{2}{ln}\left({n}\right)\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\right\} \\ $$$$=\frac{\mathrm{1}}{{n}}\left\{\:\gamma\:\:+\mathrm{2}{ln}\left({n}\right)\right\} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} ={lim}_{{n}\rightarrow+\infty} \left(\:\frac{\gamma}{{n}}\:+\mathrm{2}\frac{{ln}\left({n}\right)}{{n}}\right)=\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
$$\gamma\:{is}\:{the}\:{costant}\:{of}\:{Euler}. \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
$${f}\left({a}\right)=\frac{−\gamma\:+{ln}\left({a}\right)}{{a}} \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
$${u}_{{n}} =\frac{−\gamma\:+\mathrm{2}{ln}\left({n}\right)}{{n}} \\ $$