Question Number 72024 by mathmax by abdo last updated on 23/Oct/19
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 31/Oct/19
$${A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right){dx}\:\Rightarrow{A}_{{n}} =\int_{{R}} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right)\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx} \\ $$$${let}\:{f}_{{n}} \left({x}\right)=\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {ln}\left(\mathrm{1}+{x}\right)\chi_{\left[\mathrm{0},{n}\left[\right.\right.} \left({x}\right){dx}\:\:{we}\:\:{have} \\ $$$${f}_{{n}} \:\rightarrow^{{cs}} \:\:\:{f}\left({x}\right)={e}^{−{x}} {ln}\left(\mathrm{1}+{x}\right)\:\:\:{on}\left[\mathrm{0},+\infty\left[\:\:\:{and}\:\mid{f}_{{n}} \left({x}\right)\mid<{f}\left({x}\right)\right.\right. \\ $$$${theorem}\:{of}\:{convergence}\:{dominee}\:{give}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\int_{{R}} {lim}_{{n}\rightarrow+\infty} {f}_{{n}} \left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$$=_{\mathrm{1}+{x}={t}} \:\:\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{−\left({t}−\mathrm{1}\right)} {ln}\left({t}\right){dt}\:={e}\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt} \\ $$$$={e}\:\left(\:\int_{\mathrm{1}} ^{\mathrm{0}} \:{e}^{−{t}} {ln}\left({t}\right){dt}\:+\int_{\mathrm{0}} ^{+\infty} {e}^{−{t}} {ln}\left({t}\right){dt}\right) \\ $$$$=−{e}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{t}} {ln}\left({t}\right){dt}\:+{e}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left({t}\right){dt} \\ $$$$=−{e}\gamma\:−{e}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{t}} {ln}\left({t}\right){dt}\:…..{be}\:{continued}… \\ $$$$ \\ $$