Question Number 28242 by abdo imad last updated on 22/Jan/18
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {lnxdx}\:\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
$${let}\:{put}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} {lnxdx} \\ $$$$=\:\int_{{R}\:} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} \chi_{\left.\right]\mathrm{0},{n}\left[\right.} \left({x}\right)\:{lnxdx}\:\:.\:{the}\:{sequence}\:{of}\:{functions} \\ $$$${f}_{{n}} \left({x}\right)=\:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} \chi_{\left.\right]\mathrm{0},{n}\left[\right.} \left({x}\right){ln}\left({x}\right)\:\:{c}.{s}.\:{to}\:{f}\left({x}\right)=\:{e}^{−{x}} {lnx}\:{on} \\ $$$$\left.\right]\mathrm{0},+\infty\left[\:\:{also}\:{we}\:{have}\:\mid{f}_{{n}} \left({x}\right)\mid\:\leqslant\:{e}^{−{x}} \:\:\:\forall\:{x}\in\right]\mathrm{0},{n}\left[\:{thoreme}\:{of}\right. \\ $$$${convergence}\:{dominee}\:{give}\:{lim}_{{n}\rightarrow+\infty} {I}_{{n}\:} ={lim}_{{n}\rightarrow+\propto} \int_{{R}} {f}_{{n}} \left({x}+{dx}\right. \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {lnxdx}\:\:.{the}\:{ch}.\:\frac{{x}}{{n}}={t}\:{give} \\ $$$${I}_{{n}} =\:\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} \left({ln}\left({n}\right)+{lnt}\right){dt} \\ $$$$={n}\:{ln}\left({n}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {dt}\:\:+\:\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {lntdt} \\ $$$$={ln}\left({n}\right)\left[−\left(\mathrm{1}−{t}\right)^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt} \\ $$$$={ln}\left({n}\right)\:+\:\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt}\:\:\:\:{but}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt}=\left(\left[\:\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} \right){lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:\:\:\left({look}\:{that}\:{lim}_{{t}\rightarrow\mathrm{0}} \left(\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} \right){lnt}=\mathrm{0}\right) \\ $$$${the}\:{ch}.\:\mathrm{1}−{t}={x}\:{give} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+…{x}^{{n}−\mathrm{1}} \right){dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {x}^{{k}} \right){dx} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} {dx}=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}=−\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}=−{H}_{{n}} \:{so} \\ $$$${I}_{{n}} =\:{ln}\left({n}\right)−{H}_{{n}} =\:−\left(\:{H}_{{n}} \:\:−{ln}\left({n}\right)\right)_{{n}\rightarrow+\propto\:\:\:} \rightarrow−\gamma\: \\ $$$${so}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \:{ln}\left({x}\right){dx}=−\gamma\:\:\:\:\:\:\left({the}\:{costant}\:{number}\:{of}\:{Euler}\right) \\ $$