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 = ∫_0 ^n (1−(x/n))^(n−1) lnxdx = ∫_(R ) (1−(x/n))^(n−1) χ_(]0,n[) (x) lnxdx . the sequence of functions f_n (x)= (1−(x/n))^(n−1) χ_(]0,n[) (x)ln(x) c.s. to f(x)= e^(−x) lnx on ]0,+∞[ also we have ∣f_n (x)∣ ≤ e^(−x) ∀ x∈]0,n[ thoreme of convergence dominee give lim_(n→+∞) I_(n ) =lim_(n→+∝) ∫_R f_n (x+dx = ∫_0 ^∞ e^(−x) lnxdx .the ch. (x/n)=t give I_n = n∫_0 ^1 (1−t)^(n−1) (ln(n)+lnt)dt =n ln(n)∫_0 ^1 (1−t)^(n−1) dt + ∫_0 ^1 n(1−t)^(n−1) lntdt =ln(n)[−(1−t)^n ]_0 ^1 +∫_0 ^1 n(1−t)^(n−1) ln(t)dt =ln(n) + ∫_0 ^1 n(1−t)^(n−1) ln(t)dt but by parts ∫_0 ^1 n(1−t)^(n−1) ln(t)dt=([ 1−(1−t)^n )lnt]_0 ^1 −∫_0 ^1 ((1−(1−t)^n )/t)dt = −∫_0 ^1 ((1−(1−t)^n )/t)dt (look that lim_(t→0) (1−(1−t)^n )lnt=0) the ch. 1−t=x give −∫_0 ^1 ((1−(1−t)^n )/t)dt = −∫_0 ^1 ((1−x^n )/(1−x))dx =−∫_0 ^1 (1+x+x^2 +...x^(n−1) )dx =−∫_0 ^1 (Σ_(k=0) ^(n−1) x^k )dx =−Σ_(k=0) ^(n−1) ∫_0 ^1 x^k dx=−Σ_(k=0) ^(n−1) (1/(k+1))=−Σ_(k=1) ^n (1/k)=−H_n so I_n = ln(n)−H_n = −( H_n −ln(n))_(n→+∝ ) →−γ so ∫_0 ^∞ e^(−x) ln(x)dx=−γ (the costant number of Euler)](https://www.tinkutara.com/question/Q28276.png)
$${let}\:{put}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} {lnxdx} \\ $$$$=\:\int_{{R}\:} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} \chi_{\left.\right]\mathrm{0},{n}\left[\right.} \left({x}\right)\:{lnxdx}\:\:.\:{the}\:{sequence}\:{of}\:{functions} \\ $$$${f}_{{n}} \left({x}\right)=\:\left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}−\mathrm{1}} \chi_{\left.\right]\mathrm{0},{n}\left[\right.} \left({x}\right){ln}\left({x}\right)\:\:{c}.{s}.\:{to}\:{f}\left({x}\right)=\:{e}^{−{x}} {lnx}\:{on} \\ $$$$\left.\right]\mathrm{0},+\infty\left[\:\:{also}\:{we}\:{have}\:\mid{f}_{{n}} \left({x}\right)\mid\:\leqslant\:{e}^{−{x}} \:\:\:\forall\:{x}\in\right]\mathrm{0},{n}\left[\:{thoreme}\:{of}\right. \\ $$$${convergence}\:{dominee}\:{give}\:{lim}_{{n}\rightarrow+\infty} {I}_{{n}\:} ={lim}_{{n}\rightarrow+\propto} \int_{{R}} {f}_{{n}} \left({x}+{dx}\right. \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {lnxdx}\:\:.{the}\:{ch}.\:\frac{{x}}{{n}}={t}\:{give} \\ $$$${I}_{{n}} =\:\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} \left({ln}\left({n}\right)+{lnt}\right){dt} \\ $$$$={n}\:{ln}\left({n}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {dt}\:\:+\:\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {lntdt} \\ $$$$={ln}\left({n}\right)\left[−\left(\mathrm{1}−{t}\right)^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt} \\ $$$$={ln}\left({n}\right)\:+\:\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt}\:\:\:\:{but}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {n}\left(\mathrm{1}−{t}\right)^{{n}−\mathrm{1}} {ln}\left({t}\right){dt}=\left(\left[\:\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} \right){lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:\:\:\left({look}\:{that}\:{lim}_{{t}\rightarrow\mathrm{0}} \left(\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} \right){lnt}=\mathrm{0}\right) \\ $$$${the}\:{ch}.\:\mathrm{1}−{t}={x}\:{give} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}−\left(\mathrm{1}−{t}\right)^{{n}} }{{t}}{dt}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+…{x}^{{n}−\mathrm{1}} \right){dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {x}^{{k}} \right){dx} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} {dx}=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}=−\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}=−{H}_{{n}} \:{so} \\ $$$${I}_{{n}} =\:{ln}\left({n}\right)−{H}_{{n}} =\:−\left(\:{H}_{{n}} \:\:−{ln}\left({n}\right)\right)_{{n}\rightarrow+\propto\:\:\:} \rightarrow−\gamma\: \\ $$$${so}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \:{ln}\left({x}\right){dx}=−\gamma\:\:\:\:\:\:\left({the}\:{costant}\:{number}\:{of}\:{Euler}\right) \\ $$