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Question Number 28242 by abdo imad last updated on 22/Jan/18

find the value of ∫_0 ^∞ e^(−x) lnxdx .

findthevalueof∫0∞e−xlnxdx.

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)

letputIn=∫0n(1−xn)n−1lnxdx=∫R(1−xn)n−1χ]0,n[(x)lnxdx.thesequenceoffunctionsfn(x)=(1−xn)n−1χ]0,n[(x)ln(x)c.s.tof(x)=e−xlnxon]0,+∞[alsowehave∣fn(x)∣⩽e−x∀x∈]0,n[thoremeofconvergencedomineegivelimn→+∞In=limn→+∝∫Rfn(x+dx=∫0∞e−xlnxdx.thech.xn=tgiveIn=n∫01(1−t)n−1(ln(n)+lnt)dt=nln(n)∫01(1−t)n−1dt+∫01n(1−t)n−1lntdt=ln(n)[−(1−t)n]01+∫01n(1−t)n−1ln(t)dt=ln(n)+∫01n(1−t)n−1ln(t)dtbutbyparts∫01n(1−t)n−1ln(t)dt=([1−(1−t)n)lnt]01−∫011−(1−t)ntdt=−∫011−(1−t)ntdt(lookthatlimt→0(1−(1−t)n)lnt=0)thech.1−t=xgive−∫011−(1−t)ntdt=−∫011−xn1−xdx=−∫01(1+x+x2+...xn−1)dx=−∫01(∑k=0n−1xk)dx=−∑k=0n−1∫01xkdx=−∑k=0n−11k+1=−∑k=1n1k=−HnsoIn=ln(n)−Hn=−(Hn−ln(n))n→+∝→−γso∫0∞e−xln(x)dx=−γ(thecostantnumberofEuler)

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