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Question Number 38198 by maxmathsup by imad last updated on 22/Jun/18

we give ∫_0 ^∞ e^(−x) ln(x)dx=−γ 1) calculate f(a)= ∫_0 ^∞ e^(−ax) ln(x)dx with a>0 2) let u_n = ∫_0 ^∞ e^(−nx) ln((x/n))dx find lim_(n→+∞) u_n

$${we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} {ln}\left({x}\right){dx}=−\gamma \\ $$ $$\left.\mathrm{1}\right)\:{calculate}\:\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ax}} {ln}\left({x}\right){dx}\:\:{with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right)\:{let}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}} {ln}\left(\frac{{x}}{{n}}\right){dx}\:\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} \\ $$

Commented byabdo.msup.com last updated on 24/Jun/18

1) changement ax=t give f(a)= ∫_0 ^∞ e^(−t) ln((t/a))(dt/a) = (1/a){ ∫_0 ^∞ e^(−t) ln(t)dt −ln(a) ∫_0 ^∞ e^(−t) dt} =(1/a){ γ −ln(a) [ −e^(−t) ]_0 ^(+∞) } = ((γ +ln(a))/a) f(a)=((γ +ln(a))/a) with a>0

$$\left.\mathrm{1}\right)\:{changement}\:{ax}={t}\:{give} \\ $$ $${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left(\frac{{t}}{{a}}\right)\frac{{dt}}{{a}} \\ $$ $$=\:\frac{\mathrm{1}}{{a}}\left\{\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left({t}\right){dt}\:−{ln}\left({a}\right)\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\right\} \\ $$ $$=\frac{\mathrm{1}}{{a}}\left\{\:\gamma\:\:−{ln}\left({a}\right)\:\left[\:−{e}^{−{t}} \right]_{\mathrm{0}} ^{+\infty} \right\} \\ $$ $$=\:\frac{\gamma\:+{ln}\left({a}\right)}{{a}} \\ $$ $${f}\left({a}\right)=\frac{\gamma\:+{ln}\left({a}\right)}{{a}}\:\:{with}\:{a}>\mathrm{0} \\ $$

Commented byabdo.msup.com last updated on 24/Jun/18

2) changement nx=t give u_n =∫_0 ^∞ e^(−t) ln( (t/n^2 )) (dt/n) =(1/n){ ∫_0 ^∞ e^(−t) ln(t)dt −2ln(n)∫_0 ^∞ e^(−t) dt} =(1/n){ γ +2ln(n)} lim_(n→+∞) u_n =lim_(n→+∞) ( (γ/n) +2((ln(n))/n))=0

$$\left.\mathrm{2}\right)\:{changement}\:{nx}={t}\:{give} \\ $$ $${u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} \:{ln}\left(\:\frac{{t}}{{n}^{\mathrm{2}} }\right)\:\frac{{dt}}{{n}} \\ $$ $$=\frac{\mathrm{1}}{{n}}\left\{\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left({t}\right){dt}\:\:−\mathrm{2}{ln}\left({n}\right)\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\right\} \\ $$ $$=\frac{\mathrm{1}}{{n}}\left\{\:\gamma\:\:+\mathrm{2}{ln}\left({n}\right)\right\} \\ $$ $${lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} ={lim}_{{n}\rightarrow+\infty} \left(\:\frac{\gamma}{{n}}\:+\mathrm{2}\frac{{ln}\left({n}\right)}{{n}}\right)=\mathrm{0} \\ $$

Commented bymath khazana by abdo last updated on 24/Jun/18

γ is the costant of Euler.

$$\gamma\:{is}\:{the}\:{costant}\:{of}\:{Euler}. \\ $$

Commented bymath khazana by abdo last updated on 24/Jun/18

f(a)=((−γ +ln(a))/a)

$${f}\left({a}\right)=\frac{−\gamma\:+{ln}\left({a}\right)}{{a}} \\ $$

Commented bymath khazana by abdo last updated on 24/Jun/18

u_n =((−γ +2ln(n))/n)

$${u}_{{n}} =\frac{−\gamma\:+\mathrm{2}{ln}\left({n}\right)}{{n}} \\ $$

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