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


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

$w e g i v e \int_{0}^{\infty} e^{- x} l n (x) d x = - γ$ $1) c a l c u l a t e f (a) = \int_{0}^{\infty} e^{- a x} l n (x) d x w i t h a > 0$ $2) l e t u_{n} = \int_{0}^{\infty} e^{- n x} l n (\frac{x}{n}) d x f i n d {l i m}_{n \to + \infty} u_{n}$
Commented byabdo.msup.com last updated on 24/Jun/18
$1) changement ax=t give f(a)= ∫_0 ^∞ e^(−t) ln((t/a))(dt/a) = (1/a){ ∫_0 ^∞ e^(−t) ln(t)dt −ln(a) ∫_0 ^∞ e^(−t) dt} =(1/a){ γ −ln(a) [ −e^(−t) ]_0 ^(+∞) } = ((γ +ln(a))/a) f(a)=((γ +ln(a))/a) with a>0$
$1) c h a n g e m e n t a x = t g i v e$ $f (a) = \int_{0}^{\infty} e^{- t} l n (\frac{t}{a}) \frac{d t}{a}$ $= \frac{1}{a} {\int_{0}^{\infty} e^{- t} l n (t) d t - l n (a) \int_{0}^{\infty} e^{- t} d t}$ $= \frac{1}{a} {γ - l n (a) {[- e^{- t}]}_{0}^{+ \infty}}$ $= \frac{γ + l n (a)}{a}$ $f (a) = \frac{γ + l n (a)}{a} w i t h a > 0$
Commented byabdo.msup.com last updated on 24/Jun/18
$2) changement nx=t give u_n =∫_0 ^∞ e^(−t) ln( (t/n^2 )) (dt/n) =(1/n){ ∫_0 ^∞ e^(−t) ln(t)dt −2ln(n)∫_0 ^∞ e^(−t) dt} =(1/n){ γ +2ln(n)} lim_(n→+∞) u_n =lim_(n→+∞) ( (γ/n) +2((ln(n))/n))=0$
$2) c h a n g e m e n t n x = t g i v e$ $u_{n} = \int_{0}^{\infty} e^{- t} l n (\frac{t}{n^{2}}) \frac{d t}{n}$ $= \frac{1}{n} {\int_{0}^{\infty} e^{- t} l n (t) d t - 2 l n (n) \int_{0}^{\infty} e^{- t} d t}$ $= \frac{1}{n} {γ + 2 l n (n)}$ ${l i m}_{n \to + \infty} u_{n} = {l i m}_{n \to + \infty} (\frac{γ}{n} + 2 \frac{l n (n)}{n}) = 0$
Commented bymath khazana by abdo last updated on 24/Jun/18

$γ i s t h e c o s t a n t o f E u l e r .$
Commented bymath khazana by abdo last updated on 24/Jun/18

$f (a) = \frac{- γ + l n (a)}{a}$
Commented bymath khazana by abdo last updated on 24/Jun/18

$u_{n} = \frac{- γ + 2 l n (n)}{n}$
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