calculate A_n =∫_0 ^∞ e^(−nx) ln(1+x)dx with n natural ≥1


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Question Number 72392 by mathmax by abdo last updated on 28/Oct/19

$c a l c u l a t e A_{n} = \int_{0}^{\infty} e^{- n x} l n (1 + x) d x w i t h n n a t u r a l ⩾ 1$
Commented by mathmax by abdo last updated on 07/Nov/19
$A_n =∫_0 ^∞ e^(−nx) ln(1+x)dx =_(1+x=t) ∫_1 ^(+∞) e^(−n(t−1)) lnt dt =e^n ∫_1 ^(+∞) e^(−nt) ln(t)dt =_(nt=u) e^n ∫_n ^(+∞) e^(−u) ln((u/n))(du/n) =(e^n /n) ∫_n ^(+∞) e^(−u) {lnu−ln(n)}du =(e^n /n)∫_n ^(+∞) e^(−u) ln(u)−((e^n ln(n))/n) ∫_n ^(+∞) e^(−u) du =(e^n /n) (∫_n ^0 e^(−u) ln(u)du+∫_0 ^∞ e^(−u) lnudu)−((e^n ln(n))/n)[−e^(−u) ]_n ^(+∞) =(e^n /n){−γ−∫_0 ^n e^(−u) lnu du}−((e^n ln(n))/n)×e^(−n) =−γ (e^n /n)−(e^n /n) ∫_0 ^n e^(−u) lnu du −((ln(n))/n) ....be continued....$
$A_{n} = \int_{0}^{\infty} e^{- n x} l n (1 + x) d x =_{1 + x = t} \int_{1}^{+ \infty} e^{- n (t - 1)} l n t d t$ $= e^{n} \int_{1}^{+ \infty} e^{- n t} l n (t) d t =_{n t = u} e^{n} \int_{n}^{+ \infty} e^{- u} l n (\frac{u}{n}) \frac{d u}{n}$ $= \frac{e^{n}}{n} \int_{n}^{+ \infty} e^{- u} {l n u - l n (n)} d u$ $= \frac{e^{n}}{n} \int_{n}^{+ \infty} e^{- u} l n (u) - \frac{e^{n} l n (n)}{n} \int_{n}^{+ \infty} e^{- u} d u$ $= \frac{e^{n}}{n} (\int_{n}^{0} e^{- u} l n (u) d u + \int_{0}^{\infty} e^{- u} l n u d u) - \frac{e^{n} l n (n)}{n} {[- e^{- u}]}_{n}^{+ \infty}$ $= \frac{e^{n}}{n} {- γ - \int_{0}^{n} e^{- u} l n u d u} - \frac{e^{n} l n (n)}{n} \times e^{- n}$ $= - γ \frac{e^{n}}{n} - \frac{e^{n}}{n} \int_{0}^{n} e^{- u} l n u d u - \frac{l n (n)}{n} . . . . b e c o n t i n u e d . . . .$
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