f(x)=(1/(lnx)) −(1/(x−1)) 1) lim_(x→1) f(x)=(1/2) 2) ∫


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Question Number 95397 by ~blr237~ last updated on 25/May/20

$f (x) = \frac{1}{l n x} - \frac{1}{x - 1}$ $1) \lim_{x \to 1} f (x) = \frac{1}{2}$ $2) \int_{0}^{1} f (x) d x = γ$
Commented by Mikael_786 last updated on 25/May/20
$1. lim_(x→1) {(1/(lnx))−(1/(x−1))} lim_(x→1) {(1/(x−1))(((x−1)/(lnx))−1)}=lim_(x→1) {(1/(x−1))(∫_0 ^1 x^y dy−1)} lim_(x→1) {∫_0 ^1 (x^y /(x−1))dy−∫_0 ^1 (1/(x−1))dy} lim_(x→1) {∫_0 ^1 ((x^y −1)/(x−1))dy}=∫_0 ^1 {lim_(x→1) ((x^y −1)/(x−1))}dy =∫_0 ^1 ydy=[(y^2 /2)]_0 ^1 =(1/2)$
$1 . \underset{x \to 1}{l i m} {\frac{1}{l n x} - \frac{1}{x - 1}}$ $\underset{x \to 1}{l i m} {\frac{1}{x - 1} (\frac{x - 1}{l n x} - 1)} = \underset{x \to 1}{l i m} {\frac{1}{x - 1} (\underset{0}{\int^{1}} x^{y} d y - 1)}$ $\underset{x \to 1}{l i m} {\underset{0}{\int^{1}} \frac{x^{y}}{x - 1} d y - \underset{0}{\int^{1}} \frac{1}{x - 1} d y}$ $\underset{x \to 1}{l i m} {\underset{0}{\int^{1}} \frac{x^{y} - 1}{x - 1} d y} = \underset{0}{\int^{1}} {\underset{x \to 1}{l i m} \frac{x^{y} - 1}{x - 1}} d y$ $= \underset{0}{\int^{1}} y d y = {[\frac{y^{2}}{2}]}_{0}^{1} = \frac{1}{2}$
Commented by Mikael_786 last updated on 25/May/20
$∫_0 ^1 {(1/(lnx))−(1/(x−1))}dx let lnx=−y ⇒ x=e^(−y) dx=−e^(−y) dy ∫_∞ ^0 {(1/(−y))−(1/(e^(−y) −1))}(−e^(−y) dy) ∫_0 ^∞ {(e^(−y) /(1−e^(−y) ))−(e^(−y) /y)}dy=∫_0 ^∞ (e^(−y) /(1−e^(−y) ))dy−∫_0 ^∞ (e^(−y) /y)dy ∫_0 ^∞ (e^(−y) /(1−e^(−y) ))dy−∫_0 ^1 (e^(−y) /y)dy−∫_1 ^∞ (e^(−y) /y)dy let z=1−e^(−y) ⇒ e^(−y) =1−z −y=ln(1−z)⇒ dy=(dz/(1−z)) ∫_0 ^1 ((1−z)/z)∗(dz/(1−z))−∫_0 ^1 (e^(−y) /y)dy−∫_1 ^∞ (e^(−y) /y)dy ∫_0 ^1 (dz/z)−∫_0 ^1 (e^(−y) /y)dy−∫_1 ^∞ (e^(−y) /y)dy ∫_0 ^1 ((1−e^(−y) )/y)dy−∫_1 ^∞ (e^(−y) /y)dy = γ γ : Euler Constant$
$\underset{0}{\int^{1}} {\frac{1}{l n x} - \frac{1}{x - 1}} d x$ $l e t l n x = - y \Rightarrow x = e^{- y}$ $d x = - e^{- y} d y$ $\underset{\infty}{\int^{0}} {\frac{1}{- y} - \frac{1}{e^{- y} - 1}} (- e^{- y} d y)$ $\underset{0}{\int^{\infty}} {\frac{e^{- y}}{1 - e^{- y}} - \frac{e^{- y}}{y}} d y = \underset{0}{\int^{\infty}} \frac{e^{- y}}{1 - e^{- y}} d y - \underset{0}{\int^{\infty}} \frac{e^{- y}}{y} d y$ $\underset{0}{\int^{\infty}} \frac{e^{- y}}{1 - e^{- y}} d y - \underset{0}{\int^{1}} \frac{e^{- y}}{y} d y - \underset{1}{\int^{\infty}} \frac{e^{- y}}{y} d y$ $l e t z = 1 - e^{- y} \Rightarrow e^{- y} = 1 - z$ $- y = l n (1 - z) \Rightarrow d y = \frac{d z}{1 - z}$ $\underset{0}{\int^{1}} \frac{1 - z}{z} * \frac{d z}{1 - z} - \underset{0}{\int^{1}} \frac{e^{- y}}{y} d y - \underset{1}{\int^{\infty}} \frac{e^{- y}}{y} d y$ $\underset{0}{\int^{1}} \frac{d z}{z} - \underset{0}{\int^{1}} \frac{e^{- y}}{y} d y - \underset{1}{\int^{\infty}} \frac{e^{- y}}{y} d y$ $\underset{0}{\int^{1}} \frac{1 - e^{- y}}{y} d y - \underset{1}{\int^{\infty}} \frac{e^{- y}}{y} d y = γ$ $γ : E u l e r C o n s t a n t$
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