... advanced integral... i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=γ ii: ψ(x)=∫_0 ^( ∞) ((e^(−t) /t) −(e^(−tx) /(1−e^(−t) )))dt solution :{_(2 : ln(n) =^(easy) ∫_0 ^( 1) ((x^(n−1) −1)/(ln(x)))dx (∗∗)) ^(1: H_n = Σ_(k=1) ^n (1/k) =∫_0 ^( 1) ((1−x^n )/(1−x)) dx (∗)) (∗)−(∗∗): H_n −ln(n)=∫_0 ^1 (((1−x^n )/(1−x)) −((x^(n−1) −1)/(ln(x))))dx lim_(n→∞) (x^n )=^(0<x<1) 0 lim_(n→∞) (H_n −ln(n))=∫_0 ^( 1) ((1/(1n(x)))+(1/(1−x)))dx γ= ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx ✓ ............................. ψ(x)=^(easy) −γ+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt ψ(x)=−∫_0 ^( 1) (1/(ln(t)))+(1/(1−t))dt+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt =∫_0 ^( 1) −(1/(ln(t))) +((1−t^(x−1) −1)/(1−t))dt =−∫_0 ^( 1) (1/(ln(t)))+(t^(x−1) /(1−t)) dt=^(t=e^(−y) ) =−∫_∞ ^( 0) ((1/(−y))+(e^(−yx+y) /(1−e^(−y) )))(−e^(−y) )dy =∫_0 ^( ∞) (e^(−y) /y)−(e^(−yx) /(1−e^(−y) ))dy ∵ ψ(x)=∫_0 ^( ∞) ((e^(−y) /y)−(e^(−yx) /(1−e^(−y) )))dy ✓


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Question Number 122625 by mnjuly1970 last updated on 18/Nov/20
$... advanced integral... i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=γ ii: ψ(x)=∫_0 ^( ∞) ((e^(−t) /t) −(e^(−tx) /(1−e^(−t) )))dt solution :{_(2 : ln(n) =^(easy) ∫_0 ^( 1) ((x^(n−1) −1)/(ln(x)))dx (∗∗)) ^(1: H_n = Σ_(k=1) ^n (1/k) =∫_0 ^( 1) ((1−x^n )/(1−x)) dx (∗)) (∗)−(∗∗): H_n −ln(n)=∫_0 ^1 (((1−x^n )/(1−x)) −((x^(n−1) −1)/(ln(x))))dx lim_(n→∞) (x^n )=^(0<x<1) 0 lim_(n→∞) (H_n −ln(n))=∫_0 ^( 1) ((1/(1n(x)))+(1/(1−x)))dx γ= ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx ✓ ............................. ψ(x)=^(easy) −γ+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt ψ(x)=−∫_0 ^( 1) (1/(ln(t)))+(1/(1−t))dt+∫_0 ^( 1) ((1−t^(x−1) )/(1−t))dt =∫_0 ^( 1) −(1/(ln(t))) +((1−t^(x−1) −1)/(1−t))dt =−∫_0 ^( 1) (1/(ln(t)))+(t^(x−1) /(1−t)) dt=^(t=e^(−y) ) =−∫_∞ ^( 0) ((1/(−y))+(e^(−yx+y) /(1−e^(−y) )))(−e^(−y) )dy =∫_0 ^( ∞) (e^(−y) /y)−(e^(−yx) /(1−e^(−y) ))dy ∵ ψ(x)=∫_0 ^( ∞) ((e^(−y) /y)−(e^(−yx) /(1−e^(−y) )))dy ✓$
$. . . a d v a n c e d i n t e g r a l . . .$ $i : \int_{0}^{1} (\frac{1}{l n (x)} + \frac{1}{1 - x}) d x = γ$ $i i : ψ (x) = \int_{0}^{\infty} (\frac{e^{- t}}{t} - \frac{e^{- t x}}{1 - e^{- t}}) d t$ $s o l u t i o n : {_{2 : l n (n) \overset{e a s y}{=} \int_{0}^{1} \frac{x^{n - 1} - 1}{l n (x)} d x (* )}^{1 : H_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k} = \int_{0}^{1} \frac{1 - x^{n}}{1 - x} d x ()}$ $() - ( *) : H_{n} - l n (n) = \int_{0}^{1} (\frac{1 - x^{n}}{1 - x} - \frac{x^{n - 1} - 1}{l n (x)}) d x$ ${l i m}_{n \to \infty} (x^{n}) \overset{0 < x < 1}{=} 0$ ${l i m}_{n \to \infty} (H_{n} - l n (n)) = \int_{0}^{1} (\frac{1}{1 n (x)} + \frac{1}{1 - x}) d x$ $γ = \int_{0}^{1} (\frac{1}{l n (x)} + \frac{1}{1 - x}) d x ✓$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $ψ (x) \overset{e a s y}{=} - γ + \int_{0}^{1} \frac{1 - t^{x - 1}}{1 - t} d t$ $ψ (x) = - \int_{0}^{1} \frac{1}{l n (t)} + \frac{1}{1 - t} d t + \int_{0}^{1} \frac{1 - t^{x - 1}}{1 - t} d t$ $= \int_{0}^{1} - \frac{1}{l n (t)} + \frac{1 - t^{x - 1} - 1}{1 - t} d t$ $= - \int_{0}^{1} \frac{1}{l n (t)} + \frac{t^{x - 1}}{1 - t} d t \overset{t = e^{- y}}{=}$ $= - \int_{\infty}^{0} (\frac{1}{- y} + \frac{e^{- y x + y}}{1 - e^{- y}}) (- e^{- y}) d y$ $= \int_{0}^{\infty} \frac{e^{- y}}{y} - \frac{e^{- y x}}{1 - e^{- y}} d y$ $∵ ψ (x) = \int_{0}^{\infty} (\frac{e^{- y}}{y} - \frac{e^{- y x}}{1 - e^{- y}}) d y ✓$
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