1) find f(x) =∫_0 ^x ln(t)ln(1−t)dt with 0≤x≤1 2) find the value of ∫


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Question Number 43918 by maxmathsup by imad last updated on 17/Sep/18

$1) f i n d f (x) = \int_{0}^{x} l n (t) l n (1 - t) d t w i t h 0 ⩽ x ⩽ 1$ $2) f i n d t h e v a l u e o f \int_{0}^{1} l n (t) l n (1 - t) d t .$
Commented by maxmathsup by imad last updated on 19/Sep/18
$1) we have for 0<t≤x<1 ln^′ (1−t) =−(1/(1−t)) =−Σ_(n=0) ^∞ t^n ⇒ ln(1−t) =−Σ_(n=0) ^∞ (1/(n+1))t^(n+1) =−Σ_(n=1) ^∞ (t^n /n) ⇒ f(x) =−∫_0 ^x ln(t)(Σ_(n=1) ^∞ (t^n /n))dt =−Σ_(n=1) ^∞ (1/n) ∫_0 ^x t^n ln(t) dt by parts A_n (x)= ∫_0 ^x t^n ln(t)dt =[(1/(n+1))t^(n+1) ln(t)]_0 ^x −∫_0 ^x (1/(n+1))t^(n+1) (dt/t) =(1/(n+1)) x^(n+1) ln(x) −(1/(n+1))∫_0 ^x t^n dt = (1/(n+1))x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) ⇒ f(x) =−Σ_(n=1) ^∞ (1/n){ (1/(n+1)) x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) } = −Σ_(n=1) ^∞ (1/(n(n+1))) x^(n+1) ln(x) +Σ_(n=1) ^∞ (1/(n(n+1)^2 )) x^(n+1) =H(x)−K(x) K(x) =ln(x) Σ_(n=1) ^∞ {(1/n)−(1/(n+1))}x^(n+1) =ln(x)Σ_(n=1) ^∞ (x^(n+1) /n) −ln(x)Σ_(n=1) ^∞ (x^(n+1) /(n+1)) =xln(x) Σ_(n=1) ^∞ (x^n /n) −ln(x) Σ_(n=2) ^∞ (x^n /n) =−xln(x)ln(1−x) −ln(x)(−ln∣1−x∣−x) =−xln(x)ln∣1−x∣+ln(x)(ln∣1−x∣ +x) =(1−x)ln(x)ln∣1−x∣ +xln(x) .also we have (d/dx)K(x) =Σ_(n=1) ^∞ (1/(n(n+1))) x^n =Σ_(n=1) ^∞ {(1/n)−(1/(n+1))}x^n =Σ_(n=1) ^∞ (x^n /n) −Σ_(n=1) ^∞ (x^n /(n+1)) =−ln∣1−x∣−Σ_(n=2) ^∞ (x^(n−1) /n) =−ln∣1−x∣ −(1/x) {Σ_(n=1) ^∞ (x^n /n) −x} =−ln∣1−x∣+(1/x)ln∣1−x) +1 =((1/x)−1)ln∣1−x∣ +1 =(((1−x)ln∣1−x∣)/x) +1 ⇒ K(x) = x + ∫ (((1−x)ln∣1−x∣)/x) dx +k ....be continued...$
$1) w e h a v e f o r 0 < t ⩽ x < 1 {l n}^{^{'}} (1 - t) = - \frac{1}{1 - t} = - \sum_{n = 0}^{\infty} t^{n} \Rightarrow$ $l n (1 - t) = - \sum_{n = 0}^{\infty} \frac{1}{n + 1} t^{n + 1} = - \sum_{n = 1}^{\infty} \frac{t^{n}}{n} \Rightarrow$ $f (x) = - \int_{0}^{x} l n (t) (\sum_{n = 1}^{\infty} \frac{t^{n}}{n}) d t = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{x} t^{n} l n (t) d t$ $b y p a r t s A_{n} (x) = \int_{0}^{x} t^{n} l n (t) d t = {[\frac{1}{n + 1} t^{n + 1} l n (t)]}_{0}^{x} - \int_{0}^{x} \frac{1}{n + 1} t^{n + 1} \frac{d t}{t}$ $= \frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{n + 1} \int_{0}^{x} t^{n} d t = \frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} x^{n + 1} \Rightarrow$ $f (x) = - \sum_{n = 1}^{\infty} \frac{1}{n} {\frac{1}{n + 1} x^{n + 1} l n (x) - \frac{1}{{(n + 1)}^{2}} x^{n + 1}}$ $= - \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} x^{n + 1} l n (x) + \sum_{n = 1}^{\infty} \frac{1}{n {(n + 1)}^{2}} x^{n + 1} = H (x) - K (x)$ $K (x) = l n (x) \sum_{n = 1}^{\infty} {\frac{1}{n} - \frac{1}{n + 1}} x^{n + 1}$ $= l n (x) \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n} - l n (x) \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n + 1}$ $= x l n (x) \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - l n (x) \sum_{n = 2}^{\infty} \frac{x^{n}}{n}$ $= - x l n (x) l n (1 - x) - l n (x) (- l n ∣ 1 - x ∣ - x)$ $= - x l n (x) l n ∣ 1 - x ∣ + l n (x) (l n ∣ 1 - x ∣ + x)$ $= (1 - x) l n (x) l n ∣ 1 - x ∣ + x l n (x) . a l s o w e h a v e$ $\frac{d}{d x} K (x) = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} x^{n} = \sum_{n = 1}^{\infty} {\frac{1}{n} - \frac{1}{n + 1}} x^{n}$ $= \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n + 1} = - l n ∣ 1 - x ∣ - \sum_{n = 2}^{\infty} \frac{x^{n - 1}}{n}$ $= - l n ∣ 1 - x ∣ - \frac{1}{x} {\sum_{n = 1}^{\infty} \frac{x^{n}}{n} - x} = - l n ∣ 1 - x ∣ + \frac{1}{x} l n ∣ 1 - x) + 1$ $= (\frac{1}{x} - 1) l n ∣ 1 - x ∣ + 1 = \frac{(1 - x) l n ∣ 1 - x ∣}{x} + 1 \Rightarrow$ $K (x) = x + \int \frac{(1 - x) l n ∣ 1 - x ∣}{x} d x + k . . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 19/Sep/18
$2) ∫_0 ^1 ln(t)ln(1−t)dt =lim_(x→0) ∫_0 ^x ln(t)ln(1−t)dt =lim_(x→1) { Σ_(n=1) ^∞ (x^(n+1) /((n+1)^2 )) −Σ_(n=1) ^∞ ((x^(n+1) lnx)/(n(n+1)))} =Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=2) ^∞ (1/n^2 ) =(π^2 /6) −1 .$
$2) \int_{0}^{1} l n (t) l n (1 - t) d t = {l i m}_{x \to 0} \int_{0}^{x} l n (t) l n (1 - t) d t$ $= {l i m}_{x \to 1} {\sum_{n = 1}^{\infty} \frac{x^{n + 1}}{{(n + 1)}^{2}} - \sum_{n = 1}^{\infty} \frac{x^{n + 1} l n x}{n (n + 1)}} = \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}}$ $= \sum_{n = 2}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} - 1 .$
Commented by maxmathsup by imad last updated on 19/Sep/18

$\int_{0}^{1} l n (t) l n (1 - t) d t = {l i m}_{x \to 1} \int_{0}^{x} l n (t) l n (1 - t) d t = . . .$
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