find the value of I =∫_0 ^1 ((ln(t)ln(1−t))/t)dt


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Question Number 35628 by abdo mathsup 649 cc last updated on 21/May/18

$f i n d t h e v a l u e o f I = \int_{0}^{1} \frac{l n (t) l n (1 - t)}{t} d t$
Commented by abdo mathsup 649 cc last updated on 24/May/18
$we have ln^′ (1−t) = ((−1)/(1−t)) =−Σ_(n=0) ^∞ t^n for ∣t∣<1 ⇒ ln(1−t) =−Σ_(n=0) ^∞ (t^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (t^n /n) ⇒ ((ln(1−t))/t) = −Σ_(n=1) ^∞ (t^(n−1) /n) ⇒ I = − ∫_0 ^1 {Σ_(n=1) ^∞ (t^(n−1) /n)}ln(t)dt =−Σ_(n=1) ^∞ (1/n)∫_0 ^1 t^(n−1) ln(t)dt by parts A_n = ∫_0 ^1 t^(n−1) ln(t)dt =[(1/n)t^n ln(t)]_0 ^1 −∫_0 ^1 (1/n)t^n (dt/t) = −(1/n) ∫_0 ^1 t^(n−1) dt = −(1/n^2 ) ⇒ I = Σ_(n=1) ^∞ (1/n^3 ) I =ξ(3) the approximate value of ξ(3) is known.$
$w e h a v e {l n}^{^{'}} (1 - t) = \frac{- 1}{1 - t} = - \sum_{n = 0}^{\infty} t^{n} f o r ∣ t ∣< 1$ $\Rightarrow l n (1 - t) = - \sum_{n = 0}^{\infty} \frac{t^{n + 1}}{n + 1} = - \sum_{n = 1}^{\infty} \frac{t^{n}}{n} \Rightarrow$ $\frac{l n (1 - t)}{t} = - \sum_{n = 1}^{\infty} \frac{t^{n - 1}}{n} \Rightarrow$ $I = - \int_{0}^{1} {\sum_{n = 1}^{\infty} \frac{t^{n - 1}}{n}} l n (t) d t$ $= - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} t^{n - 1} l n (t) d t b y p a r t s$ $A_{n} = \int_{0}^{1} t^{n - 1} l n (t) d t = {[\frac{1}{n} t^{n} l n (t)]}_{0}^{1} - \int_{0}^{1} \frac{1}{n} t^{n} \frac{d t}{t}$ $= - \frac{1}{n} \int_{0}^{1} t^{n - 1} d t = - \frac{1}{n^{2}} \Rightarrow I = \sum_{n = 1}^{\infty} \frac{1}{n^{3}}$ $I = ξ (3) t h e a p p r o x i m a t e v a l u e o f ξ (3) i s k n o w n .$
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