∫_0 ^1 logxlog(1−x)dx


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Question Number 103154 by Dwaipayan Shikari last updated on 13/Jul/20

$\int_{0}^{1} l o g x l o g (1 - x) d x$
	Answered by OlafThorendsen last updated on 13/Jul/20

	$I = \int_{0}^{1} \ln x \ln (1 - x) d x$ $- \frac{1}{1 - x} = - \underset{n = 0}{\sum^{\infty}} x^{n}$ $\ln (1 - x) = - \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 1}}{n + 1} = - \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}$ $I = - \int_{0}^{1} \ln x \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} d x$ $I = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{n} \ln x d x$ $I_{n} = - \int_{0}^{1} x^{n} \ln x d x$ $I_{n} = - {[\frac{x^{n + 1}}{n + 1} \ln x]}_{0}^{1} + \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{1}{x} d x$ $I_{n} = 0 + \int_{0}^{1} \frac{x^{n}}{n + 1} d x = {[\frac{x^{n + 1}}{{(n + 1)}^{2}}]}_{0}^{1}$ $I_{n} = \frac{1}{{(n + 1)}^{2}}$ $I = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} I_{n} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 1)}^{2}}$ $I = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{{(n + 1)}^{2}})$ $\underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1}) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + . . .$ $\underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1}) = 1$ $I = 1 - \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}}$ $I = 1 - \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}}$ $I = 2 - \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}}$ $I = 2 - ζ (2) = 2 - \frac{π^{2}}{6}$
	Commented by Dwaipayan Shikari last updated on 13/Jul/20

	$G r e a t s i r!$
	Answered by bobhans last updated on 13/Jul/20
	$I = ∫_0 ^1 ln(x) ln(1−x) dx by Maclaurin series ln(1−x) = −Σ_(n=1) ^∞ (x^n /n) we obtain I=−∫_0 ^1 ln(x)Σ_(n=1) ^∞ (x^n /n) dx I = −Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^n ln(x) dx [ by parts ] { ((u = ln(x))),((dv = x^n dx )) :} I=−Σ_(n=1) ^∞ (1/n)∣((x^(n+1) /(n+1)) ln(x)−(x^(n+1) /((n+1)^2 )))∣_0 ^1 I= Σ_(n=1) ^∞ (1/(n(n+1)^2 )) [ by L′Hopital rule′s] I= Σ_(n=1) ^∞ ((1/n)−(1/(n+1))−(1/((n+1)^2 ))) the first series Σ_(n=1) ^∞ ((1/n)−(1/(n+1))) is telescoping ⇒ lim_(p→∞) Σ_(n=1) ^p ((1/n)−(1/(n+1))) = lim_(p→∞) (1−(1/(p+1))) = 1 now the second series Σ_(n=1) ^∞ (1/((n+1)^2 )) = Σ_(k=2) ^∞ (1/k^2 ) = −1+Σ_(k=1) ^∞ (1/k^2 ) = −1+(π^2 /6) . therefore I = ∫_0 ^1 ln(x) ln(1−x)dx = 1−(−1+(π^2 /6)) = 2−(π^2 /6) . ★$
	$I = \underset{0}{\int^{1}} \ln (x) \ln (1 - x) d x$ $b y M a c l a u r i n s e r i e s$ $\ln (1 - x) = - \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}$ $w e o b t a i n I = - \underset{0}{\int^{1}} \ln (x) \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} d x$ $I = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \underset{0}{\int^{1}} x^{n} \ln (x) d x [b y p a r t s]$ ${\begin{cases} u = \ln (x) \\ d v = x^{n} d x \end{cases}$ $I = - \underset{n = 1}{\sum^{\infty}} \frac{1}{n} ∣ (\frac{x^{n + 1}}{n + 1} \ln (x) - \frac{x^{n + 1}}{{(n + 1)}^{2}}) ∣_{0}^{1}$ $I = \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 1)}^{2}} [b y L^{'} H o p i t a l {r u l e}^{'} s]$ $I = \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{{(n + 1)}^{2}})$ $t h e f i r s t s e r i e s \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + 1}) i s$ $t e l e s c o p i n g \Rightarrow \lim_{p \to \infty} \underset{n = 1}{\sum^{p}} (\frac{1}{n} - \frac{1}{n + 1})$ $= \lim_{p \to \infty} (1 - \frac{1}{p + 1}) = 1$ $n o w t h e s e c o n d s e r i e s$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}} = \underset{k = 2}{\sum^{\infty}} \frac{1}{k^{2}} = - 1 + \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2}}$ $= - 1 + \frac{π^{2}}{6} .$ $t h e r e f o r e I = \underset{0}{\int^{1}} \ln (x) \ln (1 - x) d x =$ $1 - (- 1 + \frac{π^{2}}{6}) = 2 - \frac{π^{2}}{6} . ★$
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