0-1-logxlog-1-x-dx-

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}) + \dots

\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 = \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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