A-Challanging-Integral-0-1-log-x-log-1-x-1-x-dx-

Question Number 143889 by mnjuly1970 last updated on 19/Jun/21

A C h a l l a n g i n g I n t e g r a l :

Φ = \int_{0}^{1} \frac{l o g (x) . l o g (1 + x)}{1 - x} d x

Answered by Olaf_Thorendsen last updated on 19/Jun/21

Φ = \int_{0}^{1} \frac{\ln (x) . \ln (1 + x)}{1 - x} d x

Let u = 1 - x

Φ = \int_{0}^{1} \frac{\ln (1 - u) . \ln (2 - u)}{u} d u

Φ = \int_{0}^{1} \frac{\ln (2 - u)}{u} (- \underset{n = 0}{\sum^{\infty}} \frac{u^{n + 1}}{n + 1}) d u

Φ = - \int_{0}^{1} \ln (2 - u) \underset{n = 0}{\sum^{\infty}} \frac{u^{n}}{n + 1} d u

Φ = - {[\ln (2 - u) \underset{n = 0}{\sum^{\infty}} \frac{u^{n + 1}}{{(n + 1)}^{2}}]}_{0}^{1}

- \int_{0}^{1} \frac{1}{2 - u} . \underset{n = 0}{\sum^{\infty}} \frac{u^{n + 1}}{{(n + 1)}^{2}} d u

Φ = \underset{n = 0 (}{\sum^{\infty}} \frac{1}{{n + 1)}^{2}} \int_{0}^{1} \frac{u^{n + 1}}{2 - u} d u

Let I_{n} = \int_{0}^{1} \frac{u^{n}}{2 - u} d u

{2 I}_{n} - I_{n + 1} = \int_{0}^{1} \frac{2 u^{n} - u^{n + 1}}{2 - u} d u

{2 I}_{n} - I_{n + 1} = \int_{0}^{1} u^{n} d u

{2 I}_{n} - I_{n + 1} = \frac{1}{n + 1}

I_{n + 1} = {2 I}_{n} - \frac{1}{n + 1}

I_{n + 1} = 2 ({2 I}_{n - 1} - \frac{1}{n}) - \frac{1}{n + 1}

I_{n + 1} = 2^{2} I_{n - 1} - \frac{2}{n} - \frac{1}{n + 1}

I_{n + 1} = 2^{2} I_{n - 1} - \frac{2}{n} - \frac{1}{n + 1}

I_{n + 1} = 2^{3} I_{n - 2} - \frac{2^{2}}{n - 1} - \frac{2}{n} - \frac{1}{n + 1}

\dots

I_{n + 1} = 2^{n + 1} I_{0} - \underset{k = 1}{\sum^{n + 1}} \frac{2^{n + 1 - k}}{k}

I_{0} = \int_{0}^{1} \frac{u^{0}}{2 - u} d u = {[- \ln ∣ 2 - u ∣]}_{0}^{1} = \ln 2

I_{n + 1} = 2^{n + 1} \ln 2 - \underset{k = 1}{\sum^{n + 1}} \frac{2^{n + 1 - k}}{k}

Φ = \underset{n = 0 (}{\sum^{\infty}} \frac{I_{n + 1}}{{n + 1)}^{2}}

Φ = \underset{n = 0 (}{\sum^{\infty}} \frac{1}{{n + 1)}^{2}} (2^{n + 1} \ln 2 - \underset{k = 1}{\sum^{n + 1}} \frac{2^{n + 1 - k}}{k})

Φ = \underset{n = 0 (}{\sum^{\infty}} \frac{2^{n + 1}}{{n + 1)}^{2}} (\ln 2 - \underset{k = 1}{\sum^{n + 1}} \frac{2^{- k}}{k})

\underset{k = 1}{\sum^{n}} \frac{1}{2^{k} k} = \ln 2 - \frac{n - 1}{n} . \frac{LerchPhi (\frac{1}{2}, 1, n)}{2^{n + 1}}

Φ = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} (n . \frac{LerchPhi (\frac{1}{2}, 1, n + 1)}{{(n + 1)}^{3}})

\dots

Sorry, I tried to solve but I^{'} m lost \dots

Commented by mnjuly1970 last updated on 20/Jun/21

t h a n k s a l o t m r o l a f . .

Answered by mnjuly1970 last updated on 20/Jun/21

Φ := \int_{0}^{1} \frac{l o g (1 - x) l o g (2 - x)}{x} d x

:= \int_{0}^{1} \frac{l o g (1 - x) {l o g (2) + l o g (1 - \frac{x}{2})}}{x} d x

:= - l o g (2) \frac{π^{2}}{6} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} \int_{0}^{1} x^{n - 1} l o g (1 - x) d x

:= - l o g (2) . \frac{π^{2}}{2} + \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n^{2} 2^{n}}

:= - l o g (2) . \frac{π^{2}}{6} + ζ (3) - \frac{π^{2}}{12} l o g (2)

:= \frac{- π^{2}}{4} l o g (2) + ζ (3) \dots

Commented by Dwaipayan Shikari last updated on 20/Jun/21

N i c e s i r!

Commented by mnjuly1970 last updated on 20/Jun/21

t h a n k y o u s o m u c h \dots