nice-calculus-Evaluation-of-0-1-ln-x-arctan-x-dx-solution-note-1-n-1-1-n-1-2n-1-arctan-1-pi-4-note-2-n-1-1-n-1-

Question Number 128841 by mnjuly1970 last updated on 10/Jan/21

\dots n i c e c a l c u l u s \dots

E v a l u a t i o n o f :: Φ = \int_{0}^{1} l n (x) . a r c t a n (x) d x

s o l u t i o n ::

n o t e 1 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{2 n - 1} = a r c t a n (1) = \frac{π}{4}

n o t e 2 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} = l n (1 + 1) = l n (2)

n o t e 3 :: \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}} \underset{e t a f u n c t i o n}{\overset{D r i c h l e t}{=}} η (2) = \frac{π^{2}}{12}

\dots \dots s t a r t \dots \dots .

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

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

= \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{(2 n - 1)} {{[\frac{x^{2 n}}{2 n} l n (x)]}_{0}^{1} - \frac{1}{2 n} \int_{0}^{1} x^{2 n - 1} d x}

= \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n - 1) n^{2}} = \frac{1}{4} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} [\frac{2 n - (2 n - 1)}{(2 n - 1) n^{2}}]

= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n - 1) n} - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}}

= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{2 {(- 1)}^{n}}{2 n - 1} - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} + \frac{1}{4} η (2)

= \frac{- π}{4} + \frac{1}{2} l n (2) + \frac{π^{2}}{48} \dots

\dots Φ = \frac{π^{2}}{48} - \frac{π}{4} + \frac{1}{2} l n (2) \dots

\dots m . n . j u l y . 1970 \dots

Commented by Dwaipayan Shikari last updated on 10/Jan/21

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

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

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

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

= l o g (\sqrt{2}) - \frac{π}{4} - \frac{π^{2}}{48}

Commented by mnjuly1970 last updated on 11/Jan/21

t h a n k y o u m r d w a i p a y a n \dots

g r a t e f u l .

Answered by mindispower last updated on 10/Jan/21

1) b y p a r t = {[l n (x) . (x a r c t a n (x) - \frac{l n (1 + x^{2})}{2})]}_{0}^{1}

- \int_{0}^{1} a r c t a n (x) d x + \int_{0}^{1} \frac{l n (1 + x^{2})}{2 x} d x

\int_{0}^{1} a r c t a n (x) = \frac{π}{4} - \frac{l n (2)}{2}

\int_{0}^{1} \frac{l n (1 + x^{2})}{2 x} d x = \int_{0}^{1} \frac{l n (1 + x^{2})}{4 x^{2}} . d (x^{2})

= \frac{1}{4} \int_{0}^{1} \frac{l n (1 + t)}{t} d t = \frac{1}{4} \int_{0}^{1} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1} t^{n - 1}}{n}

= \frac{1}{4} \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \frac{1}{16} ζ (2) + \frac{1}{4} (\frac{3}{4}) ζ (2)

= \frac{ζ (2)}{8} = \frac{π^{2}}{48} \dots w e g e t \frac{π^{2}}{48} - \frac{π}{4} + \frac{l n (2)}{2}

2) l n (x) = \partial_{y} x^{y} ∣_{y = 0}

f (y) = \int_{0}^{1} x^{y} a r c t a n (x) d x

b y p a r t = \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{x^{y + 1}}{1 + x^{2}}

= \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{x^{y + 1} - x^{y + 3}}{1 - x^{4}} d x

x^{4} = t

= \frac{π}{4 (y + 1)} - \frac{1}{y + 1} \int_{0}^{1} \frac{t^{\frac{y + 1}{4}} - t^{\frac{y + 3}{4}}}{4 (1 - t)} . t^{- \frac{3}{4}} d t

= \frac{1}{y + 1} (\frac{π}{4} - \int_{0}^{1} \frac{t^{\frac{y - 2}{4}} - t^{\frac{y - 1}{4}}}{1 - t} d t)

Ψ (s + 1) = - γ + \int_{0}^{1} \frac{1 - x^{s}}{1 - x} d x

w e g e t \frac{1}{y + 1} (\frac{π}{4} - Ψ (\frac{y + 3}{4}) + Ψ (\frac{y + 2}{4})) = f (y)

f^{'} (y) = - \frac{1}{{(y + 1)}^{2}} (\frac{π}{4} + Ψ (\frac{y + 2}{4}) - Ψ (\frac{y + 3}{4})) + (\frac{1}{4} (Ψ^{'} (\frac{y + 2}{4}) - Ψ^{'} (\frac{y + 3}{4})) . \frac{1}{y + 1}

Commented by mnjuly1970 last updated on 11/Jan/21

t h a n k s a l o t s i r p o w e r

p o w e r f u l a n d m i g h t y a s a l w a y s

. p e a c e b e u p o n y o u \dots