advanced-calculus-evaluate-0-1-ln-x-tan-1-x-dx-m-n-1970-

Question Number 120775 by mnjuly1970 last updated on 02/Nov/20

\dots a d v a n c e d c a l c u l u s \dots

e v a l u a t e ::

Φ \overset{? ? ?}{=} \int_{0}^{1} l n (x) {t a n}^{- 1} (x) d x

\dots m . n . 1970 \dots

Answered by Dwaipayan Shikari last updated on 02/Nov/20

\int {t a n}^{- 1} x d x

= {x t a n}^{- 1} x - \int \frac{x}{1 + x^{2}}

= {x t a n}^{- 1} x - l o g (\sqrt{1 + x^{2}}) (l o g x \int_{0}^{1} {t a n}^{- 1} (x) d x = 0)

Φ = [l o g (x) \int_{0}^{1} {t a n}^{- 1} x] - \int_{0}^{1} {t a n}^{- 1} x + \frac{1}{2} \int_{0}^{1} \frac{l o g (1 + x^{2})}{x}

Φ = - [{x t a n}^{- 1} x - l o g (\sqrt{1 + x^{2}})] + \frac{1}{2} \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} x^{2 n - 1} . \frac{1}{n}

Φ = - \frac{π}{4} + \frac{1}{2} l o g (2) + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{1} x^{2 n - 1} . \frac{1}{n} d x

Φ = - \frac{π}{4} + \frac{1}{2} l o g (2) + \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{2 n^{2}}

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

Commented by mnjuly1970 last updated on 02/Nov/20

t h a n k y o u s o m u c h m r d w a i p a y a n e x c e l l e n t

Commented by Dwaipayan Shikari last updated on 03/Nov/20

W i t h p l e a s u r e

Answered by mindispower last updated on 02/Nov/20

= {[(x l n (x) - x) {t a n}^{-} (x)]}_{0}^{1} - \int_{0}^{1} \frac{x l n (x) - x}{1 + x^{2}} d x

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

- \frac{π}{4} + \frac{l n (2)}{2} - \sum_{k ⩾ 0} \int_{0}^{1} {(- 1)}^{k} x^{2 k + 1} l n (x) d x

= - \frac{π}{4} + \frac{l n (2)}{2} - \sum_{k ⩾ 0} {(- 1)}^{k} (- \frac{1}{{(2 k + 2)}^{2}})

= - \frac{π}{4} + \frac{l n (2)}{2} + \frac{1}{4} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(k + 1)}^{2}}

= - \frac{π}{4} + \frac{l n (2)}{2} + \frac{1}{4} [\frac{π^{2}}{12}]

= \frac{- 12 π + π^{2} + 24 l n (2)}{48}

Commented by mnjuly1970 last updated on 02/Nov/20

v e r y n i c e s i r m i n d s p o w e r

g r a t e f u l . .

Commented by mindispower last updated on 03/Nov/20

t h a n x w i t h e p l e a s u r n i c e d a y s i r