prove-that-0-1-x-ln-2-1-x-1-x-2-dx-1-96-ln-2-pi-2-4-ln-2-2-M-N-

Question Number 152350 by mnjuly1970 last updated on 27/Aug/21

prove that \dots .

ϕ := \int_{0}^{1} \frac{x . {l n}^{2} (1 + x)}{1 + x^{2}} d x = \frac{1}{96} l n (2) (π^{2} + 4 {l n}^{2} (2)) \dots .

M . N . .

Answered by mindispower last updated on 27/Aug/21

x \to \frac{1 - t}{1 + t} \Rightarrow d x = - \frac{2 d t}{{(1 + t)}^{2}}

\Leftrightarrow ϕ = \int_{0}^{1} \frac{\frac{1 - t}{1 + t} {l n}^{2} (\frac{2}{1 + t})}{1 + t^{2}} . d t

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

\frac{1 - t}{(1 + t) (1 + t^{2})} = \frac{1}{1 + t} - \frac{t}{1 + t^{2}}

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

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

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

2 ϕ = {l n}^{2} (2) [\frac{l n (2)}{2}] + \frac{1}{3} {l n}^{3} (2) - 2 l n (2) . \frac{{l n}^{2} (2)}{2}

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

\int_{0}^{1} \frac{t l n (1 + t)}{1 + t^{2}} = S

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

S = \int_{0}^{1} \int_{0}^{1} \frac{t^{2} d t}{(1 + t^{2}) (1 + a t)} d a

= \int_{0}^{1} \frac{a (a l n (4) - π) + 4 l n (a + 1)}{4 (a^{3} + a)}

= \int_{0}^{1} \frac{a l n (4) - π}{4 (1 + a^{2})} d a + \int_{0}^{1} \frac{l n (1 + a)}{a + a^{3}} d a

= \frac{l n (4)}{8} l n (2) - \frac{π^{2}}{16} + \int_{0}^{1} \frac{l n (1 + a)}{a} - \frac{a l n (1 + a)}{1 + a^{2}} d a

= \frac{{l n}^{2} (2)}{4} - \frac{π^{2}}{16} + \int_{0}^{1} \frac{l n (1 - (- a))}{- a} d (- a) - S

S = \frac{{l n}^{2} (2)}{8} - \frac{π^{2}}{32} - \frac{1}{2} {l i}_{2} (- 1)

= \frac{{l n}^{2} (2)}{8} + \frac{π^{2}}{96}

2 ϕ = - \frac{{l n}^{3} (2}{6} + 2 l n (2) (\frac{{l n}^{2} (2)}{8} + \frac{π^{2}}{96})

= - \frac{{l n}^{3} (2)}{6} + \frac{{l n}^{3} (2)}{4} + \frac{π^{2} l n (2)}{48} = \frac{{l n}^{3} (2)}{12} + \frac{l n (2)}{48} π^{2}

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

ϕ = \frac{l n (2)}{96} (π^{2} + 4 {l n}^{2} (2))

Commented by mindispower last updated on 29/Aug/21

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

Commented by mnjuly1970 last updated on 28/Aug/21

v e r y n i c e s i r p o w e r \dots g r a t e f u l