advanced-calculus-evaluate-2-2-0-1-2-ln-2-1-x-x-dx-1-1-0-1-2-ln-2-1-x-x-dx-M-N-1970-

Question Number 118886 by mnjuly1970 last updated on 20/Oct/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 :: {_{2 . Ω_{2} = \int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (1 + x)}{x} d x = ? ?}^{1 . Ω_{1} = \int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (1 - x)}{x} d x = ? ?}

\dots M . N . 1970 \dots

Answered by mindispower last updated on 20/Oct/20

Ω_{1} = \int_{0}^{\frac{1}{2}} \frac{l n (1 - x)}{x} l n (1 - x) d x b y p a r t

w i t h e - \int_{0}^{t} \frac{l n (1 - x)}{x} d x = {L i}_{2} (t)

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

= {l i}_{2} (\frac{1}{2}) l n (2) - A, l e t 1 - x = t i n A

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

w e h a v e {l i}_{2} (x) + {l i}_{2} (1 - x) = \frac{π^{2}}{6} - l n (x) l n (1 - x)

\Rightarrow Ω_{1} = {l i}_{2} (\frac{1}{2}) l n (2) + \int_{1}^{\frac{1}{2}} \frac{π^{2}}{6 t} d t - \int_{1}^{\frac{1}{2}} \frac{{l i}_{2} (t)}{t} d t - \int_{1}^{\frac{1}{2}} \frac{l n (t) l n (1 - t)}{t} d t

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

1 - t = x \Rightarrow

= \frac{{l n}^{3} (\frac{1}{2})}{2} - \int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (1 - x)}{2 x} = - \frac{{l n}^{3} (2)}{2} - \frac{Ω_{1}}{2}

\Rightarrow Ω_{1} = l n (2) {L i}_{2} (\frac{1}{2}) + \frac{π^{2}}{6} l n (\frac{1}{2}) - {L i}_{3} (\frac{1}{2}) + {L i}_{3} (1) + \frac{{l n}^{3} (2)}{2} + \frac{1}{2} Ω_{1}

{l i}_{2} (\frac{1}{2}) + {l i}_{2} (1 - \frac{1}{2}) = \frac{π^{2}}{6} - {l n}^{2} (2) \Rightarrow {l i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}

{l i}_{3} (1) = ζ (3)

Ω_{1} = 2 (l n (2) {L i}_{2} (\frac{1}{2}) - \frac{π^{2} l n (2)}{6} - {L i}_{3} (\frac{1}{2}) + ζ (3) + \frac{{l n}^{3} (2)}{2})

{l i}_{3} (z) + {l i}_{3} (1 - z) + {l i}_{3} (1 - \frac{1}{z}) = ζ (3) + \frac{π^{2} l n (z)}{6} + \frac{{l n}^{3} (z)}{6} - \frac{{l n}^{2} (z)}{2} l n (1 - z)

z = \frac{1}{2} \Rightarrow

2 {l i}_{3} (\frac{1}{2}) + {l i}_{3} (- 1) = ζ (3) + \frac{π^{2}}{6} l n (\frac{1}{2}) - \frac{{l n}^{3} (2)}{6} + \frac{{l n}^{3} (2)}{2}

{L i}_{3} (- 1) = \frac{1}{8} ζ (3) - \frac{7}{8} ζ (3) = - \frac{3}{4} ζ (3)

{l i}_{3} (\frac{1}{2}) = \frac{7}{8} ζ (3) - \frac{π^{2}}{12} l n (2) + \frac{{l n}^{3} (2)}{6}

Ω_{1} = 2 (l n (2) {l i}_{2} (\frac{1}{2}) - \frac{π^{2}}{6} l n (2) - {l i}_{3} (\frac{1}{2}) + ζ (3) + \frac{{l n}^{3} (2)}{6})

j u s t r e p l q c e {l i}_{2} (\frac{1}{2}), a n d {l i}_{3} (\frac{1}{2}) \dots

w e g e t n s w e r [c l o s e [f o r m

Commented by mnjuly1970 last updated on 21/Oct/20

t h a n k y o u s i r \dots

v e r y n i c e \dots

Commented by mindispower last updated on 21/Oct/20

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