prove-0-1-ln-2-x-1-x-4-dx-pi-3-32-7-8-3-

Question Number 141319 by mnjuly1970 last updated on 18/May/21

p r o v e ::

Ω := \int_{0}^{1} \frac{{l n}^{2} (x)}{1 - x^{4}} d x = \frac{π^{3}}{32} + \frac{7}{8} ζ (3) . .

Answered by qaz last updated on 17/May/21

\int_{0}^{\infty} \frac{{l n}^{2} (x)}{1 - x^{4}} d x

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

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

= \int_{0}^{1} \frac{{l n}^{2} x}{1 + x^{2}} d x

= \int_{0}^{\infty} \frac{y^{2} e^{- y}}{1 + e^{- 2 y}} d y

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

= 2 \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}}

= \frac{π^{3}}{16}

Answered by Dwaipayan Shikari last updated on 17/May/21

\int_{0}^{\infty} \frac{{l o g}^{2} (x)}{1 - x^{4}} d x

= \int_{1}^{\infty} \frac{{l o g}^{2} (x)}{1 - x^{4}} d x + \int_{0}^{1} \frac{{l o g}^{2} (x)}{1 - x^{4}} d x \frac{1}{x} = i

= \int_{0}^{1} \frac{{l o g}^{2} (u) u^{2}}{u^{4} - 1} + \int_{0}^{1} \frac{{l o g}^{2} (x)}{1 - x^{4}} d x x^{4} = m u^{4} = k

= - \frac{1}{64} \int_{0}^{1} \frac{{l o g}^{2} (k) k^{- 1 / 4}}{1 - k} + \frac{1}{64} \int_{0}^{1} \frac{{l o g}^{2} (m) m^{- 3 / 4}}{1 - m} d m

= \frac{1}{64} (ψ^{″} (\frac{3}{4}) - ψ^{″} (\frac{1}{4})) = \frac{π^{3}}{16}

ψ (1 - a) - ψ (a) = π c o t (π a)

ψ^{'} (1 - a) + ψ^{'} (a) = π^{2} {c s c}^{2} (π a)

ψ^{″} (1 - a) - ψ^{″} (a) = 2 π^{3} {c s c}^{2} (π a) c o t (π a)

ψ^{″} (\frac{3}{4}) - ψ^{″} (\frac{1}{4}) = 4 π^{3}