mathematical-analysis-evaluate-0-1-ln-2-1-x-2-x-

Question Number 137799 by mnjuly1970 last updated on 06/Apr/21

\dots . . m a t h e m a t i c a l . . \dots \dots a n a l y s i s \dots .

e v a l u a t e ::

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

Answered by Dwaipayan Shikari last updated on 06/Apr/21

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

= \frac{1}{2} \int_{0}^{1} \frac{{l o g}^{2} (1 - u)}{u} d u = {[\frac{1}{2} {l o g}^{2} (1 - u) l o g (u)]}_{0}^{1} + \int_{0}^{1} l o g (1 - u) \frac{l o g (u)}{1 - u}

= \int_{0}^{1} \frac{l o g (u) l o g (1 - u)}{u} d u

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

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = ζ (3) = 1.20206 \dots

Commented by mnjuly1970 last updated on 06/Apr/21

t h a n k s a l o t m r p a y a n

v e r y n i c e \dots .

Answered by EnterUsername last updated on 06/Apr/21

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

f (s) = \frac{1}{2} \int_{0}^{1} \frac{u^{s - 1} {l n}^{2} u}{1 - u} d u = - \frac{1}{2} ψ^{″} (s)

ϕ = f (1) = - \frac{1}{2} ψ^{″} (1) = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{2}{{(n + 1)}^{3}} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = ζ (3)

Commented by mnjuly1970 last updated on 06/Apr/21

g r a t e f u l . . t h a n k y o u v e r y m u c h . .