nice-calculus-prove-that-I-0-1-2-log-2-1-x-x-dx-3-4-1-3-log-3-2-

Question Number 123547 by mnjuly1970 last updated on 26/Nov/20

\dots . n i c e c a l c u l u s \dots .

p r o v e t h a t ::::

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

= \frac{ζ (3)}{4} - \frac{1}{3} {l o g}^{3} (2) ✓

Answered by mnjuly1970 last updated on 26/Nov/20

s o l u t i o n :

s p e c i a l v a l u e s ::

i :: {l i}_{2} (\frac{1}{2}) [\underset{t r y i t .}{\overset{w h y ? ?}{=}}] \frac{π^{2}}{12} - \frac{1}{2} {l o g}^{2} (2) ✓

i i :: {l i}_{3} (\frac{1}{2}) [\underset{p r o v e i t .}{\overset{w h y ? ?}{=}}] \frac{7}{8} ζ (3) + \frac{1}{6} {l o g}^{3} (2) - \frac{π^{2}}{12} l o g (2) ✓

i i i :: {l i}_{3} (1) = ζ (3) ✓

I \overset{1 - x = t}{=} \int_{\frac{1}{2}}^{1} \frac{{l o g}^{2} (t)}{1 - t} d t \overset{i . b . p}{=} {[- l o g (1 - t) {l o g}^{2} (t)]}_{\frac{1}{2}}^{1} + 2 \int_{\frac{1}{2}}^{1} \frac{l o g (t) l o g (1 - t)}{t} d t

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

= - {l o g}^{3} (2) + 2 {{[l o g (t) {l i}_{2} (t)]}_{\frac{1}{2}}^{1} + \int_{\frac{1}{2}}^{1} \frac{{l i}_{2} (t)}{t} d t}

= - {l o g}^{3} (2) + 2 [l o g (\frac{1}{2}) {l i}_{2} (\frac{1}{2})] + 2 {l i}_{3} (1) - 2 {l i}_{3} (\frac{1}{2})

= - {l o g}^{3} (2) - 2 l o g (2) (\frac{π^{2}}{12} - \frac{1}{2} {l o g}^{2} (2)) \underset{↫}{+}

\overset{↬}{-} \frac{7}{4} ζ (3) - \frac{1}{3} {l o g}^{3} (2) + \frac{π^{2}}{6} l o g (2) + 2 ζ (3)

= \frac{ζ (3)}{4} - \frac{1}{3} {l o g}^{3} (2) ✓ ✓

n i c e . c a l c u l u s . m . n . j u l y . 1970

Answered by mathmax by abdo last updated on 26/Nov/20

I = \int_{0}^{\frac{1}{2}} \frac{\log^{2} (1 - x)}{x} dx \Rightarrow I =_{x = \frac{t}{2}} 2 \int_{0}^{1} \frac{\log^{2} (1 - \frac{t}{2})}{t} \frac{dt}{2}

= \int_{0}^{1} \frac{\log^{2} (1 - \frac{t}{2})}{t} dt let f (a) = \int_{0}^{1} \frac{\log^{2} (1 - at)}{t} dt [with 0 < a < 1

f^{^{'}} (a) = 2 \int_{0}^{1} \frac{- t}{1 - at} \frac{\log (1 - at)}{t} dt = - 2 \int_{0}^{1} \frac{\log (1 - at)}{1 - at} dt

= - 2 \int_{0}^{1} \log (1 - at) \sum_{n = 0}^{\infty} a^{n} t^{n} dt

= - 2 \sum_{n = 0}^{\infty} a^{n} \int_{0}^{1} t^{n} \log (1 - at) dt = - 2 \sum_{n = 0}^{\infty} a^{n} U_{n}

U_{n} = \int_{0}^{1} t^{n} \log (1 - at) dt = {[\frac{t^{n + 1}}{n + 1} \log (1 - at)]}_{0}^{1} = \frac{\log (1 - a)}{n + 1}

- \int_{0}^{1} \frac{t^{n + 1}}{n + 1} \times \frac{- a}{1 - at} dt = \frac{\log (1 - a)}{n + 1} + \frac{a}{n + 1} \int_{0}^{1} \frac{t^{n + 1}}{1 - at} dt and

\int_{0}^{1} \frac{t^{n + 1}}{1 - at} dt =_{1 - at = z} \int_{1}^{1 - a} \frac{{(\frac{1 - z}{a})}^{n + 1}}{z} (- \frac{1}{a}) dz

= \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \frac{{(1 - z)}^{n + 1}}{z} dz = \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \frac{1}{z} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- z)}^{k} dz

= \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- 1)}^{k} z^{k - 1} dz

= \frac{1}{a^{n + 2}} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- 1)}^{k} {[\frac{z^{k}}{k}]}_{1 - a}^{a}

= \frac{1}{a^{n + 2}} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} \frac{{(- 1)}^{k}}{k} {a^{k} - {(1 - a)}^{k}} \dots . be continued \dots