nice-calculus-prove-that-0-1-ln-2-1-x-x-dx-3-4-m-n-

Question Number 122035 by mnjuly1970 last updated on 13/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 :

\int_{0}^{1} \frac{{l n}^{2} (1 + x)}{x} d x \overset{? ?}{=} \frac{ζ (3)}{4}

. m . n .

Answered by mindispower last updated on 14/Nov/20

\int u^{2} e^{- u}

- u^{2} e^{- u} - 2 {u e}^{- u} - 2 e^{- u}

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

= \int_{0}^{l n (2)} \sum_{n ⩾ 0} u^{2} e^{- n u} d u = \frac{{l n}^{3} (2)}{3}

+ \sum_{n ⩾ 1} \int_{0}^{l n (2)} u^{2} e^{- n u} d u = \sum_{n ⩾ 1} \frac{1}{n^{3}} \int_{0}^{n l n (2)} x^{2} e^{- x} d x

= \frac{{l n}^{3} (2)}{3} + \sum_{n ⩾ 1} \frac{1}{n^{3}} {[- x^{2} e^{- x} - 2 {x e}^{- x} - 2 e^{- x}]}_{0}^{n l n (2)}

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

= \frac{{l n}^{3} (2)}{3} - {l n}^{2} (2) \sum_{n ⩾ 1} \frac{1}{n . 2^{n}} - 2 l n (2) \sum_{n ⩾ 1} \frac{1}{n^{2} 2^{n}} - 2 \sum_{n ⩾ 1} \frac{1}{n^{3} 2^{n}} + 2 ζ (3)

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

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

= - \frac{7}{4} ζ (3) + 2 ζ (3) = \frac{1}{4} ζ (3)

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

Commented by mnjuly1970 last updated on 14/Nov/20

Commented by mnjuly1970 last updated on 14/Nov/20

t h a n k y o u m r p o w e r

Commented by mindispower last updated on 14/Nov/20

h e l l o s i r r e a l y n i c e p l e a s e s i r

h a v e y o u s o l u t i o n f o r Σ \frac{1}{n^{2} c o s (\frac{π}{2^{n}})}