nice-calculus-evaluate-0-1-x-2-ln-x-ln-1-x-dx-

Question Number 125336 by mnjuly1970 last updated on 10/Dec/20

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

e v a l u a t e :

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

Answered by Dwaipayan Shikari last updated on 10/Dec/20

- \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{2 + n} l o g (x) d x l o g (x) = t

= - \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{n} \int_{- \infty}^{0} {t e}^{(n + 3) t} d t (n + 3) t = - u

= - \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{\infty} \frac{u}{{(n + 3)}^{2}} e^{- u} d u

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

= \frac{1}{9} \sum^{\infty} \frac{1}{n} - \frac{1}{n + 3} - \frac{1}{3} \sum^{\infty} \frac{1}{{(n + 3)}^{2}}

= \frac{1}{9} (1 + \frac{1}{2} + \frac{1}{3}) - \frac{1}{3} (\frac{π^{2}}{6} - 1 - \frac{1}{4} - \frac{1}{9})

= \frac{11}{54} + \frac{49}{108} - \frac{π^{2}}{18} = \frac{71}{108} - \frac{π^{2}}{18}

Commented by mnjuly1970 last updated on 10/Dec/20

v e r y n i c e s o l u t i o n .

t h a n k s a l o t \dots

Commented by Dwaipayan Shikari last updated on 10/Dec/20

W i t h p l e a s u r e s i r!

Answered by Olaf last updated on 10/Dec/20

Let I_{n} = \int_{0}^{1} x^{n} \ln x d x, n \in N

I_{n} = {[\frac{x^{n + 1}}{n + 1} \ln x]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{d x}{x}

I_{n} = - \frac{1}{n + 1} \int_{0}^{1} x^{n} d x

I_{n} = - \frac{1}{n + 1} {[\frac{x^{n + 1}}{n + 1}]}_{0}^{1} = - \frac{1}{{(n + 1)}^{2}}

ϕ = \int_{0}^{1} x^{2} \ln x \ln (1 - x) d x

ϕ = \int_{0}^{1} x^{2} \ln x \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} d x

ϕ = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{n + 2} \ln x d x

ϕ = \underset{n = 1}{\sum^{\infty}} \frac{I_{n + 2}}{n}

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

ϕ = - \frac{1}{9} \underset{n = 1}{\sum^{\infty}} [\frac{1}{n} - \frac{1}{n + 3} - \frac{3}{{(n + 3)}^{2}}]

ϕ = - \frac{1}{9} \underset{n = 1}{\sum^{\infty}} [\frac{1}{n} - \frac{1}{n + 3}] + \frac{1}{3} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n + 3)}^{2}}

ϕ = - \frac{1}{9} [1 + \frac{1}{2} + \frac{1}{3}] + \frac{1}{3} [\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - (1 + \frac{1}{4} + \frac{1}{9})]

ϕ = - \frac{1}{9} [\frac{11}{6}] + \frac{1}{3} [\frac{π^{2}}{6} - \frac{49}{36}]

ϕ = \frac{π^{2}}{18} - \frac{71}{108}

Commented by mnjuly1970 last updated on 10/Dec/20

g r a t e f u l m r o l a f \dots