question-If-0-1-ln-2-1-x-4-x-dx-a-b-find-the-value-of-a-b-

Question Number 158293 by mnjuly1970 last updated on 02/Nov/21

You can't use 'macro parameter character #' in math mode

If, Ω = \int_{0}^{1} \frac{{l n}^{2} (1 - x^{4})}{x} d x = a ζ b)

f i n d t h e v a l u e o f, a, b .

Answered by qaz last updated on 02/Nov/21

\int_{0}^{1} \frac{\ln^{2} (1 - x^{4})}{x} dx

= \frac{1}{4} \int_{0}^{1} \frac{\ln^{2} (1 - x)}{x} dx

= \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{n} \ln^{2} xdx

= \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \frac{2!}{{(n + 1)}^{3}}

= \frac{1}{2} ζ (3)

Commented by mnjuly1970 last updated on 02/Nov/21

g r a t e f u l s i r q a z

e x c e l l e n t a s a l w a y s

Answered by mindispower last updated on 02/Nov/21

= \int_{0}^{1} \frac{{l n}^{2} (1 - x^{4}) x^{3}}{x^{4}} d x

= \int_{0}^{1} \frac{{l n}^{2} (1 - t)}{t} \frac{d t}{3}

l n (1 - t) = - u \Rightarrow d t = e^{- u} d u

= \frac{1}{3} \int_{0}^{\infty} \frac{u^{2} e^{- u}}{1 - e^{- u}} d u = \frac{1}{3} Γ (3) ζ (3) = \frac{2}{3} ζ (3)

Commented by mnjuly1970 last updated on 03/Nov/21

t h x s i r p o w e r

Answered by Ar Brandon last updated on 02/Nov/21

Ω = \int_{0}^{1} \frac{\ln^{2} (1 - x^{4})}{x} d x, x = u^{\frac{1}{4}} \Rightarrow d x = \frac{1}{4} u^{- \frac{3}{4}} d u

= \frac{1}{4} \int_{0}^{1} \frac{\ln^{2} (1 - u)}{u} d u = \frac{1}{4} \int_{0}^{1} \frac{\ln^{2} u}{1 - u} d u

= - \frac{1}{4} ψ^{(2)} (1) = \frac{2}{4} ζ (3) = \frac{1}{2} ζ (3)

Commented by mnjuly1970 last updated on 03/Nov/21

m e r c e y m r b r a n d o n