advanced-calculus-prove-that-0-1-ln-x-1-x-3-1-3-dx-pi-3-3-ln-3-pi-3-3-m-n-1970-

Question Number 120774 by mnjuly1970 last updated on 02/Nov/20

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t ::

Ω = \int_{0}^{1} \frac{l n (x)}{\sqrt[3]{1 - x^{3}}} d x \overset{? ? ?}{=} - \frac{π}{3 \sqrt{3}} (l n (3) + \frac{π}{3 \sqrt{3}})

\dots m . n . 1970 \dots

Answered by mindispower last updated on 02/Nov/20

l e t x^{3} = t \Rightarrow d x = t^{- 2 \frac{1}{3}} \frac{d t}{3}

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

Ω = \frac{1}{9} \int_{0}^{1} {(1 - t)}^{- \frac{1}{3}} t^{- 2 \frac{1}{3}} l n (t) d t

l n (t) = \frac{\partial}{\partial a} x^{a} ∣_{x = 0}

Ω = \partial_{a} \frac{1}{9} \int_{0}^{1} {(1 - t)}^{\frac{1}{3}} t^{- \frac{2}{3} + a} d t

9 Ω = \frac{\partial}{\partial a} β (\frac{2}{3}, \frac{1}{3} + a) ∣_{a = 0}

\frac{\partial}{\partial y} β (x, y) = β (x, y) [Ψ (y) - Ψ (x + y)]

\frac{\partial β}{\partial a} = β (\frac{2}{3}, \frac{1}{3} + a) [Ψ (\frac{1}{3} + a) - Ψ (1 + a)]

9 Ω = β (\frac{1}{3}, \frac{2}{3}) [Ψ (\frac{1}{3}) - Ψ (1)]

G a u s s e t h e o r e m

Ψ (\frac{p}{q}) = - γ - l n (2 q) - \frac{π}{2} c o t (\frac{p π}{q}) + 2 \underset{n = 1}{\sum^{[\frac{q - 1}{2}]}} c o s (\frac{2 π n p}{q}) l n (s i n (\frac{n π}{q}))

Ψ (\frac{1}{3}) = - γ - l n (6) - \frac{π}{2} . \frac{1}{\sqrt{3}} - l n (\frac{\sqrt{3}}{2})

= - γ - l n (3 \sqrt{3}) - \frac{π}{2 \sqrt{3}}

Ψ (1) = - γ

β (\frac{1}{3}, \frac{2}{3}) = \frac{Γ (\frac{1}{3}) Γ (\frac{2}{3})}{Γ (2)} = . Γ (\frac{1}{3}) Γ (\frac{2}{3}) = \frac{π}{s i n (\frac{π}{3})}

= \frac{2 π}{\sqrt{3}}

9 Ω = \frac{2 π}{\sqrt{3}} (- γ - l n (3 \sqrt{3}) - \frac{π}{2 \sqrt{3}} + γ)

= \frac{2 π}{\sqrt{3}} (- l n (3 \sqrt{3}) - \frac{π}{2 \sqrt{3}})

Ω = - \frac{π}{27} (- \sqrt{3} l n (27) - π) = - \frac{π}{27} (π + \sqrt{3} l n (27))

Commented by mnjuly1970 last updated on 03/Nov/20

t h a n k s a l o t s i r m i n d s p o w e r

p e a c e b e u p o n y o u \dots

Commented by mindispower last updated on 03/Nov/20

w i t h e p l e a s u r s i r h a v e a n i c e D a y