nice-calculus-0-1-ln-ln-1-x-1-3-Im-

Question Number 137474 by mnjuly1970 last updated on 03/Apr/21

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

ϕ = \int_{0}^{1} l n (\sqrt[3]{l n (\sqrt{1 - x})})

I m (ϕ) = ? ? ?

Answered by mindispower last updated on 03/Apr/21

I m (ϕ) = \frac{1}{3} \int_{0}^{1} l n (l n (x)) d x

= I m \frac{1}{3} \int_{- \infty}^{0} l n (t) e^{t} d t

= I m \frac{1}{3} \int_{0}^{\infty} l n (- x) e^{- x} d x

= \frac{1 π}{3} \int_{0}^{\infty} e^{- t} d t = \frac{π}{3} Γ (1) = \frac{π}{3}

Commented by mnjuly1970 last updated on 03/Apr/21

v e r y n i c e . . t h a n k y o u m r p o w e r \dots

Answered by Ñï= last updated on 03/Apr/21

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

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

= - \frac{1}{3} l n 2 + \frac{1}{3} \int_{- \infty}^{0} e^{u} l n u d u

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

= - \frac{1}{3} l n 2 + \frac{1}{3} \int_{0}^{\infty} e^{- u} l n (e^{i π} u) d u

= - \frac{1}{3} l n 2 + \frac{1}{3} \int_{0}^{\infty} e^{- u} (l n u + i π) d u

= - \frac{1}{3} l n 2 + \frac{1}{3} \frac{\partial}{\partial a} ∣_{a = 0} \int_{0}^{\infty} u^{a} e^{- u} d u + \frac{i π}{3}

= - \frac{1}{3} l n 2 + \frac{1}{3} \frac{\partial}{\partial a} ∣_{a = 0} Γ (a + 1) + \frac{i π}{3}

= - \frac{1}{3} l n 2 + \frac{1}{3} ψ (1) + \frac{i π}{3}

i m (ϕ) = \frac{π}{3}

- - - - - - - - - - - -

\int_{0}^{1} {(- 1)}^{x} d x = \int_{0}^{1} e^{i π x} d x = \frac{2 i}{π}

Commented by mnjuly1970 last updated on 03/Apr/21

t h a n k s a l o t . .

Answered by Dwaipayan Shikari last updated on 03/Apr/21

\frac{1}{3} \int_{0}^{1} l o g (\frac{1}{2} l o g (x)) d x

= - \frac{1}{3} l o g (2) + \frac{1}{3} \int_{0}^{1} l o g (l o g (x)) d x x = e^{- t} \Rightarrow 1 = - e^{- t} \frac{d t}{d x}

= - \frac{1}{3} l o g (2) + \frac{1}{3} \int_{0}^{\infty} e^{- t} l o g (- t) d t

= - \frac{1}{3} l o g (2) + \frac{π i}{3} - \frac{γ}{3}

I m (ϕ) = \frac{π}{3}

Commented by Dwaipayan Shikari last updated on 03/Apr/21

I f i t w a s \int_{0}^{1} l o g (\sqrt[3]{l o g (\sqrt{\frac{1}{1 - x}})}) d x t h e n t h e a n s w e r w i l l b e

= \frac{- 1}{3} l o g (2) - \frac{γ}{3}

Commented by mnjuly1970 last updated on 03/Apr/21

g r a t e f u l m r p a y a n \dots