advanced-calculus-prove-that-Re-0-pi-2-sin-3-x-ln-ln-cos-x-dx-ln-3-2-3-

Question Number 122330 by mnjuly1970 last updated on 15/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 :

R e (\int_{0}^{\frac{π}{2}} {s i n}^{3} (x) l n (l n (c o s (x))) d x)

\overset{?}{=} \frac{l n (3) - 2 γ}{3} ✓

Answered by mathmax by abdo last updated on 16/Nov/20

from where you get this sir mn \dots

Commented by mathmax by abdo last updated on 16/Nov/20

thank you sir

Commented by mnjuly1970 last updated on 16/Nov/20

h i m r m a x

f r o m

A p p p i m t e r e s t

a n d

b r i l l i a n t m a t h i n g o o g l e \dots

Answered by mindispower last updated on 16/Nov/20

l e t c o s (x) = t \Rightarrow s i n (x) d x = - d t

\int_{0}^{\frac{π}{2}} {s i n}^{2} (x) l n (l n (c o s (x))) s i n (x) d x

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

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

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

l n (- u) = l n (u) + i π

R e (Ω) = \int_{0}^{\infty} e^{- u} l n (u) d u - \int_{0}^{\infty} e^{- 3 u} l n (u) d u \dots I

3 u = y \Rightarrow d u = \frac{d y}{3}

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

Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t \Rightarrow Γ (1) = \int_{0}^{\infty} e^{- t} d t

Γ^{'} (x) = \int_{0}^{\infty} l n (t) t^{x - 1} l n (t) d t

Γ^{'} (1) = \int_{0}^{\infty} l n (t) e^{- t} d t

R e (Ω) = \frac{2}{3} Γ^{'} (1) + \frac{1}{3} Γ (1)

Γ^{'} (1) = Ψ (1) = - γ

\Leftrightarrow R e (Ω) = - \frac{2}{3} γ + \frac{1}{3} = \frac{1 - 2 γ}{3}

Commented by mnjuly1970 last updated on 16/Nov/20

t h a n k y o u s o m u c h s i r m a x

Commented by mnjuly1970 last updated on 16/Nov/20

g r a t e f u l s i r m i d s p o w e r . v e r y n i c e

s o l u t i o n \dots .