advanced-calculus-evaluation-of-0-pi-2-sin-x-log-sin-x-dx-by-using-the-euler-beta-and-gamma-function-p-1-2-2-0-pi-2-sin-2p-1-x-

Question Number 123745 by mnjuly1970 last updated on 27/Nov/20

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

e v a l u a t i o n o f :

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

b y u s i n g t h e e u l e r b e t a a n d g a m m a f u n c t i o n :

β (p, \frac{1}{2}) = 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} (x) d x

\frac{d β (p, \frac{1}{2})}{d p} = 2 \int_{0}^{\frac{π}{2}} 2 {s i n}^{2 p - 1} (x) l n (s i n (x)) d x

= 4 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} (x) l n (s i n (x)) d x

Ω = \frac{1}{4} {[\frac{d β (p, \frac{1}{2})}{d p}]}_{p = 1}

\frac{d (β (p, \frac{1}{2}))}{d p} = \sqrt{π} {[\frac{Γ^{'} (p) Γ (p + \frac{1}{2}) - Γ^{'} (p + \frac{1}{2}) Γ (p)}{Γ^{2} (p + \frac{1}{2})}]}_{p = 1}

Ω = \frac{1}{4} (\sqrt{π} [\frac{Γ^{^{'}} (1) Γ (\frac{3}{2}) - Γ^{'} (\frac{3}{2}) Γ (1)}{Γ^{2} (\frac{3}{2})}])

= \frac{\sqrt{π}}{4} [\frac{- γ (\frac{\sqrt{π}}{2}) - \frac{\sqrt{π}}{2} (2 - γ - 2 l n (2))}{\frac{π}{4}}]

= \frac{\sqrt{π}}{4} * \frac{\sqrt{π}}{2} (\frac{- γ - 2 + γ + 2 l n (2)}{\frac{π}{4}})

= l n (2) - 1

\dots m . n . j u l y . 1970 \dots