...advanced calculus... evaluation of: Ω=∫_0 ^( (π/2)) sin(x)log(sin(x))dx by using the euler beta and gamma function: β(p,(1/2))=2∫_0 ^(π/2) sin^(2p−1) (x)dx ((dβ(p,(1/2)))/dp) =2∫_0 ^(π/2) 2sin^(2p−1) (x)ln(sin(x))dx =4∫_0 ^(π/2) sin^(2p−1) (x)ln(sin(x))dx Ω=(1/4)[((dβ(p,(1/2)))/dp)]_(p=1) ((d(β(p,(1/2))))/dp)=(√π)[((Γ′(p)Γ(p+(1/2))−Γ′(p+(1/2))Γ(p))/(Γ^2 (p+(1/2))))]_(p=1) Ω=(1/4)((√π)[((Γ^′ (1)Γ((3/2))−Γ′((3/2))Γ(1))/(Γ^2 ((3/2))))]) =((√π)/4)[((−γ(((√π)/2))−((√π)/2)(2−γ−2ln(2)))/(π/4))] =((√π)/4)∗((√π)/2)(((−γ−2+γ+2ln(2))/(π/4))) =ln(2)−1 ... m.n.july.1970...


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

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t i o n o f :$ $Ω = \int_{0}^{\frac{π}{2}} s i n (x) l o g (s i n (x)) d x$ $b y u s i n g t h e e u l e r b e t a a n d g a m m a f u n c t i o n :$ $β (p, \frac{1}{2}) = 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} (x) d x$ $\frac{d β (p, \frac{1}{2})}{d p} = 2 \int_{0}^{\frac{π}{2}} 2 {s i n}^{2 p - 1} (x) l n (s i n (x)) d x$ $= 4 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} (x) l n (s i n (x)) d x$ $Ω = \frac{1}{4} {[\frac{d β (p, \frac{1}{2})}{d p}]}_{p = 1}$ $\frac{d (β (p, \frac{1}{2}))}{d p} = \sqrt{π} {[\frac{Γ^{'} (p) Γ (p + \frac{1}{2}) - Γ^{'} (p + \frac{1}{2}) Γ (p)}{Γ^{2} (p + \frac{1}{2})}]}_{p = 1}$ $Ω = \frac{1}{4} (\sqrt{π} [\frac{Γ^{^{'}} (1) Γ (\frac{3}{2}) - Γ^{'} (\frac{3}{2}) Γ (1)}{Γ^{2} (\frac{3}{2})}])$ $= \frac{\sqrt{π}}{4} [\frac{- γ (\frac{\sqrt{π}}{2}) - \frac{\sqrt{π}}{2} (2 - γ - 2 l n (2))}{\frac{π}{4}}]$ $= \frac{\sqrt{π}}{4} * \frac{\sqrt{π}}{2} (\frac{- γ - 2 + γ + 2 l n (2)}{\frac{π}{4}})$ $= l n (2) - 1$ $. . . m . n . j u l y . 1970 . . .$
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