0-pi-2-ln-2-sinx-dx-

Question Number 137894 by EnterUsername last updated on 08/Apr/21

\int_{0}^{\frac{π}{2}} \ln^{2} (sinx) dx

Answered by Dwaipayan Shikari last updated on 08/Apr/21

\int_{0}^{\frac{π}{2}} {s i n}^{a} (x) = \frac{Γ (\frac{a + 1}{2}) Γ (\frac{1}{2})}{2 Γ (\frac{a}{2} + 1)} = τ (a)

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

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

l o g (τ (a)) = l o g (Γ (\frac{a + 1}{2})) + l o g (\sqrt{π} / 2) - l o g (Γ (\frac{a}{2} + 1))

\frac{τ^{'} (a)}{τ (a)} = \frac{1}{2} ψ (\frac{a + 1}{2}) - ψ (\frac{a}{2} + 1) \Rightarrow τ^{″} (a) = \frac{τ^{'} (a)}{2} (ψ (\frac{a + 1}{2}) - ψ (\frac{a}{2} + 1)) + \frac{τ (a)}{4} (ψ^{'} (\frac{a + 1}{2}) - ψ^{'} (\frac{a}{2} + 1))

\Rightarrow τ^{″} (0) = \frac{τ^{'} (0)}{2} (- γ + γ - l o g (4)) + \frac{π}{8} (\frac{π^{2}}{2} - \frac{π^{2}}{6})

= \frac{- π}{2} l o g (2) (- l o g (4)) + \frac{π^{3}}{24} = \frac{π}{2} {l o g}^{2} (2) + \frac{π^{3}}{24}

Commented by Dwaipayan Shikari last updated on 08/Apr/21

M a y b e s o m e e r r o r . T r y i n g t o d o t h i s a g a i n

Commented by Dwaipayan Shikari last updated on 08/Apr/21

F i x e d!

Commented by Dwaipayan Shikari last updated on 08/Apr/21

I f f (α) = Γ (\frac{α}{2}) \Rightarrow f^{'} (α) = \frac{1}{2} Γ^{'} (\frac{α}{2}) = \frac{1}{2} Γ (\frac{α}{2}) ψ (\frac{α}{2})

F r o m t h e d e f i n i t i o n o f D i g a m m a f u n c t i o n

\frac{Γ^{'} (α)}{Γ (α)} = ψ (α) = - γ + \underset{n = 0}{\sum^{\infty}} \frac{1}{n + 1} - \frac{1}{n + α}

Commented by EnterUsername last updated on 08/Apr/21

\frac{π^{3}}{24} + \frac{π \ln^{2} 2}{2}

Commented by EnterUsername last updated on 08/Apr/21

E x c u s e m e p l e a s e, i f f (α) = Γ (\frac{α}{2}) t h e n w h a t i s f^{″} (α) ?

Commented by EnterUsername last updated on 08/Apr/21

T h a n k s

Commented by EnterUsername last updated on 08/Apr/21

I m e a n t f^{″} (α) i n s t e a d o f f^{'} (α)

Commented by Dwaipayan Shikari last updated on 08/Apr/21

S a m e m e t h o d

f^{'} (a) = \frac{1}{2} Γ (\frac{a}{2}) ψ (\frac{a}{2}) \Rightarrow f^{″} (a) = \frac{1}{2} . \frac{\partial}{\partial a} (Γ (\frac{a}{2}) ψ (\frac{a}{2}))

= \frac{1}{4} Γ^{'} (\frac{a}{2}) ψ (\frac{a}{2}) + \frac{1}{4} ψ^{'} (\frac{a}{2}) Γ (\frac{a}{2})

= \frac{1}{4} Γ (\frac{a}{2}) ψ^{2} (\frac{a}{2}) + \frac{1}{4} ψ^{'} (\frac{a}{2}) Γ (\frac{a}{2})

Commented by EnterUsername last updated on 08/Apr/21

T h a n k s f o r t h e i n s p i r a t i o n

Answered by EnterUsername last updated on 08/Apr/21

\int_{0}^{\frac{π}{2}} \ln^{2} (sinx) dx

f (α) = \int_{0}^{\frac{π}{2}} \sin^{α - 1} xdx = \frac{\sqrt{π} Γ (\frac{α}{2})}{2 Γ (\frac{α + 1}{2})}

\ln (f (α)) = \ln (Γ (\frac{α}{2})) - \ln (Γ (\frac{α + 1}{2})) + \ln (\frac{\sqrt{π}}{2})

\frac{f^{'} (α)}{f (α)} = \frac{Γ^{'} (\frac{α}{2})}{2 Γ (\frac{α}{2})} - \frac{Γ^{'} (\frac{α + 1}{2})}{2 Γ (\frac{α + 1}{2})} = \frac{1}{2} ψ (\frac{α}{2}) - \frac{1}{2} ψ (\frac{α + 1}{2})

\ln (f^{'} (α)) - \ln (f (α)) = \ln (ψ (\frac{α}{2}) - ψ (\frac{α + 1}{2})) - \ln 2

\frac{f^{″} (α)}{f^{'} (α)} - \frac{f^{'} (α)}{f (α)} = \frac{1}{2} \cdot \frac{ψ^{'} (\frac{α}{2}) - ψ^{'} (\frac{α + 1}{2})}{ψ (\frac{α}{2}) - ψ (\frac{α + 1}{2})}

f (1) = \frac{π}{2}, f^{'} (1) = \frac{f (1)}{2} (ψ (\frac{1}{2}) - ψ (1)) = \frac{π}{4} (- 2 \ln 2) = - \frac{π \ln 2}{2}

ψ^{'} (1) = ζ (2) = \frac{π^{2}}{6}, ψ^{'} (\frac{1}{2}) = \frac{π^{2}}{2}

\Rightarrow \frac{2 f^{″} (1)}{- π \ln 2} + \frac{π \ln 2}{π} = \frac{1}{2} \cdot \frac{\frac{π^{2}}{2} - \frac{π^{2}}{6}}{- 2 \ln 2} \Rightarrow - \frac{2 f^{″} (1)}{π \ln 2} + \ln 2 = - \frac{π^{2}}{12 \ln 2}

\Rightarrow \int_{0}^{\frac{π}{2}} \ln^{2} (sinx) dx = f^{″} (1) = \frac{π \ln^{2} 2}{2} + \frac{π^{3}}{24}