Advanced-Calculus-evaluation-the-value-of-0-pi-2-sin-2-x-ln-sin-x-dx-solution-a-0

Question Number 140588 by mnjuly1970 last updated on 09/May/21

\dots \dots . A d v a n c e d \dots . ★ ★ ★ \dots . C a l c u l u s \dots \dots .

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

ϕ := \int_{0}^{\frac{π}{2}} {s i n}^{2} (x) . l n (s i n (x)) d x

s o l u t i o n ::

ξ (a) := \int_{0}^{\frac{π}{2}} {s i n}^{2 + a} (x) d x = \frac{1}{2} β (\frac{3 + a}{2}, \frac{1}{2})

:= \frac{1}{2} (\frac{Γ (\frac{3 + a}{2}) Γ (\frac{1}{2})}{Γ (2 + \frac{a}{2})}) \dots \dots . ✓

ϕ := ξ^{'} (0) \dots \dots \dots \dots . . ✓

:= \frac{1}{2} \sqrt{π} \frac{Γ^{'} (\frac{3 + a}{2}) . Γ (2 + \frac{a}{2}) - Γ (\frac{3 + a}{2}) . Γ^{'} (2 + \frac{a}{2})}{Γ^{2} (2 + \frac{a}{2})} ∣_{a = 0}

:= \frac{1}{2} \sqrt{π} \frac{Γ^{'} (\frac{3}{2}) - Γ (\frac{3}{2}) . Γ^{'} (2)}{(Γ^{2} (2) := 1)}

:= \frac{1}{2} \sqrt{π} \frac{ψ (\frac{3}{2}) Γ (\frac{3}{2}) - Γ (\frac{3}{2}) . ψ (2)}{1}

:= \frac{\sqrt{π}}{4} {(2 - γ - 2 l n (2) - (1 - γ)}

:= \frac{\sqrt{π}}{4} (1 - l n (4)) = \sqrt{π} l n (\sqrt[4]{\frac{e}{4}})

Answered by mathmax by abdo last updated on 10/May/21

Φ = \int_{0}^{\frac{π}{2}} \sin^{2} x \log (sinx) dx let f (a) = \int_{0}^{\frac{π}{2}} \sin^{a} x dx \Rightarrow

f (a) = \int_{0}^{\frac{π}{2}} e^{alog (sinx)} dx \Rightarrow f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \log (sinx) \sin^{a} x dx \Rightarrow

f^{^{'}} (2) = \int_{0}^{\frac{π}{2}} \sin^{2} x \log (sinx) dx = Φ

we have f (a) = \int_{0}^{\frac{π}{2}} \sin^{2 (\frac{a + 1}{2}) - 1} \cos^{2 (\frac{1}{2}) - 1} x dx

= \frac{1}{2} B (\frac{a + 1}{2}, \frac{1}{2}) = \frac{1}{2} . \frac{Γ (\frac{a + 1}{2}) . Γ (\frac{1}{2})}{Γ (\frac{a + 1}{2} + \frac{1}{2})} = \frac{\sqrt{π}}{2} . \frac{Γ (\frac{a + 1}{2})}{Γ (\frac{a}{2} + 1)}

= \frac{\sqrt{π}}{2} \times \frac{Γ (\frac{a + 1}{2})}{\frac{a}{2} Γ (\frac{a}{2})} = \frac{\sqrt{π}}{a} \times \frac{Γ (\frac{a + 1}{2})}{Γ (\frac{a}{2})} \Rightarrow

f^{^{'}} (a) = - \frac{\sqrt{π}}{a^{2}} \times \frac{Γ (\frac{a + 1}{2})}{Γ (\frac{a}{2})} + \frac{\sqrt{π}}{a} \times \frac{\frac{1}{2} Γ^{^{'}} (\frac{a + 1}{2}) . Γ (\frac{a}{2}) - \frac{1}{2} Γ^{^{'}} (\frac{a}{2}) Γ (\frac{a + 1}{2})}{Γ^{2} (\frac{a}{2})}

\Rightarrow Φ = f^{^{'}} (2) = - \frac{\sqrt{π}}{4} \times \frac{Γ (\frac{3}{2})}{1} + \frac{\sqrt{π}}{4} \times \frac{Γ^{^{'}} (\frac{3}{2}) . 1 - Γ^{^{'}} (1) . Γ (\frac{3}{2})}{1^{2}}

= - \frac{\sqrt{π}}{4} . Γ (\frac{3}{2}) + \frac{\sqrt{π}}{4} \times (Γ^{^{'}} (\frac{3}{2}) + γ . Γ (\frac{3}{2}))

Γ^{^{'}} (1) = \int_{0}^{\infty} e^{- t} logt dt = - γ