∫_0 ^(π/2) ln^3 (sinx)dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 137991 by EnterUsername last updated on 08/Apr/21

$\int_{0}^{\frac{π}{2}} {l n}^{3} (s i n x) d x$
	Answered by Ar Brandon last updated on 08/Apr/21

	$f (α) = \int_{0}^{\frac{π}{2}} \sin^{α - 1} xdx = \frac{1}{2} β (\frac{α}{2}, \frac{1}{2}) = \frac{\sqrt{π} Γ (\frac{α}{2})}{2 Γ (\frac{α + 1}{2})}, f (1) = \frac{π}{2} ★$ $\ln (f (α)) = \ln (Γ (\frac{α}{2})) - \ln (Γ (\frac{α + 1}{2})) + \ln (\frac{\sqrt{π}}{2})$ $\frac{f^{'} (α)}{f (α)} = \frac{Γ^{'} (\frac{α}{2})}{2 Γ (\frac{α}{2})} - \frac{Γ^{'} (\frac{α + 1}{2})}{2 Γ (\frac{α + 1}{2})} = \frac{1}{2} ψ (\frac{α}{2}) - \frac{1}{2} ψ (\frac{α + 1}{2})$ $f^{'} (1) = \frac{π}{4} (ψ (\frac{1}{2}) - ψ (1)) = \frac{π}{4} (- γ - 2 \ln 2 + γ) = - \frac{π \ln 2}{2} ★$ $f^{″} (α) = \frac{f^{'} (α)}{2} (ψ (\frac{α}{2}) - ψ (\frac{α + 1}{2})) + \frac{f (α)}{4} (ψ^{'} (\frac{α}{2}) - ψ^{'} (\frac{α + 1}{2}))$ $f^{″} (1) = - \frac{π \ln 2}{4} (ψ (\frac{1}{2}) - ψ (1)) + \frac{π}{8} (ψ^{'} (\frac{1}{2}) - ψ^{'} (1))$ $= - \frac{π \ln 2}{4} (- γ - 2 \ln 2 + γ) + \frac{π}{8} (3 ζ (2) - ζ (2)) = \frac{π \ln^{2} 2}{2} + \frac{π^{3}}{24} ★$ $f^{‴} (α) = \frac{f^{″} (α)}{2} (ψ (\frac{α}{2}) - ψ (\frac{α + 1}{2})) + \frac{f^{'} (α)}{4} (ψ^{'} (\frac{α}{2}) - ψ^{'} (\frac{α + 1}{2}))$ $+ \frac{f^{'} (α)}{4} (ψ^{'} (\frac{α}{2}) - ψ^{'} (\frac{α + 1}{2})) + \frac{f (α)}{8} (ψ^{″} (\frac{α}{2}) - ψ^{″} (\frac{α + 1}{2}))$ $f^{‴} (1) = (\frac{π \ln^{2} 2}{4} + \frac{π^{3}}{48}) (- 2 \ln 2) - \frac{π \ln 2}{8} (\frac{π^{2}}{2} - \frac{π^{2}}{6})$ $- \frac{π \ln 2}{8} (\frac{π^{2}}{2} - \frac{π^{2}}{6}) + \frac{π}{16} (- 16 \times \frac{7}{8} ζ (3) + 2 ζ (3))$ $= - \frac{π \ln^{3} 2}{2} - \frac{π^{3} \ln 2}{24} - \frac{π^{3} \ln 2}{16} + \frac{π^{3} \ln 2}{48} - \frac{π^{3} \ln 2}{16} + \frac{π^{3} \ln 2}{48} - \frac{3 π}{4} ζ (3)$ $= - \frac{π \ln^{3} 2}{2} - \frac{π^{3} \ln 2}{8} - \frac{3 π}{4} ζ (3) ★$ $\int_{0}^{\frac{π}{2}} \ln (sinx) dx = f^{'} (1), \int_{0}^{\frac{π}{2}} \ln^{2} (sinx) dx = f^{″} (1), \int_{0}^{\frac{π}{2}} \ln^{3} (sinx) dx = f^{‴} (1)$
	Commented by Ar Brandon last updated on 08/Apr/21

	$ψ (α) = - γ + \underset{n = 0}{\sum^{\infty}} (\frac{1}{n + 1} - \frac{1}{n + α})$ $ψ^{'} (α) = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + α)}^{2}}, ψ^{″} (α) = \underset{n = 0}{\sum^{\infty}} \frac{- 2}{{(n + α)}^{3}}$ $\Rightarrow ψ^{'} (1) = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = ζ (2) = \frac{π^{2}}{6}$ $\Rightarrow ψ^{'} (\frac{1}{2}) = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{2})}^{2}} = 4 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} = 4 \times \frac{π^{2}}{8} = \frac{π^{2}}{2}$ $\underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} = 1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . .$ $= (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . .) - (\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . .)$ $= (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . .) - \frac{1}{2^{2}} (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . .)$ $= ζ (2) - \frac{1}{4} ζ (2) = \frac{3}{4} ζ (2) = \frac{π^{2}}{8}$ $ψ^{″} (1) = \underset{n = 0}{\sum^{\infty}} \frac{- 2}{{(n + 1)}^{3}} = - 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = - 2 ζ (3)$ $ψ^{″} (\frac{1}{2}) = - 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + \frac{1}{2})}^{3}} = - 16 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{3}}$ $= - 16 (1 + \frac{1}{3^{3}} + \frac{1}{5^{3}} + . . .)$ $= - 16 [(1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}} + \frac{1}{5^{3}} + . .) - \frac{1}{2^{3}} (1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}} + . . .)]$ $= - 16 \times (1 - \frac{1}{2^{3}}) ζ (3) = - 14 ζ (3)$
	Commented by Dwaipayan Shikari last updated on 08/Apr/21

	$P e a c e i n C h a o s . .$
	Commented by Ar Brandon last updated on 08/Apr/21

	🐱😊😃
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum