0-pi-2-ln-3-sinx-dx-

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}} + \dots

= (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \dots) - (\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \dots)

= (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \dots) - \frac{1}{2^{2}} (1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots)

= 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}} + \dots)

= - 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}} + \dots)]

= - 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