0-pi-2-ln-2-1-sin-t-1-sin-t-dt-

Question Number 154080 by iloveisrael last updated on 14/Sep/21

Ω = \int_{0}^{\frac{π}{2}} \ln^{2} (\frac{1 + \sin t}{1 - \sin t}) d t

Answered by mindispower last updated on 14/Sep/21

= \int_{0}^{\frac{π}{2}} 4 {l n}^{2} (\frac{1 + t g (\frac{x}{2})}{1 - t g (\frac{x}{2})}) d x

y = t g (\frac{x}{2}) \Rightarrow d x = \frac{2 d y}{1 + y^{2}}

= 8 \int_{0}^{1} \frac{{l n}^{2} (\frac{1 - y}{1 + y})}{1 + y^{2}} d y

y = \frac{1 - x}{1 + x} \Rightarrow d y = \frac{- 2 d x}{{(1 + x)}^{2}}

Ω = 8 \int_{0}^{1} \frac{{l n}^{2} (x)}{1 + {(\frac{1 - x}{1 + x})}^{2}} . \frac{2 d x}{{(1 + x)}^{2}} = 8 \int_{0}^{1} \frac{{l n}^{2} (x)}{1 + x^{2}} d x

= 8 \int_{0}^{1} \sum_{n ⩾ 0} {(- x^{2})}^{n} {l n}^{2} (x) d x

= 8 \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{1} x^{2 n} {l n}^{2} (x) d x

\int_{0}^{1} x^{2 n} {l n}^{2} (x) d x = \int_{0}^{\infty} t^{2} e^{- (2 n + 1) t} d t

\frac{1}{{(2 n + 1)}^{3}} \int_{0}^{\infty} t^{2} e^{- t} = \frac{2}{{(1 + 2 n)}^{3}}

Ω = 16 \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(1 + 2 n)}^{3}}

= 16 \sum_{n ⩾ 0} (\frac{1}{{(1 + 4 n)}^{3}} - \frac{1}{{(4 n + 3)}^{3}})

= \frac{16}{64} \sum_{n ⩾ 0} (\frac{1}{{(n + \frac{1}{4})}^{3}} - \frac{1}{{(n + \frac{3}{4})}^{3}})

= \frac{16}{64} (Ψ^{3} (\frac{1}{4}) - Ψ^{3} (\frac{3}{4}))