find-A-x-0-pi-2-ln-1-xsin-2-d-with-x-lt-1-

Question Number 72397 by mathmax by abdo last updated on 28/Oct/19

f i n d A (x) = \int_{0}^{\frac{π}{2}} l n (1 - {x s i n}^{2} θ) d θ w i t h ∣ x ∣< 1

Commented by mathmax by abdo last updated on 28/Oct/19

a t f o r m o f s e r i e w e h a v e {l n}^{^{'}} (1 - u) = \frac{- 1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n} \Rightarrow

l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} + c (c = 0)

= - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} \Rightarrow l n (1 - {x s i n}^{2} θ) = - \sum_{n = 1}^{\infty} \frac{x^{n} {s i n}^{2 n} θ}{n} \Rightarrow

A (x) = - \sum_{n = 1}^{\infty} (\int_{0}^{\frac{π}{2}} {s i n}^{2 n} θ d θ) \frac{x^{n}}{n} = - \sum_{n = 1}^{\infty} \frac{W_{n}}{n} x^{n}

w i t h W_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{2 n} d θ (w a l l i s i n t e g r a l)

Commented by mathmax by abdo last updated on 28/Oct/19

l e t t r y t h e p a r a m e t r i c m e t h o d w e h a v e

A (x) = \int_{0}^{\frac{π}{2}} l n (1 - x \frac{1 - c o s (2 θ)}{2}) d θ = \int_{0}^{\frac{π}{2}} l n (2 - x + x c o s θ) d θ - \frac{π}{2} l n (2)

l e t f (x) = \int_{0}^{\frac{π}{2}} l n (2 - x + x c o s (2 θ)) d θ \Rightarrow f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{- 1 + c o s (2 θ)}{2 - x + x c o s (2 θ)} d θ

=_{2 θ = t} \frac{1}{2} \int_{0}^{π} \frac{- 1 + c o s t}{2 - x + x c o s t} d t =_{t a n (\frac{t}{2}) = u} \frac{1}{2} \int_{0}^{\infty} \frac{\frac{1 - u^{2}}{1 + u^{2}} - 1}{2 - x + x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \frac{1}{2} \int_{0}^{\infty} \frac{- 4 u}{{(1 + u^{2})}^{2} {2 - x + x \frac{1 - u^{2}}{1 + u^{2}}}} d u

= - 2 \int_{0}^{\infty} \frac{u}{(2 - x) {(1 + u^{2})}^{2} + x (1 - u^{4})} d u

= - 2 \int_{0}^{\infty} \frac{u}{(2 - x) (u^{4} + 2 u^{2} + 1) + x - {x u}^{4}}

= - 2 \int_{0}^{\infty} \frac{u d u}{(2 - 2 x) u^{4} + (4 - 2 x) u^{2} + 2}

= - \int_{0}^{\infty} \frac{u d u}{(1 - x) u^{4} + (2 - x) u^{2} + 1} = \int_{0}^{\infty} \frac{u d u}{(x - 1) u^{4} + (x - 2) u^{2} - 1}

l e t d e c o m p o s e F (u) = \frac{u}{(x - 1) u^{4} + (x - 2) u^{2} - 1}

(x - 1) u^{4} + (x - 2) u^{2} - 1 = 0 \Rightarrow (x - 1) z^{2} + (x - 2) z - 1 = 0

w i t h z = u^{2}

Δ = {(x - 2)}^{2} - 4 (x - 1) (- 1) = x^{2} - 4 x + 4 + 4 x - 4 = x^{2} \Rightarrow

z_{1} = \frac{- x + 2 + ∣ x ∣}{2 (x - 1)} a n d z_{2} = \frac{- x + 2 - ∣ x ∣}{2 (x - 1)}

c a s e 1 x ⩾ 0 a n d x \neq 1 \Rightarrow z_{1} = \frac{1}{x - 1} a n d z_{2} = \frac{- 2 x + 2}{2 (x - 1)} = - 1 \Rightarrow

F (u) = \frac{u}{(x - 1 (z - z_{1}) (z - z_{2})} = \frac{u}{(x - 1) (u^{2} - \frac{1}{x - 1}) (u^{2} + 1)}

i f x > 1 \Rightarrow F (u) = \frac{u}{(x - 1) (u - \frac{1}{\sqrt{x - 1}}) (u + \frac{1}{\sqrt{x - 1}}) (u^{2} + 1)}

= \frac{a}{u - \frac{1}{\sqrt{x - 1}}} + \frac{b}{u + \frac{1}{\sqrt{x - 1}}} + \frac{c u + d}{u^{2} + 1} \dots b e c o n t i n u e d \dots

Commented by mind is power last updated on 29/Oct/19

this will worck sir since A (0) = 0, nice worck

Commented by mathmax by abdo last updated on 29/Oct/19

w e h a v e A (0) = 0