let-F-x-0-pi-ln-1-xcos-d-with-x-lt-1-find-F-x-

Question Number 32365 by prof Abdo imad last updated on 23/Mar/18

l e t F (x) = \int_{0}^{π} l n (1 + x c o s θ) d θ . w i t h ∣ x ∣< 1

f i n d F (x) .

Commented by prof Abdo imad last updated on 25/Mar/18

w e h a v e F^{^{'}} (x) = \int_{0}^{π} \frac{c o s θ}{1 + x c o s θ} d θ a n d f o r x \neq o

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

= \frac{π}{x} - \frac{1}{x} \int_{0}^{π} \frac{d θ}{1 + x c o s θ} b u t c h . t a n (\frac{θ}{2}) = t g i v e

\int_{0}^{π} \frac{d θ}{1 + x c o s θ} = \int_{0}^{\infty} \frac{1}{1 + x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \frac{1}{1 + x} \int_{0}^{\infty} \frac{2 d t}{1 + \frac{1 - x}{1 + x} t^{2}}

=_{u = \sqrt{\frac{1 - x}{1 + x}} t} \frac{1}{1 + x} \int_{0}^{\infty} \frac{2 d t}{1 + u^{2}} \frac{\sqrt{1 + x}}{\sqrt{1 - x}} d u

= \frac{2}{\sqrt{1 - x^{2}}} \frac{π}{2} = \frac{π}{\sqrt{1 - x^{2}}}

\frac{d F}{d x} (x) = \frac{π}{x} - \frac{π}{x \sqrt{1 - x^{2}}} \Rightarrow

F (x) = π l n ∣ x ∣ - π \int \frac{d x}{x \sqrt{1 - x^{2}}} + λ c h . x = s i n t g i v e

\int \frac{d x}{x \sqrt{1 - x^{2}}} = \int \frac{c o s t}{s i n t c o s t} d t = \int \frac{d t}{s i n t}

c h . t a n (\frac{t}{2}) = u \Rightarrow \int \frac{d t}{s i n t} = \int \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int \frac{d u}{u} = l n ∣ u ∣ = l n ∣ t a n (\frac{t}{2}) ∣= l n ∣ t a n (\frac{a r c s i n x}{2}) ∣

F (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{a r c s i n x}{2}) ∣ + λ

λ = F (1) = \int_{0}^{π} l n (1 + c o s θ) d θ

Commented by prof Abdo imad last updated on 25/Mar/18

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

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

=_{\frac{θ}{2} = t} π l n (2) + 2 \int_{0}^{\frac{π}{2}} l n (c o s t) 2 d t

= π l n (2) + 4 \int_{0}^{\frac{π}{2}} l n (c o s t) d t

= π l n (2) + 4 (- \frac{π l n 2}{2}) = - π l n (2) f i n a l l y

F (x) = π l n ∣ x ∣ - π l n ∣ t a n (\frac{a r c s i n x}{2}) ∣ - π l n (2) .

Commented by prof Abdo imad last updated on 25/Mar/18

F (0) = 0