let-f-x-0-1-ln-1-xt-2-1-t-2-dt-1-find-a-xplicit-form-of-f-x-2-developp-f-at-integr-serie-3-find-the-value-of-0-1-ln-1-t-2-1-t-2-dt-4-find-the-value-of-0-1-ln-1-2t-2

Question Number 48497 by maxmathsup by imad last updated on 24/Nov/18

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

1) f i n d a x p l i c i t f o r m o f f (x)

2) d e v e l o p p f a t i n t e g r s e r i e

3) f i n d t h e v a l u e o f \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t

4) f i n d t h e v a l u e o f \int_{0}^{1} \frac{l n (1 + 2 t^{2})}{1 + t^{2}} d t

Commented by maxmathsup by imad last updated on 26/Nov/18

1) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{(1 + {x t}^{2}) (1 + t^{2})} d t l e t d e c o m p o s e F (t) = \frac{t^{2}}{(1 + {x t}^{2}) (1 + t^{2})}

c a s e 1 i f x > 0 \Rightarrow F (x) = \frac{a t + b}{(1 + {x t}^{2})} + \frac{c t + d}{1 + t^{2}}

F (- x) = F (x) \Rightarrow \frac{- a t + b}{1 + {x t}^{2}} + \frac{- c t + d}{1 + t^{2}} = \frac{a t + b}{1 + {x t}^{2}} + \frac{c t + d}{1 + t^{2}} \Rightarrow

a = 0 a n d c = 0 \Rightarrow F (x) = \frac{b}{1 + {x t}^{2}} + \frac{d}{1 + t^{2}}

{l i m}_{t \to + \infty} t^{2} F (t) = \frac{1}{x} = \frac{b}{x} + d \Rightarrow 1 = b + x d

F (0) = 0 = b + d \Rightarrow d = - b \Rightarrow 1 = b - x b = (1 - x) b \Rightarrow b = \frac{1}{1 - x} a n d d = - \frac{1}{1 - x}

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

b u t \int_{0}^{1} \frac{d t}{1 + {x t}^{2}} =_{\sqrt{x} t = u} \int_{0}^{\sqrt{x}} \frac{d u}{\sqrt{x} (1 + u^{2})} = \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) \Rightarrow

f^{^{'}} (x) = \frac{a r c t a n (\sqrt{x})}{(1 - x) \sqrt{x}} - \frac{π}{4 (1 - x)} \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n (\sqrt{t})}{(1 - t) \sqrt{t}} d t + \frac{π}{4} l n ∣ 1 - x ∣

c h a n g e m e n t \sqrt{t} = u g i v e \int_{0}^{x} \frac{a r c t a n (\sqrt{t})}{\sqrt{t} (1 - t)} d t = \int_{0}^{\sqrt{x}} \frac{a r c t a n (u)}{u (1 - u^{2})} (2 u) d u

= 2 \int_{0}^{\sqrt{x}} \frac{a r c t a n (u)}{1 - u^{2}} d u \Rightarrow f (x) = 2 \int_{0}^{\sqrt{x}} \frac{a r c t a n (u)}{1 - u^{2}} d u + \frac{π}{4} l n ∣ 1 - x ∣ (x \neq 1)

Commented by maxmathsup by imad last updated on 26/Nov/18

c a s e 2 x < 0 \Rightarrow F (x) = \frac{t^{2}}{1 - {(\sqrt{- x} t)}^{2} (1 + t^{2})} = - \frac{t^{2}}{(\sqrt{- x} t - 1) (\sqrt{- x} t + 1) (t^{2} + 1)}

= \frac{a}{(\sqrt{- x} t - 1)} + \frac{b}{\sqrt{- x} t + 1} + \frac{c t + d}{t^{2} + 1} \Rightarrow f (x) = \frac{a}{\sqrt{- x}} l n ∣ \sqrt{- x} t + 1 ∣ + \frac{b}{\sqrt{- x}} l n ∣ \sqrt{- x} t + 1)

+ \frac{c}{2} l n (t^{2} + 1) + d a r c t a n t + c \dots .

Commented by maxmathsup by imad last updated on 26/Nov/18

2) w e h a v e f (x) = \int_{0}^{1} l n (1 + {x t}^{2}) (\sum_{n = 0}^{\infty} {(- 1)}^{n} t^{2 n}) d t

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} t^{2 n} l n (1 + {x t}^{2}) d t = \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} b y p a r t s

A_{n} = \int_{0}^{1} t^{2 n} l n (1 + {x t}^{2}) d t = {[\frac{1}{2 n + 1} t^{2 n + 1} l n (1 + {x t}^{2})]}_{t = 0}^{1} - \int_{0}^{1} \frac{1}{2 n + 1} t^{2 n + 1} \frac{2 x t}{1 + {x t}^{2}} d t

= \frac{l n (1 + x)}{2 n + 1} - \frac{2 x}{2 n + 1} \int_{0}^{1} \frac{t^{2 n + 2}}{1 + {x t}^{2}} d t b u t

\int_{0}^{1} \frac{t^{2 n + 2}}{1 + {x t}^{2}} d t =_{\sqrt{x} t = u} \int_{0}^{\sqrt{x}} \frac{u^{2 n + 2}}{x^{n + 1} (1 + u^{2})} \frac{d u}{\sqrt{x}}

= \frac{1}{x^{n + 1} \sqrt{x}} \int_{0}^{\sqrt{x}} \frac{u^{2 n + 2}}{u^{2} + 1} d u a n d c h a n g e m e n t u = t a n θ g i v e

\int_{0}^{\sqrt{x}} \frac{u^{2 n + 2}}{u^{2} + 1} d u = \int_{0}^{a r c t a n (\sqrt{x})} \frac{{t a n}^{2 n + 2} θ}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) = \int_{0}^{a r c t a n (\sqrt{x})} {t a n}^{2 n + 2} θ d θ

b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 26/Nov/18

3) l e t I = \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t c h a n g e m e n t t = t a n θ g i v e

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

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

Commented by maxmathsup by imad last updated on 26/Nov/18

4) \int_{0}^{1} \frac{l n (1 + 2 t^{2})}{1 + t^{2}} d t = f (2) = 2 \int_{0}^{\sqrt{2}} \frac{a r c t a n (u)}{1 - u^{2}} d u \dots b e c o n t i n u e d \dots