let-f-x-0-1-ln-1-xt-2-1-t-2-dt-1-find-a-simple-form-of-f-x-2-calculate-0-1-ln-1-t-2-1-t-2-dt-3-calculate-0-1-ln-1-2t-2-1-t-2-dt-

Question Number 41762 by math khazana by abdo last updated on 12/Aug/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 s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t

3) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + 2 t^{2})}{1 + t^{2}} d t

Commented by math khazana by abdo last updated on 13/Aug/18

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})}

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

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

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

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

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

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

d = 1 - \frac{x}{x - 1} = \frac{- 1}{x - 1} \Rightarrow

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

f^{^{'}} (x) = \frac{1}{x - 1} \int_{0}^{1} \frac{d t}{1 + t^{2}} - \frac{1}{x - 1} \int_{0}^{1} \frac{d x}{1 + {x t}^{2}}

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

c a s e 1 x > 0 c h a n g e m e n t t \sqrt{x} = u g i v e

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

f^{^{'}} (x) = \frac{π}{4 (x - 1)} - \frac{a r c t a n (\sqrt{x})}{(x - 1) \sqrt{x}} \Rightarrow

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

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

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

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

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

Commented by math khazana by abdo last updated on 13/Aug/18

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

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

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

= \frac{1}{2 \sqrt{- x}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\sqrt{- x}} = \frac{1}{2 \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ \Rightarrow

f^{^{'}} (x) = \frac{π}{4 (x - 1)} - \frac{1}{2 (x - 1) \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ \Rightarrow

f (x) = \frac{π}{4} l n ∣ x - 1 ∣ - \int \frac{1}{2 (x - 1) \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ d x + c

Commented by math khazana by abdo last updated on 13/Aug/18

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

A =_{t = t a n θ} \int_{0}^{\frac{π}{4}} \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}{4}} l n (\frac{1}{{c o s}^{2} θ}) d θ = - 2 \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ

l e t H = \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ a n d K = \int_{0}^{\frac{π}{4}} l n (s i n θ) d θ

w e h a v e H =_{θ = \frac{π}{2} - t} - \int_{\frac{π}{2}}^{\frac{π}{4}} l n (s i n t) d t

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

= - K - \frac{π}{2} l n (2) \Rightarrow H + K = - \frac{π}{2} l n (2)

K - H = \int_{0}^{\frac{π}{4}} l n (s i n θ) d θ - \int_{0}^{\frac{π}{4}} l n (c o s θ) d θ

= \int_{0}^{\frac{π}{4}} l n (t a n θ) d θ =_{t a n θ = u} \int_{0}^{1} \frac{l n (u)}{1 + u^{2}} d u

= \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n}) l n (u) d u

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

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

- \int_{0}^{1} \frac{1}{(2 n + 1)} u^{2 n} d u = - \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

K - H = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(2 n + 1)}^{2}} = λ_{0} \Rightarrow

2 K = - \frac{π}{2} l n (2) + λ_{0} \Rightarrow K = - \frac{π}{4} l n (2) + \frac{λ_{0}}{2}

H = - \frac{π}{2} l n (2) - K = - \frac{π}{4} l n (2) - \frac{λ_{0}}{2}