let-f-x-0-1-ln-1-xt-2-dt-with-x-lt-1-1-find-f-x-2-calculate-0-1-ln-2-t-2-dt-

Question Number 37343 by math khazana by abdo last updated on 12/Jun/18

l e t f (x) = \int_{0}^{1} l n (1 + {x t}^{2}) d t w i t h ∣ x ∣< 1

1) f i n d f (x)

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

Commented by math khazana by abdo last updated on 15/Jun/18

i f - 1 < x < 0 w e g e t

f (x) = \frac{1}{x} - \frac{1}{x} \int_{0}^{1} \frac{d t}{1 - (- x) t^{2}} a n d c h a n v e m e n t

t \sqrt{- x} = u g i v e

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

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

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

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

= \frac{1}{x} - \frac{1}{2 x \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ \Rightarrow

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

c = f (- 1) = \int_{0}^{1} l n (1 - t^{2}) d t \Rightarrow

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

+ \int_{0}^{1} l n (1 - t^{2}) d t . w i t h - 1 < x < 0 \dots .

Commented by math khazana by abdo last updated on 15/Jun/18

1) w e h a v e f^{^{'}} {(x)}^{} = \int_{0}^{1} \frac{t^{2}}{1 + {x t}^{2}} d t s o i f x \neq 0

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

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

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

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

c = f (1) = \int_{0}^{1} l n (1 + t^{2}) d t c h a n g . \sqrt{t} = u g i v e

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

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

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

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

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

= \frac{π}{2} - \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} + 2 {[l n (\frac{u}{\sqrt{1 + u^{2}}})]}_{1}^{\sqrt{x}}

= \frac{π}{2} - 2 \frac{a r c t a n (\sqrt{x})}{\sqrt{x}} + 2 l n (\frac{\sqrt{x}}{\sqrt{1 + x}}) \Rightarrow

f (x) = l n (x) - \frac{π}{2} + \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} - 2 l n (\frac{\sqrt{x}}{\sqrt{1 + x}})

+ \int_{0}^{1} l n (1 + t^{2}) d t

f (x) = l n (x) - \frac{π}{2} + \frac{2 a r c t a n (\sqrt{x})}{\sqrt{x}} - l n x + l n (1 + x)

+ \int_{0}^{1} l n (1 + t^{2}) d t . a n d b y p a r t s

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

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

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

f (x) = \frac{2 a r t a n (\sqrt{x})}{\sqrt{x}} + l n (1 + x) + l n (2) - 2

w i t h 0 < x < 1

Commented by math khazana by abdo last updated on 15/Jun/18

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

= l n (2) + f (\frac{1}{2})

= l n (2) + \frac{2 a r c t a n (\frac{1}{\sqrt{2}})}{\frac{1}{\sqrt{2}}} + l n (\frac{3}{2}) + l n (2) - 2

= l n (2) + l n (3) + 2 \sqrt{2} (\frac{π}{2} - a r c t a n (\sqrt{2})) - 2

= π \sqrt{2} - 2 \sqrt{2} a r c t a n (\sqrt{2}) + l n (6) - 2 .