find-f-x-0-1-ln-1-xt-2-dt-with-x-gt-0-2-give-thevalue-of-0-1-ln-1-t-2-dt-and-0-1-ln-1-2t-2-dt-

Question Number 31502 by abdo imad last updated on 09/Mar/18

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

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

Commented by abdo imad last updated on 15/Mar/18

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

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

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

λ = f (1) = \int_{0}^{1} l n (1 + t^{2}) d t s o

f (x) = l n (x) - \int_{1}^{x} \frac{a r c t a n (\sqrt{t)}}{t \sqrt{t}} d t + \int_{0}^{1} l n (1 + t^{2}) d t c h . \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^{3}} (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{d u}{u (1 + u^{2})})

= 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})} b u t

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

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

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

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

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

Commented by abdo imad last updated on 15/Mar/18

2) l e t p u t I = \int_{0}^{1} l n (1 + t^{2}) d t w e h a v e f o r ∣ x ∣< 1

{l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow

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

I = \int_{0}^{1} (\sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{t^{2 n}}{n}) d t

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \frac{1}{2 n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (2 n + 1)}

\frac{1}{2} I = \sum_{n = 1}^{\infty} (\frac{1}{2 n} - \frac{1}{2 n + 1}) {(- 1)}^{n - 1}

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}

= \frac{1}{2} l n (2) + \frac{π}{4} - 1 \Rightarrow I = l n (2) + \frac{π}{2} - 2

Commented by abdo imad last updated on 15/Mar/18

w e h a v e p r o v e d t h a t \int_{0}^{1} l n (1 + t^{2}) d t = l n (2) + \frac{π}{2} - 2 \Rightarrow

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

l e t t a k e x = 2 w e o b t a i n

\int_{0}^{1} l n (1 + 2 t^{2}) d t = l n (2) + \frac{a r c t a n (\sqrt{2})}{2 \sqrt{2}} - 2 l n (\frac{\sqrt{2}}{\sqrt{3}}) .