find-f-x-0-arctan-xt-1-t-2-dt-with-x-real-

Question Number 53284 by maxmathsup by imad last updated on 20/Jan/19

f i n d f (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t w i t h x r e a l .

Commented by prof Abdo imad last updated on 21/Jan/19

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

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

= \int_{0}^{\infty} \frac{u}{(u^{2} + 1)) (u^{2} + x^{2})} d u l e t d e c o m p o s e

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

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

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

= \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} + x^{2}} \Rightarrow b = d = 0 \Rightarrow

F (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + x^{2}}

{l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow

F (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + x^{2}}

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

(x^{2} - 1) a = 1 \Rightarrow a = \frac{1}{x^{2} - 1} (w e s u p p o s e x \neq \bar{+} 1) \Rightarrow

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

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

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

= \frac{l n ∣ x ∣}{x^{2} - 1} l e t s u p p o s e x > 1 \Rightarrow

f^{^{'}} (x) = \frac{l n (x)}{x^{2} - 1} \Rightarrow f (x) = \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t + λ

λ = f (1) = \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2}} d t b y p a r t s

λ = {[{a r c t a n}^{2} (t)]}_{0}^{\infty} - \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2}} d t

= \frac{π^{2}}{4} - λ \Rightarrow 2 λ = \frac{π^{2}}{4} \Rightarrow λ = \frac{π^{2}}{8} \Rightarrow

f (x) = \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t + \frac{π^{2}}{8}

Commented by prof Abdo imad last updated on 21/Jan/19

l e t d e t e t m i n e \int_{1}^{x} \frac{l n (t)}{t^{2} - 1} d t c h s n g . t = \frac{1}{u} g i v e

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

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

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

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

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

= \sum_{n = 0}^{\infty} \int_{\frac{1}{x}}^{1} u^{n} l n (u) d u = \sum_{n = 0}^{\infty} A_{n}

b y p a r t s A_{n} = {[\frac{1}{n + 1} u^{n + 1} l n (u)]}_{\frac{1}{x}}^{1} - \int_{\frac{1}{x}}^{1} \frac{u^{n}}{n + 1} d u

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

= \frac{l n (x)}{(n + 1) x^{n + 1}} - \frac{1}{{(n + 1)}^{2}} {1 - \frac{1}{x^{n + 1}}} \Rightarrow

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

- \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} x^{n + 1}}

\sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} = ξ (2) = \frac{π^{2}}{6}

l e t f i n d \sum_{n = 0}^{\infty} \frac{1}{(n + 1) x^{n + 1}} b e c o n t i n u e d \dots