find-f-0-arctan-x-1-x-2-dx-with-gt-0-

Question Number 50683 by maxmathsup by imad last updated on 18/Dec/18

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

Commented by Abdo msup. last updated on 21/Dec/18

c h a n g e m e n t \sqrt{λ} x = t g i v e

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

a n d \int_{0}^{\infty} \frac{a r c t a n (\sqrt{λ} t)}{1 + t^{2}} d t = W (\sqrt{λ}) w i t h

W (x) = \int_{0}^{\infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t (x > 0) l e t d e t e r m i n e W (x)

W^{^{'}} (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 d u}{(u^{2} + x^{2}) (u^{2} + 1)} l e t d e v o m p o s e

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

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

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

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

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

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

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

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

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

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

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

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

\int_{0}^{\infty} \frac{u d u}{1 + u^{2}} d u = \frac{1}{2} l n (1 + u^{2}) a l s o

\int_{0}^{\infty} \frac{u d u}{u^{2} + x^{2}} d u =_{u = x α} \int_{0}^{\infty} \frac{x α}{x^{2} α^{2} + x^{2}} x d α

= \int_{0}^{\infty} \frac{α d α}{α^{2} + 1} \Rightarrow \int_{0}^{\infty} F (u) d u = 0 \Rightarrow W^{^{'}} (x) = 0 \Rightarrow

W (x) = c = W (1) = \int_{0}^{\infty} \frac{a r c t a n t}{1 + t^{2}} d t c h a n g e m e n t

t = \frac{1}{u} g i v e W (1) = - \int_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n u}{1 + \frac{1}{u^{2}}} \frac{- d u}{u^{2}}

= \int_{0}^{\infty} \frac{\frac{π}{2} - a r c t a n u}{u^{2} + 1} d u

= \frac{π}{2} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} - \int_{0}^{\infty} \frac{a r c t a n u}{1 + u^{2}} d u = \frac{π^{2}}{4} - W (1) \Rightarrow

2 W (1) = \frac{π^{2}}{4} \Rightarrow W (1) = \frac{π^{2}}{8} \Rightarrow f (λ) = \frac{π^{2}}{8 \sqrt{λ}} .