find-the-value-of-0-arctan-x-1-x-4-dx-

Question Number 52418 by Abdo msup. last updated on 07/Jan/19

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n (x)}{1 + x^{4}} d x

Commented by maxmathsup by imad last updated on 08/Jan/19

l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t x)}{1 + x^{4}} d x \Rightarrow f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(1 + x^{2} t^{2}) (1 + x^{4})} d x

=_{x t = u} \int_{0}^{\infty} \frac{u}{t (1 + u^{2}) (1 + \frac{u^{4}}{t^{4}})} \frac{d u}{t} = \int_{0}^{\infty} \frac{u}{(1 + u^{2}) (t^{2} + \frac{u^{4}}{t^{2}})} d u

= t^{2} \int_{0}^{\infty} \frac{u}{(u^{2} + 1) (u^{4} + t^{4})} d u l e t d e c o m p o s e F (u) = \frac{u}{(u^{2} + 1) (u^{4} + t^{4})}

F (u) = \frac{u}{(u^{2} + 1) ({(u^{2} + t^{2})}^{2} - 2 u^{2} t^{2})} = \frac{u}{(u^{2} + 1) (u^{2} + t^{2} - \sqrt{2} t u) (u^{2} + t^{2} + \sqrt{2} t u)}

= \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} - \sqrt{2} t u + t^{2}} + \frac{e u + f}{u^{2} + \sqrt{2} t u + t^{2}}

F (- u) = - F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + \sqrt{2} t u + t^{2}} + \frac{- e u + f}{u^{2} - \sqrt{2} t u + t^{2}}

= \frac{- a u - b}{u^{2} + 1} + \frac{- c u - d}{u^{2} - \sqrt{2} t u + t^{2}} + \frac{- e u - f}{u^{2} + \sqrt{2} t u + t^{2}} \Rightarrow

b = 0 a n d c = e a n d f = - d \Rightarrow

F (u) = \frac{a u}{u^{2} + 1} + \frac{c u + d}{u^{2} - \sqrt{2} t u + t^{2}} + \frac{c u - d}{u^{2} + \sqrt{2} t u + t^{2}}

{l i m}_{u \to + \infty} u F (u) = 0 = a + 2 c \Rightarrow c = - \frac{a}{2} \Rightarrow

F (u) = \frac{a u}{u^{2} + 1} - \frac{1}{2} \frac{a u - 2 d}{u^{2} - \sqrt{2} t u + t^{2}} - \frac{1}{2} \frac{a u + 2 d}{u^{2} + \sqrt{2} t u + t^{2}}

F (1) = \frac{1}{2 (1 + t^{4})} = \frac{a}{2} - \frac{1}{2} \frac{a - 2 d}{1 + t^{2} - \sqrt{2} t} - \frac{1}{2} \frac{a + 2 d}{1 + t^{2} + \sqrt{2} t}

F (- 1) = \frac{- 1}{2 (1 + t^{4})} - \frac{1}{2} \frac{a - 2 d}{1 + t^{2} + \sqrt{2} t} - \frac{1}{2} \frac{- a + 2 d}{1 + t^{2} - \sqrt{2} t}

\dots . b e c o n t i n u e d \dots