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Question Number 40829 by math khazana by abdo last updated on 28/Jul/18

l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t x)}{x^{3} + 8} d x

1) f i n d a s i m p l e f o r m o f f (t)

2) c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (x)}{x^{3} + 8} d x .

Answered by maxmathsup by imad last updated on 29/Jul/18

1) w e h a v e f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(1 + t^{2} x^{2}) (x^{3} + 8)} d x =_{t x = α} \int_{0}^{\infty} \frac{1}{(1 + α^{2}) (\frac{α^{3}}{t^{3}} + 8)} \frac{α}{t} \frac{d α}{t}

= \frac{1}{t^{2}} \int_{0}^{\infty} t^{3} \frac{α d α}{(1 + α^{2}) (α^{3} + 8 t^{3})} = t \int_{0}^{\infty} \frac{x d x}{(x^{2} + 1) (x^{3} + 8 t^{3})} l e t d e c o m p o s e

F (x) = \frac{x}{(x^{2} + 1) (x^{3} + 8 t^{3})}

F (x) = \frac{x}{(x^{2} + 1) (x + 2 t) (x^{2} - 2 x t + 4 t^{2})} = \frac{a}{x + 2 t} + \frac{b x + c}{x^{2} + 1} + \frac{d x + e}{x^{2} - 2 x t + 4 t^{2}}

a = {l i m}_{x \to - 2 t} (x + 2 t) F (x) = \frac{- 2 t}{(4 t^{2} + 1) (12 t^{2})} = \frac{- 1}{6 t (4 t^{2} + 1)}

{l i m}_{x \to + \infty} x F (x) = 0 = a + b + d \Rightarrow b + d = - a

F (o) = 0 = \frac{a}{2 t} + c + \frac{e}{4 t^{2}} \Rightarrow 2 a t + 4 t^{2} c + e = 0 \dots . b e c o n t i n u e d \dots