let x>0 and F(x)= ∫_0 ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt 1) find a simple form of F(x) 2)find the value of ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt 3)find the value of ∫


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Question Number 38211 by prof Abdo imad last updated on 22/Jun/18

$l e t x > 0 a n d F (x) = \int_{0}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{1 + t^{2}} d t$ $1) f i n d a s i m p l e f o r m o f F (x)$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n (2 t^{2})}{1 + t^{2}} d t$ $3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n (3 t^{2})}{1 + t^{2}} d t .$
Commented bymath khazana by abdo last updated on 26/Jun/18
$1) we have F(x)=(1/2) ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt but ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt= π{ arctan((√(2x)) +1) +arctan((√(2x)) −1) −arctanx} ⇒ F(x)=(π/2){ arctan((√(2x)) +1)+arctan((√(2x)) −1)−arctan(x)} 2) ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt =F(2) =(π/2){arctan(3) +arctan(1) −arctan(2)} =(π/2){ arctan(3)−arctan(2)}+(π^2 /8) 3) ∫_0 ^∞ ((arctan(3t^2 ))/(1+t^2 ))dt =F(3) =(π/2){ arctan((√6) +1)+arctan((√6)−1)arctan(3)}.$
$1) w e h a v e F (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{1 + t^{2}} d t b u t$ $\int_{- \infty}^{+ \infty} \frac{a r c t a n ({x t}^{2})}{1 + t^{2}} d t =$ $π {a r c t a n (\sqrt{2 x} + 1) + a r c t a n (\sqrt{2 x} - 1) - a r c t a n x}$ $\Rightarrow$ $F (x) = \frac{π}{2} {a r c t a n (\sqrt{2 x} + 1) + a r c t a n (\sqrt{2 x} - 1) - a r c t a n (x)}$ $2) \int_{0}^{\infty} \frac{a r c t a n (2 t^{2})}{1 + t^{2}} d t = F (2)$ $= \frac{π}{2} {a r c t a n (3) + a r c t a n (1) - a r c t a n (2)}$ $= \frac{π}{2} {a r c t a n (3) - a r c t a n (2)} + \frac{π^{2}}{8}$ $3) \int_{0}^{\infty} \frac{a r c t a n (3 t^{2})}{1 + t^{2}} d t = F (3)$ $= \frac{π}{2} {a r c t a n (\sqrt{6} + 1) + a r c t a n (\sqrt{6} - 1) a r c t a n (3)} .$
Commented bymath khazana by abdo last updated on 26/Jun/18

$F (3) = \frac{π}{2} {a r c t a n (\sqrt{6} + 1) + a r c t a n (\sqrt{6} - 1) - a r c t a n (3)}$
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