let-x-gt-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-0-arctan-3t-2-1-t-2

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 by math khazana by abdo last updated on 26/Jun/18

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 by math 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)}