calculate-0-arctan-2x-x-2-3-dx-

Question Number 74484 by mathmax by abdo last updated on 24/Nov/19

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

Commented by mathmax by abdo last updated on 26/Nov/19

l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t x)}{x^{2} + 3} d x w i t h t ⩾ 0

f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(1 + t^{2} x^{2}) (x^{2} + 3)} d x =_{t x = u} \int_{0}^{\infty} \frac{u}{t (1 + u^{2}) (\frac{u^{2}}{t^{2}} + 3)} \frac{d u}{t}

= \int_{0}^{\infty} \frac{u d u}{(u^{2} + 1) (u^{2} + 3 t^{2})} d e c o m p s i t i o n o f F (u) = \frac{1}{(u^{2} + 1) (u^{2} + 3 t^{2})}

F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + 3 t^{2}}

F (- u) = F (u) \Rightarrow \frac{- a u + b}{u^{2} + 1} + \frac{- c u + d}{u^{2} + 3 t^{2}} = F (u) \Rightarrow a = c = 0 \Rightarrow

F (u) = \frac{b}{u^{2} + 1} + \frac{d}{u^{2} + 3 t^{2}}

{l i m}_{u \to + \infty} u^{2} F (u) = 0 = b + d \Rightarrow d = - b \Rightarrow F (u) = \frac{b}{u^{2} + 1} - \frac{b}{u^{2} + 3 t^{2}}

F (0) = \frac{1}{3 t^{2}} = b - \frac{b}{3 t^{2}} = (\frac{3 t^{2} - 1}{3 t^{2}}) b \Rightarrow b = \frac{1}{3 t^{2} - 1} \Rightarrow

F (u) = \frac{1}{3 t^{2} - 1} {\frac{1}{u^{2} + 1} - \frac{1}{u^{2} + 3 t^{2}}} \Rightarrow

f^{^{'}} (t) = \frac{1}{3 t^{2} - 1} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} - \frac{1}{3 t^{2} - 1} \int_{0}^{\infty} \frac{d u}{u^{2} + 3 t^{2}}

\int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{π}{2}

\int_{0}^{\infty} \frac{d u}{u^{2} + 3 t^{2}} =_{u = \sqrt{3} t z} \int_{0}^{\infty} \frac{\sqrt{3} t d z}{3 t^{2} (1 + z^{2})} = \frac{1}{\sqrt{3} t} \times \frac{π}{2} \Rightarrow

f^{^{'}} (t) = \frac{π}{2 (3 t^{2} - 1)} - \frac{π}{2 \sqrt{3} t (3 t^{2} - 1)} \Rightarrow

f (t) = \frac{π}{2} \int \frac{d t}{3 t^{2} - 1} - \frac{π}{2 \sqrt{3}} \int \frac{d t}{t (3 t^{2} - 1)} + c \dots . b e c o n t i n u e d \dots .