find-f-0-arctan-x-x-2-3-2-dx-with-real-

Question Number 72012 by mathmax by abdo last updated on 23/Oct/19

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

Commented by mathmax by abdo last updated on 07/Nov/19

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

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

= \frac{α^{4}}{α^{2}} \int_{0}^{\infty} \frac{t d t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}} = α^{2} \int_{0}^{\infty} \frac{t d t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}}

l e t d e c o m p o s e F (t) = \frac{t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}}

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

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

+ \frac{- e t - f}{{(t^{2} + 3 α^{2})}^{2}} \Rightarrow b = d = f = 0 \Rightarrow

F (t) = \frac{a t}{t^{2} + 1} + \frac{c t}{t^{2} + 3 α^{2}} + \frac{e t}{{(t^{2} + 3 α^{2})}^{2}}

{l i m}_{t \to + \infty} t F (t) = 0 = a + c + e \Rightarrow e = - a - c \Rightarrow

F (t) = \frac{a t}{t^{2} + 1} + \frac{c t}{t^{2} + 3 α^{2}} - \frac{(a + c) t}{{(t^{2} + 3 α^{2})}^{2}}

F (1) = \frac{1}{2 {(1 + 3 α^{2})}^{2}} = \frac{a}{2} + \frac{c}{1 + 3 α^{2}} - \frac{a + c}{{(1 + 3 α^{2})}^{2}} \Rightarrow

\frac{1}{2} = \frac{1}{2} {(1 + 3 α^{2})}^{2} a + (1 + 3 α^{2}) c - a - c \Rightarrow

1 = {(1 + 3 α^{2})}^{2} a + 2 (1 + 3 α^{2}) c - 2 a - 2 c \Rightarrow

{{(1 + 3 α^{2})}^{2} - 2} a + {2 (1 + 3 α^{2}) - 2} c = 1

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