find f(α) =∫_0 ^∞ ((arctan(αx))/((x^2 +3)^2 ))dx with α real.


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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
$we have f(α)=∫_0 ^∞ ((arctan(αx))/((x^2 +3)^2 ))dx ⇒ f^′ (α) =∫_0 ^∞ (x/((1+α^2 x^2 )(x^2 +3)^2 ))dx =_(αx =t) ∫_0 ^∞ (t/(α(1+t^2 )((t^2 /α^2 ) +3)^2 ))(dt/α) =(α^4 /α^2 )∫_0 ^∞ ((tdt)/((t^2 +1)(t^2 +3α^2 )^2 )) =α^(2 ) ∫_0 ^∞ ((tdt)/((t^2 +1)(t^2 +3α^2 )^2 )) let decompose F(t) =(t/((t^2 +1)(t^2 +3α^2 )^2 )) F(t)=((at +b)/(t^(2 ) +1)) +((ct+d)/(t^2 +3α^2 )) +((et +f)/((t^2 +3α^2 )^2 )) F(−t)=F(t) ⇒((−at+b)/(t^2 +1))+((−ct+d)/(t^2 +3α^2 )) +((−et+f)/((t^2 +3α^2 )^2 )) =((−at−b)/(t^2 +1))+((−ct−d)/(t^2 +3α^2 )) +((−et−f)/((t^2 +3α^2 )^2 )) ⇒b=d=f=0 ⇒ F(t) =((at)/(t^2 +1)) +((ct)/(t^(2 ) +3α^2 )) +((et)/((t^2 +3α^2 )^2 )) lim_(t→+∞) tF(t) =0 =a+c+e ⇒e =−a−c ⇒ F(t)=((at)/(t^2 +1)) +((ct)/(t^2 +3α^2 ))−(((a+c)t)/((t^2 +3α^2 )^2 )) F(1) =(1/(2(1+3α^2 )^2 )) =(a/2) +(c/(1+3α^2 ))−((a+c)/((1+3α^2 )^2 )) ⇒ (1/2) =(1/2)(1+3α^2 )^2 a +(1+3α^2 )c−a−c ⇒ 1 =(1+3α^2 )^2 a+2(1+3α^2 )c−2a−2c ⇒ {(1+3α^2 )^2 −2}a +{2(1+3α^2 )−2}c =1 ....be continued...$
$w e h a v e f (α) = \int_{0}^{\infty} \frac{a r c t a n (α x)}{{(x^{2} + 3)}^{2}} d x \Rightarrow$ $f^{^{'}} (α) = \int_{0}^{\infty} \frac{x}{(1 + α^{2} x^{2}) {(x^{2} + 3)}^{2}} d x =_{α x = t} \int_{0}^{\infty} \frac{t}{α (1 + t^{2}) {(\frac{t^{2}}{α^{2}} + 3)}^{2}} \frac{d t}{α}$ $= \frac{α^{4}}{α^{2}} \int_{0}^{\infty} \frac{t d t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}} = α^{2} \int_{0}^{\infty} \frac{t d t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}}$ $l e t d e c o m p o s e F (t) = \frac{t}{(t^{2} + 1) {(t^{2} + 3 α^{2})}^{2}}$ $F (t) = \frac{a t + b}{t^{2} + 1} + \frac{c t + d}{t^{2} + 3 α^{2}} + \frac{e t + f}{{(t^{2} + 3 α^{2})}^{2}}$ $F (- t) = F (t) \Rightarrow \frac{- a t + b}{t^{2} + 1} + \frac{- c t + d}{t^{2} + 3 α^{2}} + \frac{- e t + f}{{(t^{2} + 3 α^{2})}^{2}} = \frac{- a t - b}{t^{2} + 1} + \frac{- c t - d}{t^{2} + 3 α^{2}}$ $+ \frac{- e t - f}{{(t^{2} + 3 α^{2})}^{2}} \Rightarrow b = d = f = 0 \Rightarrow$ $F (t) = \frac{a t}{t^{2} + 1} + \frac{c t}{t^{2} + 3 α^{2}} + \frac{e t}{{(t^{2} + 3 α^{2})}^{2}}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + c + e \Rightarrow e = - a - c \Rightarrow$ $F (t) = \frac{a t}{t^{2} + 1} + \frac{c t}{t^{2} + 3 α^{2}} - \frac{(a + c) t}{{(t^{2} + 3 α^{2})}^{2}}$ $F (1) = \frac{1}{2 {(1 + 3 α^{2})}^{2}} = \frac{a}{2} + \frac{c}{1 + 3 α^{2}} - \frac{a + c}{{(1 + 3 α^{2})}^{2}} \Rightarrow$ $\frac{1}{2} = \frac{1}{2} {(1 + 3 α^{2})}^{2} a + (1 + 3 α^{2}) c - a - c \Rightarrow$ $1 = {(1 + 3 α^{2})}^{2} a + 2 (1 + 3 α^{2}) c - 2 a - 2 c \Rightarrow$ ${{(1 + 3 α^{2})}^{2} - 2} a + {2 (1 + 3 α^{2}) - 2} c = 1$ $. . . . b e c o n t i n u e d . . .$
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