find ∫_0 ^∞ ((arctan(3x))/(x^2 +x+1))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 87527 by mathmax by abdo last updated on 04/Apr/20

$f i n d \int_{0}^{\infty} \frac{a r c t a n (3 x)}{x^{2} + x + 1} d x$
Commented by mathmax by abdo last updated on 05/Apr/20
$let f(t) =∫_0 ^∞ ((arctan(tx))/(x^2 +x+1))dx with t>0 we have f^′ (t) =∫_0 ^∞ (x/((1+t^2 x^2 )(x^2 +x+1)))dx =_(tx=u) ∫_0 ^∞ (u/(t(1+u^2 )((u^2 /t^2 )+(u/t)+1)))(du/t) =∫_0 ^∞ ((udu)/((1+u^2 )(u^2 +tu +t^2 ))) u^2 +tu +t^2 →Δ=t^2 −4t^2 =−3t^2 <0 ⇒ F(u) =(u/((u^2 +1)(u^2 +tu +t^2 ))) =((au+b)/(u^2 +1)) +((cu +d)/(u^2 +tu +t^2 )) lim_(u→+∞) uF(u) =0 =a+c ⇒c=−a ⇒ F(u) =((au+b)/(u^2 +1)) +((−au +d)/(u^2 +tu +t^2 )) F(0) =0 =b+d ⇒d=−b ⇒F(u) =((au+b)/(u^2 +1))−((au+b)/(u^2 +tu +t^2 )) F(1) =(1/(2(t^2 +t+1))) =((a+b)/2)−((a+b)/(t^2 +t+1)) ⇒ (1/2) =((a+b)/2)(t^2 +t+1)−(a+b) ⇒1 =(a+b)(t^2 +t+1)−2a−2b ⇒ (t^2 +t−1)a +(t^2 +t−1)b =1 F(−1)=((−1)/(2(t^2 −t+1))) =((−a+b)/2) +((a+b)/(t^2 −t +1)) ⇒ −1 =(t^2 −t+1)(−a+b) +2(a+b) ⇒ 1 =(t^2 −t+1)(a−b) −2a−2b ⇒ 1 =(t^2 −t−1)a −(t^2 −t−1)b we get the system { (((t^2 +t−1)a +(t^2 +t−1)b =1)),(((t^2 −t−1)a−(t^2 −t−1)b =1)) :} Δ =−(t^2 +t−1)(t^2 −t−1)−(t^2 −t−1)(t^2 +t−1) =−2((t^2 −1)^2 −t^2 ) =−2(t^4 −2t^2 +1−t^2 )=−2(t^4 −3t^2 +1) Δ_a = determinant (((1 t^2 +t−1)),((1 −t^2 +t+1)))=−t^2 +t+1−t^2 −t +1 =−2t^2 +2 Δ_b = determinant (((t^2 +t−1 1)),((t^2 −t−1 1)))=t^2 +t−1−t^2 +t+1 =2t ⇒ a =((−2t^2 +2)/(−2(t^4 −3t^2 +1))) =((t^2 −1)/(t^4 −3t^2 +1)) b =((2t)/(−2(t^4 −3t^2 +1))) =((−t)/(t^4 −3t^2 +1)) ....be continued...$
$l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t x)}{x^{2} + x + 1} d x w i t h t > 0$ $w e h a v e f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(1 + t^{2} x^{2}) (x^{2} + x + 1)} d x$ $=_{t x = u} \int_{0}^{\infty} \frac{u}{t (1 + u^{2}) (\frac{u^{2}}{t^{2}} + \frac{u}{t} + 1)} \frac{d u}{t} = \int_{0}^{\infty} \frac{u d u}{(1 + u^{2}) (u^{2} + t u + t^{2})}$ $u^{2} + t u + t^{2} \to Δ = t^{2} - 4 t^{2} = - 3 t^{2} < 0 \Rightarrow$ $F (u) = \frac{u}{(u^{2} + 1) (u^{2} + t u + t^{2})} = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + t u + t^{2}}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (u) = \frac{a u + b}{u^{2} + 1} + \frac{- a u + d}{u^{2} + t u + t^{2}}$ $F (0) = 0 = b + d \Rightarrow d = - b \Rightarrow F (u) = \frac{a u + b}{u^{2} + 1} - \frac{a u + b}{u^{2} + t u + t^{2}}$ $F (1) = \frac{1}{2 (t^{2} + t + 1)} = \frac{a + b}{2} - \frac{a + b}{t^{2} + t + 1} \Rightarrow$ $\frac{1}{2} = \frac{a + b}{2} (t^{2} + t + 1) - (a + b) \Rightarrow 1 = (a + b) (t^{2} + t + 1) - 2 a - 2 b \Rightarrow$ $(t^{2} + t - 1) a + (t^{2} + t - 1) b = 1$ $F (- 1) = \frac{- 1}{2 (t^{2} - t + 1)} = \frac{- a + b}{2} + \frac{a + b}{t^{2} - t + 1} \Rightarrow$ $- 1 = (t^{2} - t + 1) (- a + b) + 2 (a + b) \Rightarrow$ $1 = (t^{2} - t + 1) (a - b) - 2 a - 2 b \Rightarrow$ $1 = (t^{2} - t - 1) a - (t^{2} - t - 1) b w e g e t t h e s y s t e m$ ${\begin{cases} (t^{2} + t - 1) a + (t^{2} + t - 1) b = 1 \\ (t^{2} - t - 1) a - (t^{2} - t - 1) b = 1 \end{cases}$ $Δ = - (t^{2} + t - 1) (t^{2} - t - 1) - (t^{2} - t - 1) (t^{2} + t - 1)$ $= - 2 ({(t^{2} - 1)}^{2} - t^{2}) = - 2 (t^{4} - 2 t^{2} + 1 - t^{2}) = - 2 (t^{4} - 3 t^{2} + 1)$ $Δ_{a} = \| \begin{matrix} 1 t^{2} + t - 1 \\ 1 - t^{2} + t + 1 \end{matrix} \| = - t^{2} + t + 1 - t^{2} - t + 1 = - 2 t^{2} + 2$ $Δ_{b} = \| \begin{matrix} t^{2} + t - 1 1 \\ t^{2} - t - 1 1 \end{matrix} \| = t^{2} + t - 1 - t^{2} + t + 1 = 2 t \Rightarrow$ $a = \frac{- 2 t^{2} + 2}{- 2 (t^{4} - 3 t^{2} + 1)} = \frac{t^{2} - 1}{t^{4} - 3 t^{2} + 1}$ $b = \frac{2 t}{- 2 (t^{4} - 3 t^{2} + 1)} = \frac{- t}{t^{4} - 3 t^{2} + 1} . . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum