calculate ∫_(−∞) ^(+∞) ((arctan(2x+1))/((x^2 +3)^2 ))dx


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Question Number 77995 by mathmax by abdo last updated on 12/Jan/20

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 x + 1)}{{(x^{2} + 3)}^{2}} d x$
Commented by msup trace by abdo last updated on 13/Jan/20
$I =∫_(−∞) ^(+∞) ((arctan(2x+1))/((x^2 +3)^2 ))dx ⇒ I=_(x=(√3)t) ∫_(−∞) ^(+∞) ((arctan(2(√3)t+1))/(9(t^2 +1)^2 ))(√3)dt =((√3)/9) ∫_(−∞) ^(+∞) ((arctan(2(√3)t+1))/((t^2 +1)^2 ))dt let ϕ(z)=((arctan(2(√3)z+1))/((z^2 +1)^2 )) ⇒ ϕ(z)=((arctan(2(√3)z +1))/((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {((arctan(2(√3)z +1))/((z+i)^2 ))}^((1)) =lim_(z→i) ((((2(√3))/(1+(2(√3)z+1)^2 ))(z+i)^2 −2(z+i)arctan(2(√3)z +1))/((z+i)^4 )) =lim_(z→i) ((((2(√3)(z+i))/(1+(2(√3)z +1)^2 ))−2arctan(2(√3)z+1))/((z+i)^3 )) =((((4i(√3))/(1+(2(√3)i +1)^2 ))−2arctan(2(√3)i+1))/(−8i)) =−((√3)/2)×(1/(1−12+4(√3)i +1))+(1/(4i))arctan(2(√3)i+1) =((−(√3))/(2(−10+4(√3)i))) +(1/(4i))arctan(2(√3)i+1) =((√3)/(4(5−2(√3)i)))−(i/4) arctan(2(√3)i+1)$
$I = \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 x + 1)}{{(x^{2} + 3)}^{2}} d x \Rightarrow$ $I =_{x = \sqrt{3} t} \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 \sqrt{3} t + 1)}{9 {(t^{2} + 1)}^{2}} \sqrt{3} d t$ $= \frac{\sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{a r c t a n (2 \sqrt{3} t + 1)}{{(t^{2} + 1)}^{2}} d t$ $l e t φ (z) = \frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z^{2} + 1)}^{2}} \Rightarrow$ $φ (z) = \frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z - i)}^{2} {(z + i)}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{\frac{2 \sqrt{3}}{1 + {(2 \sqrt{3} z + 1)}^{2}} {(z + i)}^{2} - 2 (z + i) a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{\frac{2 \sqrt{3} (z + i)}{1 + {(2 \sqrt{3} z + 1)}^{2}} - 2 a r c t a n (2 \sqrt{3} z + 1)}{{(z + i)}^{3}}$ $= \frac{\frac{4 i \sqrt{3}}{1 + {(2 \sqrt{3} i + 1)}^{2}} - 2 a r c t a n (2 \sqrt{3} i + 1)}{- 8 i}$ $= - \frac{\sqrt{3}}{2} \times \frac{1}{1 - 12 + 4 \sqrt{3} i + 1} + \frac{1}{4 i} a r c t a n (2 \sqrt{3} i + 1)$ $= \frac{- \sqrt{3}}{2 (- 10 + 4 \sqrt{3} i)} + \frac{1}{4 i} a r c t a n (2 \sqrt{3} i + 1)$ $= \frac{\sqrt{3}}{4 (5 - 2 \sqrt{3} i)} - \frac{i}{4} a r c t a n (2 \sqrt{3} i + 1)$
Commented by msup trace by abdo last updated on 13/Jan/20

$w e h a v e a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) \Rightarrow$ $a r c t a n (2 \sqrt{3} i + 1) = \frac{1}{2 i} l n (\frac{1 + i (2 \sqrt{3} i + 1)}{1 - i (2 \sqrt{3} i + 1)})$ $= \frac{1}{2 i} l n (\frac{1 - 2 \sqrt{3} + i}{1 + 2 \sqrt{3} - i}) a l s o$ $1 - 2 \sqrt{3} + i = \sqrt{{(1 - 2 \sqrt{3})}^{2} + 1} e^{i a r c t a n (\frac{1}{1 - 2 \sqrt{3}})}$ $1 + 2 \sqrt{3} - i = \sqrt{{(1 + 2 \sqrt{3})}^{2} + 1} e^{- i a r c t a n (\frac{1}{1 + 2 \sqrt{3}})}$ $\Rightarrow l n (\frac{. . .}{. . .}) = \frac{1}{2} l n ({(1 - 2 \sqrt{3})}^{2} + 1) + i a r c t a n (\frac{1}{1 - 2 \sqrt{3}})$ $- \frac{1}{2} l n ({(1 + 2 \sqrt{3})}^{2} + 1) + i a r c t a n (\frac{1}{1 + 2 \sqrt{3}})$
Commented by mathmax by abdo last updated on 13/Jan/20

$\Rightarrow a r c t a n (2 \sqrt{3} i + 1) = \frac{1}{4 i} l n (\frac{1 - 4 \sqrt{3} + 12 + 1}{1 + 4 \sqrt{3} + 12 + 1})$ $+ \frac{1}{2} {a r c t a n (\frac{1}{1 - 2 \sqrt{3}}) + a r c t a n (\frac{1}{1 + 2 \sqrt{3}})}$ $= \frac{1}{4 i} l n (\frac{7 - 2 \sqrt{3}}{7 + 2 \sqrt{3}}) + \frac{1}{2} {\frac{π}{2} - a r c t a n (2 \sqrt{3} + 1) - (\frac{π}{2} - a r c t a n (2 \sqrt{3} - 1)}$ $= \frac{1}{4 i} l n (\frac{7 - 2 \sqrt{3}}{7 + 2 \sqrt{3}}) + \frac{1}{2} (a r c t a n (2 \sqrt{3} - 1) - a r c t a n (2 \sqrt{3} + 1))$
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