calculate ∫_(−∞) ^(+∞) ((arctan(x^2 −3))/((x^2 +x+1)^2 ))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 78625 by mathmax by abdo last updated on 19/Jan/20

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{a r c t a n (x^{2} - 3)}{{(x^{2} + x + 1)}^{2}} d x$
Commented by mathmax by abdo last updated on 19/Jan/20
$let W(z)=((arctan(z^2 −3))/((z^2 +z+1)^2 )) poles of W? z^2 +z+1 =0 →Δ =1−4 =−3 ⇒z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) z_2 =e^(−((2iπ)/3)) ⇒W(z) =((arctan(z^2 −3))/((z−e^((i2π)/3) )^2 (z−e^(−((2iπ)/3)) )^2 )) ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,e^(i((2π)/3)) ) Res(W,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){(z−e^((i2π)/3) )^2 W(z)}^((1)) =lim_(z→e^((i2π)/3) ) { ((arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) ((((2z)/(1+(z^2 −3)^2 ))×(z−e^(−((2iπ)/3)) )^2 −2(z−e^(−((2iπ)/3)) )arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^4 )) =lim_(z→e^((i2π)/3) ) ((((2z(z−e^(−i((2π)/3)) ))/(1+(z^2 −3)^2 ))−2arctan(z^2 −3))/((z−e^(−((2iπ)/3)) )^3 )) =((((2e^(i((2π)/3)) (2isin(((2π)/3)))/(1+(e^((i4π)/3) −3)^2 ))−2arctan(e^((i4π)/3) −3))/((2isin(((2π)/3)))^3 )) =((((2i(√3)e^(i((2π)/3)) )/(1+(e^((i4π)/3) −3)^2 ))−2arctan(e^((i4π)/3) −3))/((i(√3))^3 )) ....be continued...$
$l e t W (z) = \frac{a r c t a n (z^{2} - 3)}{{(z^{2} + z + 1)}^{2}} p o l e s o f W ?$ $z^{2} + z + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}}$ $z_{2} = e^{- \frac{2 i π}{3}} \Rightarrow W (z) = \frac{a r c t a n (z^{2} - 3)}{{(z - e^{\frac{i 2 π}{3}})}^{2} {(z - e^{- \frac{2 i π}{3}})}^{2}}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, e^{i \frac{2 π}{3}})$ $R e s (W, e^{\frac{i 2 π}{3}}) = {l i m}_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i 2 π}{3}})}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} {\frac{a r c t a n (z^{2} - 3)}{{(z - e^{- \frac{2 i π}{3}})}^{2}}}^{(1)}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} \frac{\frac{2 z}{1 + {(z^{2} - 3)}^{2}} \times {(z - e^{- \frac{2 i π}{3}})}^{2} - 2 (z - e^{- \frac{2 i π}{3}}) a r c t a n (z^{2} - 3)}{{(z - e^{- \frac{2 i π}{3}})}^{4}}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} \frac{\frac{2 z (z - e^{- i \frac{2 π}{3}})}{1 + {(z^{2} - 3)}^{2}} - 2 a r c t a n (z^{2} - 3)}{{(z - e^{- \frac{2 i π}{3}})}^{3}}$ $= \frac{\frac{2 e^{i \frac{2 π}{3}} (2 i s i n (\frac{2 π}{3})}{1 + {(e^{\frac{i 4 π}{3}} - 3)}^{2}} - 2 a r c t a n (e^{\frac{i 4 π}{3}} - 3)}{{(2 i s i n (\frac{2 π}{3}))}^{3}}$ $= \frac{\frac{2 i \sqrt{3} e^{i \frac{2 π}{3}}}{1 + {(e^{\frac{i 4 π}{3}} - 3)}^{2}} - 2 a r c t a n (e^{\frac{i 4 π}{3}} - 3)}{{(i \sqrt{3})}^{3}} . . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum