find tbe value of ∫_(−∞) ^(+∞) ((x−3)/((x^2 +1)(x^2 −x +2)^2 )) dx


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Question Number 57231 by maxmathsup by imad last updated on 31/Mar/19

$f i n d t b e v a l u e o f \int_{- \infty}^{+ \infty} \frac{x - 3}{(x^{2} + 1) {(x^{2} - x + 2)}^{2}} d x$
Commented by maxmathsup by imad last updated on 02/Apr/19
$residus method let ϕ(z)=((z−3)/((z^2 +1)(z^2 −z +2)^2 )) poles of ϕ? z^2 −z +2 =0 →Δ =1−4(2)=−7 =(i(√7))^2 ⇒z_1 =((1+i(√7))/2) and z_2 =((1−i(√7))/2) ⇒ϕ(z) =((z−3)/((z−i)(z+i)(z−z_1 )^2 (z−z_2 )^2 )) so the poles of ϕ are +^− i(simples) and ((1+^− (√7))/2)(doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i) +Res(ϕ,z_1 )} Res(ϕ,i) =lim_(z→i) (z−i)ϕ(z) =((i−3)/((2i)(i−z_1 )^2 (i−z_1 ^− )^2 )) =((i−3)/((2i)(−1−i(z_1 +z_1 ^− ))^2 )) =((i−3)/((2i)(1+i)^2 )) =((i−3)/((2i)(1+2i−1))) =((i−3)/(−4)) =((−i+3)/4) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {((z−3)/((z^2 +1)(z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) (((z^2 +1)(z−z_2 )^2 −(z−3){2z(z−z_2 )^2 +2(z−z_2 )(z^2 +1)})/((z^2 +1)^2 (z−z_2 )^4 )) =lim_(z→z_1 ) (((z^2 +1)(z−z_2 )−(z−3){ 2z(z−z_2 )+2z^2 +2})/((z^2 +1)^2 (z−z_2 )^3 )) =(((z_1 ^2 +1)(z_1 −z_2 ) −(z_1 −3){2z_1 (z_1 −z_2 )+2z_1 ^2 +2})/((z_1 ^2 +1)(z_1 −z_2 )^2 )) =(((1+z_1 ^2 )(i(√7)) −(z_1 −3){2z_1 (i(√7))+2z_1 ^2 +2})/((i(√7))^2 (1+z_1 ^2 ))) 1+z_1 ^2 =1+(1/4)(1+i(√7))^2 =1+(1/4)(1+2i(√7) −7) =1+(1/4)(−6+2i(√7)) =((−2+2i(√7))/4) =((−1+i(√7))/2) ....be continued...$
$r e s i d u s m e t h o d l e t φ (z) = \frac{z - 3}{(z^{2} + 1) {(z^{2} - z + 2)}^{2}} p o l e s o f φ ?$ $z^{2} - z + 2 = 0 \to Δ = 1 - 4 (2) = - 7 = {(i \sqrt{7})}^{2} \Rightarrow z_{1} = \frac{1 + i \sqrt{7}}{2}$ $a n d z_{2} = \frac{1 - i \sqrt{7}}{2} \Rightarrow φ (z) = \frac{z - 3}{(z - i) (z + i) {(z - z_{1})}^{2} {(z - z_{2})}^{2}} s o t h e p o l e s o f φ a r e$ $\bar{+} i (s i m p l e s) a n d \frac{1 \bar{+} \sqrt{7}}{2} (d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, z_{1})}$ $R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{i - 3}{(2 i) {(i - z_{1})}^{2} {(i - {\bar{z}}_{1})}^{2}}$ $= \frac{i - 3}{(2 i) {(- 1 - i (z_{1} + {\bar{z}}_{1}))}^{2}} = \frac{i - 3}{(2 i) {(1 + i)}^{2}} = \frac{i - 3}{(2 i) (1 + 2 i - 1)} = \frac{i - 3}{- 4} = \frac{- i + 3}{4}$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{z - 3}{(z^{2} + 1) {(z - z_{2})}^{2}}}^{(1)}$ $= {l i m}_{z \to z_{1}} \frac{(z^{2} + 1) {(z - z_{2})}^{2} - (z - 3) {2 z {(z - z_{2})}^{2} + 2 (z - z_{2}) (z^{2} + 1)}}{{(z^{2} + 1)}^{2} {(z - z_{2})}^{4}}$ $= {l i m}_{z \to z_{1}} \frac{(z^{2} + 1) (z - z_{2}) - (z - 3) {2 z (z - z_{2}) + 2 z^{2} + 2}}{{(z^{2} + 1)}^{2} {(z - z_{2})}^{3}}$ $= \frac{(z_{1}^{2} + 1) (z_{1} - z_{2}) - (z_{1} - 3) {2 z_{1} (z_{1} - z_{2}) + 2 z_{1}^{2} + 2}}{(z_{1}^{2} + 1) {(z_{1} - z_{2})}^{2}}$ $= \frac{(1 + z_{1}^{2}) (i \sqrt{7}) - (z_{1} - 3) {2 z_{1} (i \sqrt{7}) + 2 z_{1}^{2} + 2}}{{(i \sqrt{7})}^{2} (1 + z_{1}^{2})}$ $1 + z_{1}^{2} = 1 + \frac{1}{4} {(1 + i \sqrt{7})}^{2} = 1 + \frac{1}{4} (1 + 2 i \sqrt{7} - 7) = 1 + \frac{1}{4} (- 6 + 2 i \sqrt{7})$ $= \frac{- 2 + 2 i \sqrt{7}}{4} = \frac{- 1 + i \sqrt{7}}{2} . . . . b e c o n t i n u e d . . .$
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