calculate ∫_0 ^(+∞) ((3x^2 −2)/((x^2 +1)( x^2 −2i)^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 62805 by mathmax by abdo last updated on 25/Jun/19

$c a l c u l a t e \int_{0}^{+ \infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x$
Commented by mathmax by abdo last updated on 26/Jun/19
$let I =∫_0 ^∞ ((3x^2 −2)/((x^2 +1)(x^2 −2i)^2 )) dx⇒2I =∫_(−∞) ^(+∞) ((3x^2 −2)/((x^2 +1)(x^2 −2i)^2 ))dx let w(z) =((3z^2 −2)/((z^2 +1)(z^2 −2i)^2 )) ⇒w(z) =((3z^2 −2)/((z−i)(z+i)(z−(√(2i)))^2 (z+(√(2i)))^2 )) =((3z^2 −2)/((z−i)(z+i)(z−(√2)e^((iπ)/4) )^2 (z+(√2)e^((iπ)/4) )^2 )) so the poles of w are +^− i and +^− (√2)e^((iπ)/4) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ { Res(w,i)+Res(w,(√2)e^((iπ)/4) )} Res(w,i) =lim_(z→i) (z−i)w(z) =((−5)/(2i(−1−2i)^2 )) =((−5)/(2i(2i+1)^2 )) =((−5)/(2i(−4+4i +1))) =((−5)/(2i(−3+4i))) Res(w,(√2)e^((iπ)/4) ) =lim_(z→(√2)e^((iπ)/4) ) (1/((2−1)!)){ (z−(√2)e^((iπ)/4) )^2 w(z)}^((1)) =lim_(z→(√2)e^((iπ)/4) ) {((3z^2 −2)/((z^2 +1)(z+(√2)e^((iπ)/4) )^2 ))}^((1)) =lim_(z→(√2)e^((iπ)/4) ) {((6z(z^2 +1)(z+(√2)e^((iπ)/4) )^2 −(3z^2 −2){2z(z+(√2)e^((iπ)/4) }^2 +2(z^2 +1)(z+(√2)e^((iπ)/4) ))/((z^2 +1)^2 (z+(√2)e^((iπ)/4) )^4 ))} =lim_(z→(√2)e^((iπ)/4) ) {((6z(z^2 +1)(z+(√2)e^((iπ)/4) )−(3z^2 −2){2z(z+(√2)e^((iπ)/4) )+2(z^2 +1)})/((z^2 +1)^2 (z+(√2)e^((iπ)/4) )^3 ))} ...be continued...$
$l e t I = \int_{0}^{\infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{3 x^{2} - 2}{(x^{2} + 1) {(x^{2} - 2 i)}^{2}} d x l e t$ $w (z) = \frac{3 z^{2} - 2}{(z^{2} + 1) {(z^{2} - 2 i)}^{2}} \Rightarrow w (z) = \frac{3 z^{2} - 2}{(z - i) (z + i) {(z - \sqrt{2 i})}^{2} {(z + \sqrt{2 i})}^{2}}$ $= \frac{3 z^{2} - 2}{(z - i) (z + i) {(z - \sqrt{2} e^{\frac{i π}{4}})}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{2}} s o t h e p o l e s o f w a r e \bar{+} i a n d \bar{+} \sqrt{2} e^{\frac{i π}{4}}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π {R e s (w, i) + R e s (w, \sqrt{2} e^{\frac{i π}{4}})}$ $R e s (w, i) = {l i m}_{z \to i} (z - i) w (z) = \frac{- 5}{2 i {(- 1 - 2 i)}^{2}} = \frac{- 5}{2 i {(2 i + 1)}^{2}} = \frac{- 5}{2 i (- 4 + 4 i + 1)}$ $= \frac{- 5}{2 i (- 3 + 4 i)}$ $R e s (w, \sqrt{2} e^{\frac{i π}{4}}) = {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - \sqrt{2} e^{\frac{i π}{4}})}^{2} w (z)}^{(1)}$ $= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} {\frac{3 z^{2} - 2}{(z^{2} + 1) {(z + \sqrt{2} e^{\frac{i π}{4}})}^{2}}}^{(1)}$ $= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} {\frac{6 z (z^{2} + 1) {(z + \sqrt{2} e^{\frac{i π}{4}})}^{2} - (3 z^{2} - 2) {2 z {(z + \sqrt{2} e^{\frac{i π}{4}}}}^{2} + 2 (z^{2} + 1) (z + \sqrt{2} e^{\frac{i π}{4}})}{{(z^{2} + 1)}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{4}}}$ $= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{4}}} {\frac{6 z (z^{2} + 1) (z + \sqrt{2} e^{\frac{i π}{4}}) - (3 z^{2} - 2) {2 z (z + \sqrt{2} e^{\frac{i π}{4}}) + 2 (z^{2} + 1)}}{{(z^{2} + 1)}^{2} {(z + \sqrt{2} e^{\frac{i π}{4}})}^{3}}}$ $. . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum