let A =∫_0 ^(+∞) (dx/((x^2 −i)^2 )) ( i^2 =−1) 1) calculate A 2) let R =Re(A) and I =Im(A) find the value of R and I .


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Question Number 62141 by maxmathsup by imad last updated on 15/Jun/19

$l e t A = \int_{0}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}} (i^{2} = - 1)$ $1) c a l c u l a t e A$ $2) l e t R = R e (A) a n d I = I m (A)$ $f i n d t h e v a l u e o f R a n d I .$
Commented by maxmathsup by imad last updated on 16/Jun/19
$1) we have 2A =∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 )) let w(z) =(1/((z^2 −i)^2 )) ⇒ w(z) =(1/((z−(√i))^2 (z+(√i))^2 )) =(1/((z−e^((iπ)/4) )^2 (z +e^((iπ)/4) )^2 )) so the poles of w are +^− e^((iπ)/4) (doubles) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,e^((iπ)/4) ) Res(w,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){(z−e^((iπ)/4) )^2 w(z)}^((1)) =lim_(z→e^((iπ)/4) ) { (z+e^((iπ)/4) )^(−2) }^((1)) =lim_(z→e^((iπ)/4) ) −2(z+e^((iπ)/4) )^(−3) =−2 (2 e^((iπ)/4) )^(−3) =−2.2^(−3) .e^((−3iπ)/4) =−(1/4){ cos(((3π)/4))−isin(((3π)/4))} =−(1/4){−cos((π/4))−isin((π/4))} =(1/4)((√2)/2) +i(1/4)((√2)/2) =((√2)/8) +i((√2)/8) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ {((√2)/8) +i((√2)/8)} =((iπ(√2))/4) −((π(√2))/4) =2A ⇒ A =−((π(√2))/8) +((iπ(√2))/8) 2) we have A = ∫_0 ^∞ (dx/((x^2 −i)^2 )) =∫_0 ^∞ (((x^2 +i)^2 )/((x^2 −i)^2 (x^2 +i)^2 )) dx =∫_0 ^∞ ((x^4 +2ix^2 −1)/((x^4 +1)^2 )) dx =∫_0 ^∞ ((x^4 −1 +2ix^2 )/(x^8 +2x^4 +1))dx = ∫_0 ^∞ ((x^4 −1)/(x^8 +2x^4 +1)) dx +i ∫_0 ^∞ ((2x^2 )/(x^8 +2x^4 +1)) dx ⇒ R =∫_0 ^∞ ((x^4 −1)/(x^8 +2x^4 +1)) and I = ∫_0 ^∞ ((2x^2 )/(x^8 +2x^4 +1)) dx ⇒ R =−((π(√2))/8) and I =((π(√2))/8) .$
$1) w e h a v e 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}} l e t w (z) = \frac{1}{{(z^{2} - i)}^{2}} \Rightarrow$ $w (z) = \frac{1}{{(z - \sqrt{i})}^{2} {(z + \sqrt{i})}^{2}} = \frac{1}{{(z - e^{\frac{i π}{4}})}^{2} {(z + e^{\frac{i π}{4}})}^{2}} s o t h e p o l e s o f w a r e \bar{+} e^{\frac{i π}{4}} (d o u b l e s)$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, e^{\frac{i π}{4}})$ $R e s (w, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{4}})}^{2} w (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{4}}} {{(z + e^{\frac{i π}{4}})}^{- 2}}^{(1)} = {l i m}_{z \to e^{\frac{i π}{4}}} - 2 {(z + e^{\frac{i π}{4}})}^{- 3}$ $= - 2 {(2 e^{\frac{i π}{4}})}^{- 3} = - 2 . 2^{- 3} . e^{\frac{- 3 i π}{4}} = - \frac{1}{4} {c o s (\frac{3 π}{4}) - i s i n (\frac{3 π}{4})}$ $= - \frac{1}{4} {- c o s (\frac{π}{4}) - i s i n (\frac{π}{4})} = \frac{1}{4} \frac{\sqrt{2}}{2} + i \frac{1}{4} \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{8} + i \frac{\sqrt{2}}{8} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π {\frac{\sqrt{2}}{8} + i \frac{\sqrt{2}}{8}} = \frac{i π \sqrt{2}}{4} - \frac{π \sqrt{2}}{4} = 2 A \Rightarrow$ $A = - \frac{π \sqrt{2}}{8} + \frac{i π \sqrt{2}}{8}$ $2) w e h a v e A = \int_{0}^{\infty} \frac{d x}{{(x^{2} - i)}^{2}} = \int_{0}^{\infty} \frac{{(x^{2} + i)}^{2}}{{(x^{2} - i)}^{2} {(x^{2} + i)}^{2}} d x$ $= \int_{0}^{\infty} \frac{x^{4} + 2 {i x}^{2} - 1}{{(x^{4} + 1)}^{2}} d x = \int_{0}^{\infty} \frac{x^{4} - 1 + 2 {i x}^{2}}{x^{8} + 2 x^{4} + 1} d x$ $= \int_{0}^{\infty} \frac{x^{4} - 1}{x^{8} + 2 x^{4} + 1} d x + i \int_{0}^{\infty} \frac{2 x^{2}}{x^{8} + 2 x^{4} + 1} d x \Rightarrow$ $R = \int_{0}^{\infty} \frac{x^{4} - 1}{x^{8} + 2 x^{4} + 1} a n d I = \int_{0}^{\infty} \frac{2 x^{2}}{x^{8} + 2 x^{4} + 1} d x \Rightarrow$ $R = - \frac{π \sqrt{2}}{8} a n d I = \frac{π \sqrt{2}}{8} .$
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