let give I = ∫_0 ^∞ (dx/((x^2 +i)^2 )) 1) extract Re(I) and Im(I) 2) find the value of I 3) calculate Re(I) and Im(I) .


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Question Number 36167 by abdo mathsup 649 cc last updated on 29/May/18

$l e t g i v e I = \int_{0}^{\infty} \frac{d x}{{(x^{2} + i)}^{2}}$ $1) e x t r a c t R e (I) a n d I m (I)$ $2) f i n d t h e v a l u e o f I$ $3) c a l c u l a t e R e (I) a n d I m (I) .$
Commented by maxmathsup by imad last updated on 22/Aug/18

$1) w e h a v e I = \int_{0}^{\infty} \frac{{(x^{2} - i)}^{2}}{{(x^{2} + i)}^{2} {(x^{2} - i)}^{2}} d x$ $= \int_{0}^{\infty} \frac{x^{4} - 2 {i x}^{2} - 1}{{(x^{4} + 1)}^{2}} d x = \int_{0}^{\infty} \frac{x^{4} - 1}{{(x^{4} + 1)}^{2}} d x + i \int_{0}^{\infty} \frac{- 2 x^{2}}{{(x^{4} + 1)}^{2}} d x \Rightarrow$ $R e (I) = \int_{0}^{\infty} \frac{x^{4} - 1}{{(x^{4} + 1)}^{2}} d x a n d I m (I) = \int_{0}^{\infty} \frac{- 2 x^{2}}{{(x^{4} + 1)}^{2}} d x$
Commented by maxmathsup by imad last updated on 22/Aug/18
$2) we have 2I =∫_(−∞) ^(+∞) (dx/((x^2 +i)^2 )) let consider the complex function ϕ(z) = (1/((z^2 +i)^2 )) ⇒ϕ(z) = (1/((z^2 −((√(−i)))^2 )^2 )) =(1/((z−(√(−i)))^2 (z+(√(−i)))^2 )) = (1/((z−e^(−((iπ)/4)) )^2 (z +e^(−((iπ)/4)) )^2 )) so the poles of ϕ are +^− e^(−((iπ)/4)) (doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,−e^(−((iπ)/4)) ) but Res(ϕ,e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (1/((2−1)!)){(z+e^(−((iπ)/4)) )^2 ϕ(z)}^((1)) =lim_(z→−e^(−((iπ)/4)) ) {(z−e^(−((iπ)/4)) )^(−2) }^((1)) =lim_(z→−e^(−((iπ)/4)) ) −2 (z−e^(−((iπ)/4)) )^(−3) =−2(−2 e^(−((iπ)/4)) )^(−3) =((−2)/((−2)^3 )) e^(i((3π)/4)) =(1/4) e^((i3π)/4) =(1/4){cos(((3π)/4)) +isin(((3π)/4))} =(1/4){−(1/(√2)) +(i/(√2))} ⇒∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/4){−(1/(√2)) +(i/(√2))}=((iπ)/2){−(1/(√2)) +(i/(√2))} =2I ⇒ I =((iπ)/4){−(1/(√2)) +(i/(√2))} =−(π/(4(√2))) −(π/(4(√2))) i 3) we have I =Re(I) +iIm(I) ⇒Re(I) =−(π/(4(√2))) and Im(I) =−(π/(4(√2)))$
$2) w e h a v e 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + i)}^{2}} l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{1}{{(z^{2} + i)}^{2}} \Rightarrow φ (z) = \frac{1}{{(z^{2} - {(\sqrt{- i})}^{2})}^{2}} = \frac{1}{{(z - \sqrt{- i})}^{2} {(z + \sqrt{- i})}^{2}}$ $= \frac{1}{{(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}} s o t h e p o l e s o f φ a r e \bar{+} e^{- \frac{i π}{4}} (d o u b l e s) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, - e^{- \frac{i π}{4}}) b u t$ $R e s (φ, e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z + e^{- \frac{i π}{4}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to - e^{- \frac{i π}{4}}} {{(z - e^{- \frac{i π}{4}})}^{- 2}}^{(1)} = {l i m}_{z \to - e^{- \frac{i π}{4}}} - 2 {(z - e^{- \frac{i π}{4}})}^{- 3}$ $= - 2 {(- 2 e^{- \frac{i π}{4}})}^{- 3} = \frac{- 2}{{(- 2)}^{3}} e^{i \frac{3 π}{4}} = \frac{1}{4} e^{\frac{i 3 π}{4}} = \frac{1}{4} {c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4})}$ $= \frac{1}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = \frac{i π}{2} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = 2 I \Rightarrow$ $I = \frac{i π}{4} {- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = - \frac{π}{4 \sqrt{2}} - \frac{π}{4 \sqrt{2}} i$ $3) w e h a v e I = R e (I) + i I m (I) \Rightarrow R e (I) = - \frac{π}{4 \sqrt{2}} a n d I m (I) = - \frac{π}{4 \sqrt{2}}$
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