calculate ∫_0 ^(+∞) (dx/((x^2 −i)^2 ))


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Question Number 49967 by maxmathsup by imad last updated on 12/Dec/18

$c a l c u l a t e \int_{0}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}}$
Commented by Abdo msup. last updated on 13/Dec/18
$let A =∫_0 ^∞ (dx/((x^2 −i)^2 )) ⇒ A =(1/2)∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 )) let consider the complex function ϕ(z)=(1/((z^2 −i)^2 )) we have ϕ(z) = (1/((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 )) so the poles of ϕ are +^− e^((iπ)/4) (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/4) ) but Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)) { (z−e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/4) ) { (1/((z+e^((iπ)/4) )^2 ))}^((1)) =lim_(z→e^((iπ)/4) ) −2(((z+e^((iπ)/4) ))/((z+e^((iπ)/4) )^4 )) =lim_(z→ e^((iπ)/4) ) ((−2)/((z+e^((iπ)/4) )^3 )) =((−2)/((2e^((iπ)/4) )^3 )) =((−1)/(4 e^(i((3π)/4)) )) =((−1)/(4 e^(i(π−(π/4))) )) =(1/4) e^((iπ)/4) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(1/4) e^((iπ)/4) =((iπ)/2)( (1/(√2)) +(i/(√2))) =((iπ)/(2(√2))) −(π/(2(√2))) = 2A . remark we have ∫_(−∞) ^∞ (dx/((x^2 −i)^2 )) =∫_(−∞) ^∞ (((x^2 +i)^2 )/((x^4 +1)^2 )) dx =∫_(−∞) ^∞ ((x^4 +2ix^2 −1)/((x^4 +1)^2 )) dx =−(π/(2(√2))) +i(π/(2(√2))) ∫_(−∞) ^∞ ((x^4 −1)/((x^4 +1)^2 )) dx =−(π/(2(√2))) and ∫_(−∞) ^∞ ((2x^2 )/((x^4 +1)^2 )) dx =(π/(2(√2))) .$
$l e t A = \int_{0}^{\infty} \frac{d x}{{(x^{2} - i)}^{2}} \Rightarrow A = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - i)}^{2}} l e t c o n s i d e r$ $t h e c o m p l e x f u n c t i o n φ (z) = \frac{1}{{(z^{2} - i)}^{2}} w e h a v e$ $φ (z) = \frac{1}{{(z - e^{\frac{i π}{4}})}^{2} {(z + e^{\frac{i π}{4}})}^{2}} s o t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}}$ $(d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i π}{4}}) b u t$ $R e s (φ, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{4}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{4}}} {\frac{1}{{(z + e^{\frac{i π}{4}})}^{2}}}^{(1)} = {l i m}_{z \to e^{\frac{i π}{4}}} - 2 \frac{(z + e^{\frac{i π}{4}})}{{(z + e^{\frac{i π}{4}})}^{4}}$ $= {l i m}_{z \to e^{\frac{i π}{4}}} \frac{- 2}{{(z + e^{\frac{i π}{4}})}^{3}} = \frac{- 2}{{(2 e^{\frac{i π}{4}})}^{3}} = \frac{- 1}{4 e^{i \frac{3 π}{4}}} = \frac{- 1}{4 e^{i (π - \frac{π}{4})}} = \frac{1}{4} e^{\frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{4} e^{\frac{i π}{4}} = \frac{i π}{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = \frac{i π}{2 \sqrt{2}} - \frac{π}{2 \sqrt{2}} = 2 A .$ $r e m a r k w e h a v e \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} - i)}^{2}}$ $= \int_{- \infty}^{\infty} \frac{{(x^{2} + i)}^{2}}{{(x^{4} + 1)}^{2}} d x = \int_{- \infty}^{\infty} \frac{x^{4} + 2 {i x}^{2} - 1}{{(x^{4} + 1)}^{2}} d x = - \frac{π}{2 \sqrt{2}} + i \frac{π}{2 \sqrt{2}}$ $\int_{- \infty}^{\infty} \frac{x^{4} - 1}{{(x^{4} + 1)}^{2}} d x = - \frac{π}{2 \sqrt{2}} a n d \int_{- \infty}^{\infty} \frac{2 x^{2}}{{(x^{4} + 1)}^{2}} d x = \frac{π}{2 \sqrt{2}} .$
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