find the value of ∫_0 ^∞ (dx/((2x^2 +1)^2 ))


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Question Number 35440 by prof Abdo imad last updated on 19/May/18

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{{(2 x^{2} + 1)}^{2}}$
Commented by prof Abdo imad last updated on 19/May/18
$let put I = ∫_0 ^∞ (dx/((2x^2 +1)^2 )) 2I = ∫_(−∞) ^(+∞) (dx/((2x^2 +1)^2 )) =(1/4) ∫_(−∞) ^(+∞) (dx/((x^2 +(1/2))^2 )) let condider tbe complex function ϕ(z) = (1/((z^2 +(1/2))^2 )) we have ϕ(z) = (1/((z−(i/(√2)))^2 (z +(i/(√2)))^2 )) so the?polrs of ϕ are (i/(√2)) and ((−i)/(√2)) (doubles) and Residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/(√2))) Res(ϕ,(i/(√2))) =lim_(z→(i/(√2))) { (z−(i/(√2)))^2 ϕ(z)}^′ =lim_(z→(i/(√2))) { (z+(i/(√2)))^(−2) }^′ =lim_(z→(i/(√2))) −2(z +(i/(√2)))^(−3) =−2 (((2i)/(√2)))^(−3) =−2((√2) i)^(−3) =2 ((√2))^(−3) .(1/i) = (1/(i(√2))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(i(√2))) = π(√2) 2I =(1/4) π(√2) ⇒ I = ((π(√2))/8) = ((2π)/(8(√2))) ⇒ I = (π/(4(√2))) .$
$l e t p u t I = \int_{0}^{\infty} \frac{d x}{{(2 x^{2} + 1)}^{2}}$ $2 I = \int_{- \infty}^{+ \infty} \frac{d x}{{(2 x^{2} + 1)}^{2}} = \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + \frac{1}{2})}^{2}}$ $l e t c o n d i d e r t b e c o m p l e x f u n c t i o n$ $φ (z) = \frac{1}{{(z^{2} + \frac{1}{2})}^{2}} w e h a v e$ $φ (z) = \frac{1}{{(z - \frac{i}{\sqrt{2}})}^{2} {(z + \frac{i}{\sqrt{2}})}^{2}} s o t h e ? p o l r s o f φ a r e$ $\frac{i}{\sqrt{2}} a n d \frac{- i}{\sqrt{2}} (d o u b l e s) a n d R e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{\sqrt{2}})$ $R e s (φ, \frac{i}{\sqrt{2}}) = {l i m}_{z \to \frac{i}{\sqrt{2}}} {{(z - \frac{i}{\sqrt{2}})}^{2} φ (z)}^{^{'}}$ $= {l i m}_{z \to \frac{i}{\sqrt{2}}} {{(z + \frac{i}{\sqrt{2}})}^{- 2}}^{^{'}}$ $= {l i m}_{z \to \frac{i}{\sqrt{2}}} - 2 {(z + \frac{i}{\sqrt{2}})}^{- 3}$ $= - 2 {(\frac{2 i}{\sqrt{2}})}^{- 3} = - 2 {(\sqrt{2} i)}^{- 3} = 2 {(\sqrt{2})}^{- 3} . \frac{1}{i}$ $= \frac{1}{i \sqrt{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{i \sqrt{2}} = π \sqrt{2}$ $2 I = \frac{1}{4} π \sqrt{2} \Rightarrow I = \frac{π \sqrt{2}}{8} = \frac{2 π}{8 \sqrt{2}} \Rightarrow$ $I = \frac{π}{4 \sqrt{2}} .$
Commented by prof Abdo imad last updated on 19/May/18

$I = \int_{0}^{\infty} \frac{d x}{{(2 x^{2} + 1)}^{2}} l e t u s e t h e c h a n g e m e n t$ $x = \frac{1}{\sqrt{2}} t a n t \Rightarrow I = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{1 + {t a n}^{2} t}{{(1 + {t a n}^{2})}^{2}} d t$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {t a n}^{2} t} = \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} (\frac{1 + c o s (2 t)}{2}) d t$ $= \frac{1}{2 \sqrt{2}} \frac{π}{2} + \frac{1}{4 \sqrt{2}} {[s i n (2 t)]}_{0}^{\frac{π}{2}}$ $★ I = \frac{π}{4 \sqrt{2}} ★$
	Answered by ajfour last updated on 19/May/18

	$I = \int_{0}^{\infty} \frac{d x}{{(2 x^{2} + 1)}^{2}}$ $l e t x = \frac{1}{\sqrt{2}} \tan θ \Rightarrow d x = \frac{1}{\sqrt{2}} \sec^{2} θ d θ$ $\Rightarrow I = \frac{1}{\sqrt{2}} \int \cos^{2} θ d θ$ $I = \frac{1}{2 \sqrt{2}} \int_{0}^{π / 2} (1 + \cos 2 θ) d θ$ $= \frac{1}{2 \sqrt{2}} (θ + \frac{\sin 2 θ}{2}) ∣_{0}^{π / 2} = \frac{π}{4 \sqrt{2}} .$
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