find U_n =∫_0 ^∞ ((cos(nx))/((3+nx^2 )^2 ))dx with n integr


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Question Number 72016 by mathmax by abdo last updated on 23/Oct/19

$f i n d U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{(3 + {n x}^{2})}^{2}} d x w i t h n i n t e g r$
Commented by mathmax by abdo last updated on 01/Nov/19
$we have U_n =∫_0 ^∞ ((cos(nx))/((nx^2 +3)^2 ))dx =_((√n)x=(√3)u) ∫_0 ^∞ ((cos(n(((√3)u)/(√n))))/((3u^2 +3)^2 )) (((√3)du)/(√n)) =((√3)/(9(√n))) ∫_0 ^∞ ((cos((√(3n))u))/((u^2 +1)^2 ))du =((√3)/(18(√n))) ∫_(−∞) ^(+∞) ((cos((√(3n))u))/((u^2 +1)^2 ))du =((√3)/(18(√n))) Re( ∫_(−∞) ^(+∞) (e^(i(√(3n))u) /((u^2 +1)^2 )))du let W(z)=(e^(i(√(3n))z) /((z^2 +1)^2 )) ⇒ W(z)=(e^(i(√(3n))z) /((z−i)^2 (z+i)^2 )) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,i) Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(i(√(3n))z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((i(√(3n))e^(i(√(3n))z) (z+i)^2 −2(z+i)e^(i(√(3n))z) )/((z+i)^4 )) =lim_(z→i) (((i(√(3n))(z+i)−2)e^(i(√(3n))z) )/((z+i)^3 )) =(((−2 (√(3n))−2)e^(−(√(3n))) )/((2i)^3 )) =(((−2(√(3n))−2)e^(−(√(3n))) )/(−8i)) =(((1+(√(3n)))e^(−(√(3n))) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((1+(√(3n)))e^(−(√(3n))) )/(4i)) =(π/2)(1+(√(3n)))e^(−(√(3n))) ⇒U_n =((√3)/(18(√n)))×(π/2)(1+(√(3n)))e^(−(√(3n))) =((π(√3))/(36(√n)))(1+(√(3n)))e^(−(√(3n)))$
$w e h a v e U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{({n x}^{2} + 3)}^{2}} d x =_{\sqrt{n} x = \sqrt{3} u} \int_{0}^{\infty} \frac{c o s (n \frac{\sqrt{3} u}{\sqrt{n}})}{{(3 u^{2} + 3)}^{2}} \frac{\sqrt{3} d u}{\sqrt{n}}$ $= \frac{\sqrt{3}}{9 \sqrt{n}} \int_{0}^{\infty} \frac{c o s (\sqrt{3 n} u)}{{(u^{2} + 1)}^{2}} d u = \frac{\sqrt{3}}{18 \sqrt{n}} \int_{- \infty}^{+ \infty} \frac{c o s (\sqrt{3 n} u)}{{(u^{2} + 1)}^{2}} d u$ $= \frac{\sqrt{3}}{18 \sqrt{n}} R e (\int_{- \infty}^{+ \infty} \frac{e^{i \sqrt{3 n} u}}{{(u^{2} + 1)}^{2}}) d u l e t W (z) = \frac{e^{i \sqrt{3 n} z}}{{(z^{2} + 1)}^{2}} \Rightarrow$ $W (z) = \frac{e^{i \sqrt{3 n} z}}{{(z - i)}^{2} {(z + i)}^{2}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i \sqrt{3 n} z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{i \sqrt{3 n} e^{i \sqrt{3 n} z} {(z + i)}^{2} - 2 (z + i) e^{i \sqrt{3 n} z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(i \sqrt{3 n} (z + i) - 2) e^{i \sqrt{3 n} z}}{{(z + i)}^{3}} = \frac{(- 2 \sqrt{3 n} - 2) e^{- \sqrt{3 n}}}{{(2 i)}^{3}}$ $= \frac{(- 2 \sqrt{3 n} - 2) e^{- \sqrt{3 n}}}{- 8 i} = \frac{(1 + \sqrt{3 n}) e^{- \sqrt{3 n}}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times \frac{(1 + \sqrt{3 n}) e^{- \sqrt{3 n}}}{4 i} = \frac{π}{2} (1 + \sqrt{3 n}) e^{- \sqrt{3 n}}$ $\Rightarrow U_{n} = \frac{\sqrt{3}}{18 \sqrt{n}} \times \frac{π}{2} (1 + \sqrt{3 n}) e^{- \sqrt{3 n}}$ $= \frac{π \sqrt{3}}{36 \sqrt{n}} (1 + \sqrt{3 n}) e^{- \sqrt{3 n}}$
Commented by mathmax by abdo last updated on 01/Nov/19

$n ⩾ 1$
	Answered by mind is power last updated on 23/Oct/19

	$u = nx$ $U_{n} = n \int_{0}^{+ \infty} \frac{\cos (u)}{{(3 n + u^{2})}^{2}} du$ $withe residu bee continued$
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