let f(t) = ∫_0 ^∞ ((cos(tx))/((2+x^2 )^2 ))dx 1) find a simple form of f(t) 2) calculate ∫


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Question Number 36203 by prof Abdo imad last updated on 30/May/18

$l e t f (t) = \int_{0}^{\infty} \frac{c o s (t x)}{{(2 + x^{2})}^{2}} d x$ $1) f i n d a s i m p l e f o r m o f f (t)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (3 x)}{{(2 + x^{2})}^{2}} d x$
Commented by prof Abdo imad last updated on 01/Jun/18
$we have 2f(t) = ∫_(−∞) ^(+∞) ((cos(tx))/((2+x^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(itx) /((2+x^2 )^2 ))dx) let introduce the complex?function ϕ(z) = (e^(itz) /((2+z^2 )^2 )) ϕ(z) = (e^(itz) /((z−i(√2))^2 (z +i(√2))^2 )) so the poles of ϕ are i(√2) and −i(√2) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res( ϕ,i(√2)) Res(ϕ,i(√2)) =lim_(z→i(√2)) { (z−i(√2))^2 ϕ(z)}^′ =lim_(z→i(√2)) { (e^(itz) /((z+i(√2))^2 ))}^′ =lim_(z→i(√2)) { ((it e^(itz) (z+i(√2))^2 −2(z+i(√2))e^(itz) )/((z +i(√2))^4 ))} =lim_(z→i(√2)) e^(itz) ((it(z +i(√2)) −2)/((z +i(√2))^3 )) =e^(it(i(√2))) ((it(2i(√2)) −2)/((2i(√2))^3 )) = ((−2t(√2)−2)/(−8i(2(√2)))) e^(−t(√2)) = ((t(√2) +1)/(8i(√2))) e^(−t(√2)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((1+t(√2))/(8i(√2))) e^(−t(√2)) = (π/(4(√2)))(1+t(√2)) e^(−t(√2)) but we have 2f(t)= Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ ★ f(t) = (π/(8(√2))) (1+t(√2))e^(−t(√2)) ★ 2) we have ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx =f(3) ⇒ ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx = (π/(8(√2)))( 1+3(√2))e^(−3(√2)) .$
$w e h a v e 2 f (t) = \int_{- \infty}^{+ \infty} \frac{c o s (t x)}{{(2 + x^{2})}^{2}} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i t x}}{{(2 + x^{2})}^{2}} d x) l e t i n t r o d u c e t h e$ $c o m p l e x ? f u n c t i o n φ (z) = \frac{e^{i t z}}{{(2 + z^{2})}^{2}}$ $φ (z) = \frac{e^{i t z}}{{(z - i \sqrt{2})}^{2} {(z + i \sqrt{2})}^{2}} s o t h e p o l e s o f φ a r e$ $i \sqrt{2} a n d - i \sqrt{2} (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{2})$ $R e s (φ, i \sqrt{2}) = {l i m}_{z \to i \sqrt{2}} {{(z - i \sqrt{2})}^{2} φ (z)}^{^{'}}$ $= {l i m}_{z \to i \sqrt{2}} {\frac{e^{i t z}}{{(z + i \sqrt{2})}^{2}}}^{^{'}}$ $= {l i m}_{z \to i \sqrt{2}} {\frac{i t e^{i t z} {(z + i \sqrt{2})}^{2} - 2 (z + i \sqrt{2}) e^{i t z}}{{(z + i \sqrt{2})}^{4}}}$ $= {l i m}_{z \to i \sqrt{2}} e^{i t z} \frac{i t (z + i \sqrt{2}) - 2}{{(z + i \sqrt{2})}^{3}}$ $= e^{i t (i \sqrt{2})} \frac{i t (2 i \sqrt{2}) - 2}{{(2 i \sqrt{2})}^{3}}$ $= \frac{- 2 t \sqrt{2} - 2}{- 8 i (2 \sqrt{2})} e^{- t \sqrt{2}} = \frac{t \sqrt{2} + 1}{8 i \sqrt{2}} e^{- t \sqrt{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1 + t \sqrt{2}}{8 i \sqrt{2}} e^{- t \sqrt{2}}$ $= \frac{π}{4 \sqrt{2}} (1 + t \sqrt{2}) e^{- t \sqrt{2}} b u t w e h a v e$ $2 f (t) = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow$ $★ f (t) = \frac{π}{8 \sqrt{2}} (1 + t \sqrt{2}) e^{- t \sqrt{2}} ★$ $2) w e h a v e \int_{0}^{\infty} \frac{c o s (3 x)}{{(2 + x^{2})}^{2}} d x = f (3) \Rightarrow$ $\int_{0}^{\infty} \frac{c o s (3 x)}{{(2 + x^{2})}^{2}} d x = \frac{π}{8 \sqrt{2}} (1 + 3 \sqrt{2}) e^{- 3 \sqrt{2}} .$
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