calculate f(t)=∫_0 ^∞ ((cos(tx))/((1+tx^2 )^2 )) dx with t≥0 2) find the values of ∫_0 ^∞ ((cos(2x))/((1+2x^2 )^2 ))dx and ∫


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Question Number 38470 by maxmathsup by imad last updated on 25/Jun/18

$c a l c u l a t e f (t) = \int_{0}^{\infty} \frac{c o s (t x)}{{(1 + {t x}^{2})}^{2}} d x w i t h t ⩾ 0$ $2) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{c o s (2 x)}{{(1 + 2 x^{2})}^{2}} d x a n d \int_{0}^{\infty} \frac{c o s x}{{(2 + x^{2})}^{2}} d x$
Commented by math khazana by abdo last updated on 26/Jun/18

$t > 0$
Commented by math khazana by abdo last updated on 27/Jun/18
$we have 2f(x)=∫_(−∞) ^(+∞) ((cos(tx))/((1+tx^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(itx) /((1+tx^2 )^2 ))dx) let consider ϕ(z) = (e^(itz) /((tz^2 +1)^2 )) we have ϕ(z)= (e^(itz) /(t^2 (z^2 +(1/t))^2 )) = (e^(itz) /(t^2 (z−(i/(√t)))^2 (z+(i/(√t)))^2 )) the poles of ϕ are (i/(√t)) and ((−i)/(√t)) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/(√t))) but Res(ϕ,(i/(√t)))=lim_(z→(i/(√t))) (1/((2−1)!)){ (z−(i/(√t)))^2 ϕ(z)}^((1)) =lim_(z→(i/(√t))) (1/t^2 ){ (e^(itz) /((z+(i/(√t)))^2 ))}^((1)) =lim_(z→(i/(√t))) (1/t^2 ){ ((it e^(itz) (z+(i/(√t)))^2 −2(z+(i/(√t)))e^(itz) )/((z+(i/(√t)))^4 ))} =lim_(z→(i/(√t))) (1/t^2 ){ ((it e^(itz) (z+(i/(√t))) −2 e^(itz) )/((z+(i/(√t)))^3 ))} =(1/t^2 )((it e^(it(i/(√t))) (((2i)/(√t)))−2 e^(it((i/(√t)))) )/((((2i)/(√t)))^3 ))=(1/t^2 ) ((−2(√t)e^(−(√t)) −2 e^(−(√t)) )/((−8i)/(t(√t)))) =((t(√t))/t^2 ) (((1+(√t))e^(−(√t)) )/(4i)) =((√t)/(4ti))(1+(√t))e^(−(√t)) =(((t+(√t))e^(−(√t)) )/(4ti)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((t+(√t))e^(−(√t)) )/(4it)) =(π/(2t)) (t+(√t))e^(−(√t)) 2f(x)= Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ f(x)= (π/(4t))(t+(√t))e^(−(√t)) (t>0) 2) ∫_0 ^∞ ((cos(2x))/((1+2x^2 )^2 ))dx=f(2) =(π/8)(2+(√2))e^(−(√2))$
$w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{c o s (t x)}{{(1 + {t x}^{2})}^{2}} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i t x}}{{(1 + {t x}^{2})}^{2}} d x) l e t c o n s i d e r$ $φ (z) = \frac{e^{i t z}}{{({t z}^{2} + 1)}^{2}} w e h a v e$ $φ (z) = \frac{e^{i t z}}{t^{2} {(z^{2} + \frac{1}{t})}^{2}} = \frac{e^{i t z}}{t^{2} {(z - \frac{i}{\sqrt{t}})}^{2} {(z + \frac{i}{\sqrt{t}})}^{2}}$ $t h e p o l e s o f φ a r e \frac{i}{\sqrt{t}} a n d \frac{- i}{\sqrt{t}} (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{\sqrt{t}}) b u t$ $R e s (φ, \frac{i}{\sqrt{t}}) = {l i m}_{z \to \frac{i}{\sqrt{t}}} \frac{1}{(2 - 1)!} {{(z - \frac{i}{\sqrt{t}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to \frac{i}{\sqrt{t}}} \frac{1}{t^{2}} {\frac{e^{i t z}}{{(z + \frac{i}{\sqrt{t}})}^{2}}}^{(1)}$ $= {l i m}_{z \to \frac{i}{\sqrt{t}}} \frac{1}{t^{2}} {\frac{i t e^{i t z} {(z + \frac{i}{\sqrt{t}})}^{2} - 2 (z + \frac{i}{\sqrt{t}}) e^{i t z}}{{(z + \frac{i}{\sqrt{t}})}^{4}}}$ $= {l i m}_{z \to \frac{i}{\sqrt{t}}} \frac{1}{t^{2}} {\frac{i t e^{i t z} (z + \frac{i}{\sqrt{t}}) - 2 e^{i t z}}{{(z + \frac{i}{\sqrt{t}})}^{3}}}$ $= \frac{1}{t^{2}} \frac{i t e^{i t \frac{i}{\sqrt{t}}} (\frac{2 i}{\sqrt{t}}) - 2 e^{i t (\frac{i}{\sqrt{t}})}}{{(\frac{2 i}{\sqrt{t}})}^{3}} = \frac{1}{t^{2}} \frac{- 2 \sqrt{t} e^{- \sqrt{t}} - 2 e^{- \sqrt{t}}}{\frac{- 8 i}{t \sqrt{t}}}$ $= \frac{t \sqrt{t}}{t^{2}} \frac{(1 + \sqrt{t}) e^{- \sqrt{t}}}{4 i} = \frac{\sqrt{t}}{4 t i} (1 + \sqrt{t}) e^{- \sqrt{t}}$ $= \frac{(t + \sqrt{t}) e^{- \sqrt{t}}}{4 t i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(t + \sqrt{t}) e^{- \sqrt{t}}}{4 i t} = \frac{π}{2 t} (t + \sqrt{t}) e^{- \sqrt{t}}$ $2 f (x) = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow$ $f (x) = \frac{π}{4 t} (t + \sqrt{t}) e^{- \sqrt{t}} (t > 0)$ $2) \int_{0}^{\infty} \frac{c o s (2 x)}{{(1 + 2 x^{2})}^{2}} d x = f (2) = \frac{π}{8} (2 + \sqrt{2}) e^{- \sqrt{2}}$
Commented by math khazana by abdo last updated on 27/Jun/18
$let calculate I =∫_0 ^∞ ((cosx)/((2+x^2 )^2 ))dx 2I =∫_(−∞) ^(+∞) ((cosx)/((2+x^2 )^2 ))dx=Re( ∫_(−∞) ^(+∞) (e^(ix) /((x^2 +2)^2 ))dx) let ϕ(z)= (e^(iz) /((z^2 +2)^2 )) ϕ(z)= (e^(iz) /((z−i(√2))^2 (z+i(√2))^2 )) so the polesof ϕ are i(√2) and −i(√2) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i(√2)) Res(ϕ,i(√2))=lim_(z→i(√2)) (1/((2−1)!)){ (z−i(√2))^2 ϕ(z)}^((1)) =lim_(z→i(√2)) { (e^(iz) /((z+i(√2))^2 ))}^((1)) =lim_(z→i(√2)) ((i e^(iz) (z+i(√2))^2 −2(z+i(√2))e^(iz) )/((z+i(√2))^4 )) =lim_(z→i(√2)) ((i e^(iz) (z+i(√2)) −2 e^(iz) )/((z+i(√2))^3 )) =((i e^(i(i(√2))) (2i(√2)) −2 e^(i(i(√2))) )/((2i(√2))^3 )) = ((−2(√2) e^(−(√2)) −2 e^(−(√2)) )/(−8i(2(√2)))) =(((1+(√2))e^(−(√2)) )/(8i(√2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((1+(√2))e^(−(√2)) )/(8i(√2))) = (π/(4(√2)))(1+(√2))e^(−(√2)) 2I =Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ I = (π/(8(√2)))(1+(√2))e^(−(√2)) .$
$l e t c a l c u l a t e I = \int_{0}^{\infty} \frac{c o s x}{{(2 + x^{2})}^{2}} d x$ $2 I = \int_{- \infty}^{+ \infty} \frac{c o s x}{{(2 + x^{2})}^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i x}}{{(x^{2} + 2)}^{2}} d x)$ $l e t φ (z) = \frac{e^{i z}}{{(z^{2} + 2)}^{2}}$ $φ (z) = \frac{e^{i z}}{{(z - i \sqrt{2})}^{2} {(z + i \sqrt{2})}^{2}} s o t h e p o l e s o f φ a r e$ $i \sqrt{2} a n d - i \sqrt{2} (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{2})$ $R e s (φ, i \sqrt{2}) = {l i m}_{z \to i \sqrt{2}} \frac{1}{(2 - 1)!} {{(z - i \sqrt{2})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i \sqrt{2}} {\frac{e^{i z}}{{(z + i \sqrt{2})}^{2}}}^{(1)}$ $= {l i m}_{z \to i \sqrt{2}} \frac{i e^{i z} {(z + i \sqrt{2})}^{2} - 2 (z + i \sqrt{2}) e^{i z}}{{(z + i \sqrt{2})}^{4}}$ $= {l i m}_{z \to i \sqrt{2}} \frac{i e^{i z} (z + i \sqrt{2}) - 2 e^{i z}}{{(z + i \sqrt{2})}^{3}}$ $= \frac{i e^{i (i \sqrt{2})} (2 i \sqrt{2}) - 2 e^{i (i \sqrt{2})}}{{(2 i \sqrt{2})}^{3}}$ $= \frac{- 2 \sqrt{2} e^{- \sqrt{2}} - 2 e^{- \sqrt{2}}}{- 8 i (2 \sqrt{2})} = \frac{(1 + \sqrt{2}) e^{- \sqrt{2}}}{8 i \sqrt{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(1 + \sqrt{2}) e^{- \sqrt{2}}}{8 i \sqrt{2}}$ $= \frac{π}{4 \sqrt{2}} (1 + \sqrt{2}) e^{- \sqrt{2}}$ $2 I = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow$ $I = \frac{π}{8 \sqrt{2}} (1 + \sqrt{2}) e^{- \sqrt{2}} .$
Commented by abdo.msup.com last updated on 27/Jun/18

$f (t) = \frac{π}{4 t} (t + \sqrt{t}) e^{- \sqrt{t}}$
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