let f(t) =∫_0 ^∞ ((cos^2 (tx))/((x^2 +3)^2 )) dx with t ≥0 1) give a explicit form of f(t) 2) find the value of ∫_0 ^∞ ((xsin(2tx))/((x^2 +3)^2 )) dx 3) give the values of integrals ∫_0 ^∞ (dx/((x^2 +3)^2 )) and ∫_0 ^∞ ((cos^2 (πx))/((x^2 +3)^2 ))dx 4) give the values of integrals ∫_0 ^∞ ((xsin(πx))/((x^2 +3)^2 )) and ∫_0 ^∞ ((xsin(((πx)/2)))/((x^2 +3)^2 )) dx .


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Question Number 52703 by maxmathsup by imad last updated on 11/Jan/19

$l e t f (t) = \int_{0}^{\infty} \frac{{c o s}^{2} (t x)}{{(x^{2} + 3)}^{2}} d x w i t h t ⩾ 0$ $1) g i v e a e x p l i c i t f o r m o f f (t)$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x s i n (2 t x)}{{(x^{2} + 3)}^{2}} d x$ $3) g i v e t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{{(x^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{{c o s}^{2} (π x)}{{(x^{2} + 3)}^{2}} d x$ $4) g i v e t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{x s i n (π x)}{{(x^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{x s i n (\frac{π x}{2})}{{(x^{2} + 3)}^{2}} d x .$
Commented by Abdo msup. last updated on 13/Jan/19
$channgement x=(√3)u give f(t)=∫_0 ^∞ ((cos^2 ((√3)tu))/(9(u^2 +1)^2 )) (√3)du =((√3)/9) ∫_0 ^(+∞) ((cos^2 ((√3)tu))/((u^2 +1)^2 )) du =((√3)/(18)) ∫_(−∞) ^(+∞) ((cos^2 ((√3)tu))/((u^2 +1)^2 ))du =((√3)/(18)) ∫_(−∞) ^(+∞) ((1+cos(2(√3)tu))/(2(u^2 +1)^2 ))du =((√3)/(36)) ∫_(−∞) ^(+∞) (du/((u^2 +1)^2 )) +((√3)/(36)) ∫_(−∞) ^(+∞) ((cos(2(√3)tu)du)/((u^2 +1)^2 )) ∫_(−∞) ^(+∞) (du/((u^2 +1)^2 )) =_(u=tanθ) ∫_(−(π/2)) ^(π/2) ((1+tan^2 θ)/((1+tan^2 θ)^2 ))dθ =2 ∫_0 ^(π/2) cos^2 θ dθ =2 ∫_0 ^(π/2) ((1+cos(2θ))/2) dθ =(π/2) +[(1/2)sin(2θ)]_0 ^(π/2) =(π/2) let determine ∫_(−∞) ^(+∞) ((cos(2(√3)tu))/((u^2 +1)^2 )) du =I I =Re( ∫_(−∞) ^(+∞) (e^(2i(√3)tu) /((u^2 +1)^2 )) du) let ϕ(z)= (e^(2i(√3)tz) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(2i(√3)tz) /((z−i)^2 (z+i)^2 )) the poles of ϕ are i and −i(doubles) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Re(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(2i(√3)t z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2i(√3)t e^(2i(√3)tz) (z+i)^2 −2(z+i)e^(2i(√3)tz) )/((z+i)^4 )) =lim_(z→i) (((2i(√3)t (z+i)−2)e^(2i(√3)tz) )/((z+i)^3 )) =(((2i(√3)t(2i)−2)e^(−2(√3)t) )/((2i)^3 )) =(((−4i(√3)t−2) e^(−2(√3)t) )/(−8i)) =(((2i(√3)t +1)e^(−2(√3)t) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz= 2iπ (((2i(√3)t +1)e^(−2(√3)t) )/(4i)) =(π/2)(2i(√3)t +1)e^(−2(√3)t) but I =Re(∫_(−∞) ^(+∞) ϕ(z)dz) =(π/2) e^(−2(√3)t) ⇒ f(t)=((π(√3))/(72)) +((√3)/(36)){ (π/2) e^(−2(√3)t) } ⇒ f(t) =((π(√3))/(72))(1+e^(−2(√3)t) ) .$
$c h a n n g e m e n t x = \sqrt{3} u g i v e$ $f (t) = \int_{0}^{\infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{9 {(u^{2} + 1)}^{2}} \sqrt{3} d u$ $= \frac{\sqrt{3}}{9} \int_{0}^{+ \infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u = \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u$ $= \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{1 + c o s (2 \sqrt{3} t u)}{2 {(u^{2} + 1)}^{2}} d u$ $= \frac{\sqrt{3}}{36} \int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}} + \frac{\sqrt{3}}{36} \int_{- \infty}^{+ \infty} \frac{c o s (2 \sqrt{3} t u) d u}{{(u^{2} + 1)}^{2}}$ $\int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}} =_{u = t a n θ} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{2}} d θ$ $= 2 \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ = 2 \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 θ)}{2} d θ$ $= \frac{π}{2} + {[\frac{1}{2} s i n (2 θ)]}_{0}^{\frac{π}{2}} = \frac{π}{2} l e t d e t e r m i n e$ $\int_{- \infty}^{+ \infty} \frac{c o s (2 \sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u = I$ $I = R e (\int_{- \infty}^{+ \infty} \frac{e^{2 i \sqrt{3} t u}}{{(u^{2} + 1)}^{2}} d u) l e t$ $φ (z) = \frac{e^{2 i \sqrt{3} t z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{e^{2 i \sqrt{3} t z}}{{(z - i)}^{2} {(z + i)}^{2}}$ $t h e p o l e s o f φ a r e i a n d - i (d o u b l e s) s o$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{2 i \sqrt{3} t z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{2 i \sqrt{3} t e^{2 i \sqrt{3} t z} {(z + i)}^{2} - 2 (z + i) e^{2 i \sqrt{3} t z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{(2 i \sqrt{3} t (z + i) - 2) e^{2 i \sqrt{3} t z}}{{(z + i)}^{3}}$ $= \frac{(2 i \sqrt{3} t (2 i) - 2) e^{- 2 \sqrt{3} t}}{{(2 i)}^{3}} = \frac{(- 4 i \sqrt{3} t - 2) e^{- 2 \sqrt{3} t}}{- 8 i}$ $= \frac{(2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t}}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z =$ $2 i π \frac{(2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t}}{4 i} = \frac{π}{2} (2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t} b u t$ $I = R e (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{π}{2} e^{- 2 \sqrt{3} t} \Rightarrow$ $f (t) = \frac{π \sqrt{3}}{72} + \frac{\sqrt{3}}{36} {\frac{π}{2} e^{- 2 \sqrt{3} t}} \Rightarrow$ $f (t) = \frac{π \sqrt{3}}{72} (1 + e^{- 2 \sqrt{3} t}) .$
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