let f(x) =∫_0 ^∞ ((cos(πxt))/((t^2 +3x^2 )^2 )) dt with x>0 1) find a explicit form for f(x) 2) find the value of ∫_0 ^∞ ((cos(πt))/((t^2 +3)^2 ))dt 3) let U_n =f(n) find nature of Σ U


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Question Number 57900 by maxmathsup by imad last updated on 13/Apr/19

$l e t f (x) = \int_{0}^{\infty} \frac{c o s (π x t)}{{(t^{2} + 3 x^{2})}^{2}} d t w i t h x > 0$ $1) f i n d a e x p l i c i t f o r m f o r f (x)$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (π t)}{{(t^{2} + 3)}^{2}} d t$ $3) l e t U_{n} = f (n) f i n d n a t u r e o f Σ U_{n}$
Commented bymaxmathsup by imad last updated on 17/Apr/19
$1) we have 2f(x) =∫_(−∞) ^(+∞) ((cos(πxt))/((t^2 +3x^2 )^2 ))dt =Re(∫_(−∞) ^(+∞) (e^(iπxt) /((t^2 +3x^2 )^2 ))dt) let ϕ(z) =(e^(iπxz) /((z^2 +3x^2 )^2 )) ⇒ ϕ(z) =(e^(iπxz) /((z−ix(√3))^2 (z+ix(√3))^2 )) so the poles of ϕ are +^− ix(√3) (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ix(√3)) Res(ϕ,ix(√3)) =lim_(z→ix(√3)) (1/((2−1)!)) {(z−ix(√3))^2 ϕ(z)}^((1)) =lim_(z→ix(√x)) { (e^(iπxz) /((z+ix(√3))^2 ))}^((1)) =lim_(z→ix(√3)) ((iπxe^(iπxz) (z+ix(√3))^2 −2(z+ix(√3))e^(iπxz) )/((z+ix(√3))^4 )) =lim_(z→ix(√3)) (({iπx(z+ix(√3))−2}e^(iπxz) )/((z+ix(√3))^3 )) (({iπx(2ix(√3))−2)e^(iπx(ix(√3))) )/((2ix(√3))^3 )) =(({−2π(√3)x^2 −2) e^(−π(√3)x^2 ) )/(−8i (3(√3)))) =(((π(√3)x^2 +1)e^(−π(√3)x^2 ) )/(12i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((π(√3)x^2 +1)e^(−π(√3)x^2 ) )/(12i(√3))) =(π/(6(√3))) (π(√3)x^2 +1)e^(−π(√3)x^2 ) ⇒ f(x) =(π/(12(√3)))(π(√3)x^2 +1)e^(−π(√3)x^2 ) . 2) ∫_0 ^∞ ((cos(πt))/((t^2 +3)^2 )) dt =f(1) =(π/(12(√3)))(π(√3) +1)e^(−π(√3))$
$1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{c o s (π x t)}{{(t^{2} + 3 x^{2})}^{2}} d t = R e (\int_{- \infty}^{+ \infty} \frac{e^{i π x t}}{{(t^{2} + 3 x^{2})}^{2}} d t) l e t$ $φ (z) = \frac{e^{i π x z}}{{(z^{2} + 3 x^{2})}^{2}} \Rightarrow φ (z) = \frac{e^{i π x z}}{{(z - i x \sqrt{3})}^{2} {(z + i x \sqrt{3})}^{2}} s o t h e p o l e s o f φ$ $a r e \bar{+} i x \sqrt{3} (d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i x \sqrt{3})$ $R e s (φ, i x \sqrt{3}) = {l i m}_{z \to i x \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i x \sqrt{3})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i x \sqrt{x}} {\frac{e^{i π x z}}{{(z + i x \sqrt{3})}^{2}}}^{(1)}$ $= {l i m}_{z \to i x \sqrt{3}} \frac{i π {x e}^{i π x z} {(z + i x \sqrt{3})}^{2} - 2 (z + i x \sqrt{3}) e^{i π x z}}{{(z + i x \sqrt{3})}^{4}}$ $= {l i m}_{z \to i x \sqrt{3}} \frac{{i π x (z + i x \sqrt{3}) - 2} e^{i π x z}}{{(z + i x \sqrt{3})}^{3}}$ $\frac{{i π x (2 i x \sqrt{3}) - 2) e^{i π x (i x \sqrt{3})}}{{(2 i x \sqrt{3})}^{3}} = \frac{{- 2 π \sqrt{3} x^{2} - 2) e^{- π \sqrt{3} x^{2}}}{- 8 i (3 \sqrt{3})} = \frac{(π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}}}{12 i \sqrt{3}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}}}{12 i \sqrt{3}} = \frac{π}{6 \sqrt{3}} (π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}} \Rightarrow$ $f (x) = \frac{π}{12 \sqrt{3}} (π \sqrt{3} x^{2} + 1) e^{- π \sqrt{3} x^{2}} .$ $2) \int_{0}^{\infty} \frac{c o s (π t)}{{(t^{2} + 3)}^{2}} d t = f (1) = \frac{π}{12 \sqrt{3}} (π \sqrt{3} + 1) e^{- π \sqrt{3}}$
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