let f(x) =∫_0 ^∞ ((cos(xt))/((xt^2 +i)^2 ))dx with x from R and x≠0 1) find a explicit form of f(x) 2) extract A =Re(f(x)) and B =Im(f(x)) and find its values . 3) calculate ∫_0 ^∞ ((cos(2t))/((2t^2 +i)^2 ))dt 4) let U_n =∫_0 ^∞ ((cos(nt))/((nt^2 +i)^2 ))dt .calculate lim_(n→+∞) u_n and study the convergence of Σu


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Question Number 55760 by maxmathsup by imad last updated on 03/Mar/19

$l e t f (x) = \int_{0}^{\infty} \frac{c o s (x t)}{{({x t}^{2} + i)}^{2}} d x w i t h x f r o m R a n d x \neq 0$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) e x t r a c t A = R e (f (x)) a n d B = I m (f (x)) a n d f i n d i t s v a l u e s .$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (2 t)}{{(2 t^{2} + i)}^{2}} d t$ $4) l e t U_{n} = \int_{0}^{\infty} \frac{c o s (n t)}{{({n t}^{2} + i)}^{2}} d t . c a l c u l a t e {l i m}_{n \to + \infty} u_{n}$ $a n d s t u d y t h e c o n v e r g e n c e o f Σ u_{n}$
Commented by maxmathsup by imad last updated on 04/Mar/19

$i^{2} = - 1$
Commented by maxmathsup by imad last updated on 04/Mar/19

$f (x) = \int_{0}^{\infty} \frac{c o s (x t)}{{({x t}^{2} + i)}^{2}} d t$
Commented by maxmathsup by imad last updated on 10/Mar/19
$case 1 x>0 changement xt =u give f(x)=∫_0 ^∞ ((cosu)/((x(u^2 /x^2 )+i)^2 ))(du/x) =(1/x)∫_0 ^∞ ((cosu)/(((u^2 /x)+i)^2 )) =x∫_0 ^∞ ((cosu)/((u^2 +ix)^2 )) du=x ∫_0 ^∞ (((u^2 −ix)^2 cosu)/((u^4 +x^2 )^2 ))du =x ∫_0 ^∞ (((u^2 −2ix u^2 −x^2 )cosu)/((u^4 +x^2 )^2 )) du =x ∫_0 ^∞ (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 ))du −2ix^2 ∫_0 ^∞ ((u^2 cosu)/((u^4 +x^2 )^2 )) ⇒Re(f(x))=x∫_0 ^∞ (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 ))du =I and Im(f(x)) =−2x^2 ∫_0 ^∞ ((u^2 cosu)/((u^4 +x^2 ))) du let find I we have 2I =x ∫_(−∞) ^(+∞) (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 )) du =x Re(∫_(−∞) ^(+∞) (((u^2 −x^2 )e^(iu) )/((u^4 +x^2 )^2 )) du) let ϕ(z) =(((z^2 −x^2 )e^(iz) )/((z^4 +x^2 )^2 )) ⇒ϕ(z) =(((z^2 −x^2 )e^(iz) )/((z^2 −ix)^2 (z^2 +ix)^2 )) =(((z^2 −x^2 )e^(iz) )/((z−(√x)e^(i(π/4)) )^2 (z+(√x)e^((iπ)/4) )^2 (z−(√x)e^(−((iπ)/4)) )^2 (z+(√x)e^(−((iπ)/4)) )^2 )) so the poles of ϕ are +^− (√x)e^((iπ)/4) and +^− (√x)e^(−((iπ)/4)) (double poles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√x)e^((iπ)/4) ) +Res(ϕ,−(√x)e^(−((iπ)/4)) )} Res(ϕ, (√x)e^((iπ)/4) ) =lim_(z→(√x)e^((iπ)/4) ) (1/((2−1)!)) {(z−(√x)e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→(√x)e^((iπ)/4) ) { (((z^2 −x^2 )e^(iz) )/((z +(√x)e^((iπ)/4) )^2 (z^2 +ix)^2 ))}^((1)) ....be continued... mim$
$c a s e 1 x > 0 c h a n g e m e n t x t = u g i v e f (x) = \int_{0}^{\infty} \frac{c o s u}{{(x \frac{u^{2}}{x^{2}} + i)}^{2}} \frac{d u}{x}$ $= \frac{1}{x} \int_{0}^{\infty} \frac{c o s u}{{(\frac{u^{2}}{x} + i)}^{2}} = x \int_{0}^{\infty} \frac{c o s u}{{(u^{2} + i x)}^{2}} d u = x \int_{0}^{\infty} \frac{{(u^{2} - i x)}^{2} c o s u}{{(u^{4} + x^{2})}^{2}} d u$ $= x \int_{0}^{\infty} \frac{(u^{2} - 2 i x u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u$ $= x \int_{0}^{\infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u - 2 {i x}^{2} \int_{0}^{\infty} \frac{u^{2} c o s u}{{(u^{4} + x^{2})}^{2}} \Rightarrow R e (f (x)) = x \int_{0}^{\infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u = I$ $a n d I m (f (x)) = - 2 x^{2} \int_{0}^{\infty} \frac{u^{2} c o s u}{(u^{4} + x^{2})} d u l e t f i n d I w e h a v e$ $2 I = x \int_{- \infty}^{+ \infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u = x R e (\int_{- \infty}^{+ \infty} \frac{(u^{2} - x^{2}) e^{i u}}{{(u^{4} + x^{2})}^{2}} d u)$ $l e t φ (z) = \frac{(z^{2} - x^{2}) e^{i z}}{{(z^{4} + x^{2})}^{2}} \Rightarrow φ (z) = \frac{(z^{2} - x^{2}) e^{i z}}{{(z^{2} - i x)}^{2} {(z^{2} + i x)}^{2}}$ $= \frac{(z^{2} - x^{2}) e^{i z}}{{(z - \sqrt{x} e^{i \frac{π}{4}})}^{2} {(z + \sqrt{x} e^{\frac{i π}{4}})}^{2} {(z - \sqrt{x} e^{- \frac{i π}{4}})}^{2} {(z + \sqrt{x} e^{- \frac{i π}{4}})}^{2}} s o t h e p o l e s o f φ a r e$ $\bar{+} \sqrt{x} e^{\frac{i π}{4}} a n d \bar{+} \sqrt{x} e^{- \frac{i π}{4}} (d o u b l e p o l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \sqrt{x} e^{\frac{i π}{4}}) + R e s (φ, - \sqrt{x} e^{- \frac{i π}{4}})}$ $R e s (φ, \sqrt{x} e^{\frac{i π}{4}}) = {l i m}_{z \to \sqrt{x} e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - \sqrt{x} e^{\frac{i π}{4}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to \sqrt{x} e^{\frac{i π}{4}}} {\frac{(z^{2} - x^{2}) e^{i z}}{{(z + \sqrt{x} e^{\frac{i π}{4}})}^{2} {(z^{2} + i x)}^{2}}}^{(1)} . . . . b e c o n t i n u e d . . .$ $m i m$
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