let f_n (a) =∫_(−∞) ^(+∞) ((cos(nx))/((x^2 +x +a)^2 ))dx with a≥1 1) find a explicit form of f_n (a) 2)study the convervenge of Σ f_n (a) 3) determine also g_n (a) = ∫_(−∞) ^(+∞) ((cos(nx))/((x^2 +x+a)^3 ))dx study the convergence of Σ gn(a)


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Question Number 61884 by maxmathsup by imad last updated on 10/Jun/19

$l e t f_{n} (a) = \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{2}} d x w i t h a ⩾ 1$ $1) f i n d a e x p l i c i t f o r m o f f_{n} (a)$ $2) s t u d y t h e c o n v e r v e n g e o f Σ f_{n} (a)$ $3) d e t e r m i n e a l s o g_{n} (a) = \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{3}} d x$ $s t u d y t h e c o n v e r g e n c e o f Σ g n (a)$
Commented by maxmathsup by imad last updated on 11/Jun/19
$1) we have f_n (a) =Re( ∫_(−∞) ^(+∞) (e^(inx) /((x^2 +x +a)^2 ))) let w(z) =(e^(inz) /((z^2 +z+a)^2 )) poles of w ? z^2 +z +a =0 →Δ =1−4a <0 ⇒ Δ =(i(√(4a−1)))^2 ⇒ z_1 =((−1+i(√(4a−1)))/2) and z_2 =((−1−i(√(4a−1)))/2) the poles of w are z_1 and z_2 (doubles) w(z) = (e^(inz) /((z−z_1 )^2 (z−z_2 )^2 )) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,z_1 ) Res(w,z_1 ) =lim_(z→z_1 ) (1/((2−1)!)){ (z−z_1 )^2 w(z)}^((1)) =lim_(z→z_1 ) { (e^(inz) /((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) ((in e^(inz) (z−z_2 )^2 −2(z−z_2 )e^(inz) )/((z−z_2 )^4 )) =lim_(z→z_1 ) (((in(z−z_2 )−2}e^(inz) )/((z−z_2 )^3 )) =(({in(z_1 −z_2 )−2)e^(inz_1 ) )/((z_1 −z_2 )^3 )) =(({ini(√(4a−1))−2}e^(in(((−1+i(√(4a−1)))/2))) )/((i(√(4a−1)))^3 )) =−(((n(√(4a−1)) +2)e^(−((in)/2)) e^(−(n/2)(√(4a−1))) )/(−i(4a−1)(√(4a−1)))) =(((n(√(4a−1))+2)e^(−(n/2)(√(4a−1))) )/(i(4a−1)(√(4a−1)))){ cos((n/(2 )))−i sin((n/2))} ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ (((n(√(4a−1))+2)e^(−(n/2)(√(4a−1))) )/(i(4a−1)(√(4a−1)))) {....} ⇒ f_n (a) =((2π(n(√(4a−1))+2)e^(−(n/2)(√(4a−1))) )/((4a−1)(√(4a−1))))$
$1) w e h a v e f_{n} (a) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i n x}}{{(x^{2} + x + a)}^{2}}) l e t w (z) = \frac{e^{i n z}}{{(z^{2} + z + a)}^{2}} p o l e s o f w ?$ $z^{2} + z + a = 0 \to Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow$ $z_{1} = \frac{- 1 + i \sqrt{4 a - 1}}{2} a n d z_{2} = \frac{- 1 - i \sqrt{4 a - 1}}{2} t h e p o l e s o f w a r e z_{1} a n d z_{2} (d o u b l e s)$ $w (z) = \frac{e^{i n z}}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} w (z) d z = 2 i π R e s (w, z_{1})$ $R e s (w, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} w (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{e^{i n z}}{{(z - z_{2})}^{2}}}^{(1)} = {l i m}_{z \to z_{1}} \frac{i n e^{i n z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{i n z}}{{(z - z_{2})}^{4}}$ $= {l i m}_{z \to z_{1}} \frac{(i n (z - z_{2}) - 2} e^{i n z}}{{(z - z_{2})}^{3}} = \frac{{i n (z_{1} - z_{2}) - 2) e^{{i n z}_{1}}}{{(z_{1} - z_{2})}^{3}}$ $= \frac{{i n i \sqrt{4 a - 1} - 2} e^{i n (\frac{- 1 + i \sqrt{4 a - 1}}{2})}}{{(i \sqrt{4 a - 1})}^{3}} = - \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{i n}{2}} e^{- \frac{n}{2} \sqrt{4 a - 1}}}{- i (4 a - 1) \sqrt{4 a - 1}}$ $= \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{n}{2} \sqrt{4 a - 1}}}{i (4 a - 1) \sqrt{4 a - 1}} {c o s (\frac{n}{2}) - i s i n (\frac{n}{2})} \Rightarrow$ $\int_{- \infty}^{+ \infty} w (z) d z = 2 i π \frac{(n \sqrt{4 a - 1} + 2) e^{- \frac{n}{2} \sqrt{4 a - 1}}}{i (4 a - 1) \sqrt{4 a - 1}} {. . . .} \Rightarrow$ $f_{n} (a) = \frac{2 π (n \sqrt{4 a - 1} + 2) e^{- \frac{n}{2} \sqrt{4 a - 1}}}{(4 a - 1) \sqrt{4 a - 1}}$
Commented by maxmathsup by imad last updated on 11/Jun/19

$2) w e h a v e Σ f_{n} (a) = Σ \frac{2 π \sqrt{4 a - 1}}{(4 a - 1) \sqrt{4 a - 1}} n e^{- \frac{n}{2} \sqrt{4 a - 1}} + Σ \frac{4 π}{(4 a - 1) \sqrt{4 a - 1}} e^{- \frac{n}{2} \sqrt{4 a - 1}}$ $= \frac{2 π}{(4 a - 1) \sqrt{4 a - 1}} Σ n {(e^{- \frac{\sqrt{4 a - 1}}{2}})}^{n} + \frac{4 π}{(4 a - 1) \sqrt{4 a - 1}} {(e^{- \frac{\sqrt{4 a - 1}}{2}})}^{n}$ $w e h a v e ∣ e^{- \frac{\sqrt{4 a - 1}}{2}} ∣< 1 a n d Σ x^{n}, Σ {n x}^{n} c o n v e r g s \Rightarrow Σ f_{n} c o n v e r g e s$
Commented by maxmathsup by imad last updated on 11/Jun/19

$3) w e h a v e f_{n}^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial a} (\frac{c o s (n x)}{{(x^{2} + x + a)}^{2}}) d x$ $= - \int_{- \infty}^{+ \infty} \frac{2 (x^{2} + x + a)}{{(x^{2} + x + a)}^{4}} c o s (n x) d x = - 2 \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{{(x^{2} + x + a)}^{2}} d x = - 2 g_{n} (a) \Rightarrow$ $g_{n} (a) = - \frac{1}{2} f_{n}^{^{'}} (a) r e s t t o c a l c u l a t e f^{^{'}} n (a) . . . .$
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