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Question Number 61884 by maxmathsup by imad last updated on 10/Jun/19
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)
$${let}\:{f}_{{n}} \left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} +{x}\:\:+{a}\right)^{\mathrm{2}} }{dx}\:\:\:\:{with}\:\:\:{a}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}_{{n}} \left({a}\right) \\ $$$$\left.\mathrm{2}\right){study}\:{the}\:{convervenge}\:{of}\:\Sigma\:{f}_{{n}} \left({a}\right) \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{also}\:{g}_{{n}} \left({a}\right)\:=\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{x}+{a}\right)^{\mathrm{3}} }{dx} \\ $$$${study}\:{the}\:{convergence}\:{of}\:\Sigma\:{gn}\left({a}\right) \\ $$
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))))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}_{{n}} \left({a}\right)\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{inx}} }{\left({x}^{\mathrm{2}} \:+{x}\:+{a}\right)^{\mathrm{2}} }\right)\:{let}\:{w}\left({z}\right)\:=\frac{{e}^{{inz}} }{\left({z}^{\mathrm{2}} \:+{z}+{a}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:{w}\:? \\ $$$${z}^{\mathrm{2}} \:+{z}\:+{a}\:=\mathrm{0}\:\rightarrow\Delta\:=\mathrm{1}−\mathrm{4}{a}\:\:<\mathrm{0}\:\Rightarrow\:\Delta\:=\left({i}\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:\:{the}\:{poles}\:{of}\:{w}\:{are}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} \left({doubles}\right) \\ $$$${w}\left({z}\right)\:=\:\frac{{e}^{{inz}} }{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} {w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{z}_{\mathrm{1}} \right) \\ $$$${Res}\left({w},{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} {w}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\left\{\:\frac{{e}^{{inz}} }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{{in}\:{e}^{{inz}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right){e}^{{inz}} }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\:\:\:\:\frac{\left({in}\left({z}−{z}_{\mathrm{2}} \right)−\mathrm{2}\right\}{e}^{{inz}} }{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\left\{{in}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)−\mathrm{2}\right){e}^{{inz}_{\mathrm{1}} } }{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{{ini}\sqrt{\mathrm{4}{a}−\mathrm{1}}−\mathrm{2}\right\}{e}^{{in}\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\right)} }{\left({i}\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)^{\mathrm{3}} }\:=−\frac{\left({n}\sqrt{\mathrm{4}{a}−\mathrm{1}}\:+\mathrm{2}\right){e}^{−\frac{{in}}{\mathrm{2}}} \:{e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} }{−{i}\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}} \\ $$$$=\frac{\left({n}\sqrt{\mathrm{4}{a}−\mathrm{1}}+\mathrm{2}\right){e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} }{{i}\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\left\{\:{cos}\left(\frac{{n}}{\mathrm{2}\:}\right)−{i}\:{sin}\left(\frac{{n}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left({n}\sqrt{\mathrm{4}{a}−\mathrm{1}}+\mathrm{2}\right){e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} }{{i}\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:\left\{….\right\}\:\Rightarrow \\ $$$${f}_{{n}} \left({a}\right)\:=\frac{\mathrm{2}\pi\left({n}\sqrt{\mathrm{4}{a}−\mathrm{1}}+\mathrm{2}\right){e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} }{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 11/Jun/19
2) we have Σ f_n (a) =Σ((2π(√(4a−1)))/((4a−1)(√(4a−1)))) n e^(−(n/2)(√(4a−1))) +Σ ((4π)/((4a−1)(√(4a−1)))) e^(−(n/2)(√(4a−1))) =((2π)/((4a−1)(√(4a−1)))) Σ n (e^(−((√(4a−1))/2)) )^n +((4π)/((4a−1)(√(4a−1)))) (e^(−((√(4a−1))/2)) )^n we have ∣e^(−((√(4a−1))/2)) ∣<1 and Σ x^n , Σnx^n convergs ⇒Σ f_n converges
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\Sigma\:{f}_{{n}} \left({a}\right)\:=\Sigma\frac{\mathrm{2}\pi\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:{n}\:{e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} \:+\Sigma\:\frac{\mathrm{4}\pi}{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:{e}^{−\frac{{n}}{\mathrm{2}}\sqrt{\mathrm{4}{a}−\mathrm{1}}} \\ $$$$=\frac{\mathrm{2}\pi}{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:\Sigma\:{n}\:\left({e}^{−\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}} \right)^{{n}} \:\:+\frac{\mathrm{4}\pi}{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:\left({e}^{−\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}} \right)^{{n}} \\ $$$${we}\:{have}\:\mid{e}^{−\frac{\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}} \mid<\mathrm{1}\:\:\:\:{and}\:\Sigma\:{x}^{{n}} \:\:\:,\:\Sigma{nx}^{{n}} \:{convergs}\:\Rightarrow\Sigma\:{f}_{{n}} \:{converges} \\ $$
Commented by maxmathsup by imad last updated on 11/Jun/19
3) we have f_n ^′ (a) =∫_(−∞) ^(+∞) (∂/∂a)(((cos(nx))/((x^2 +x+a)^2 )))dx =−∫_(−∞) ^(+∞) ((2(x^2 +x+a))/((x^2 +x+a)^4 )) cos(nx)dx =−2 ∫_(−∞) ^(+∞) ((cos(nx))/((x^2 +x+a)^2 )) dx =−2g_n (a) ⇒ g_n (a) =−(1/2) f_n ^′ (a) rest to calculate f^′ n(a)....
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}_{{n}} ^{'} \left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{\partial}{\partial{a}}\left(\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{x}+{a}\right)^{\mathrm{2}} }\right){dx} \\ $$$$=−\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{2}\left({x}^{\mathrm{2}} \:+{x}+{a}\right)}{\left({x}^{\mathrm{2}} \:+{x}+{a}\right)^{\mathrm{4}} }\:{cos}\left({nx}\right){dx}\:=−\mathrm{2}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{x}+{a}\right)^{\mathrm{2}} }\:{dx}\:=−\mathrm{2}{g}_{{n}} \left({a}\right)\:\Rightarrow \\ $$$${g}_{{n}} \left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\:{f}_{{n}} ^{'} \left({a}\right)\:\:\:\:{rest}\:{to}\:{calculate}\:{f}^{'} {n}\left({a}\right)…. \\ $$

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