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Question Number 38727 by maxmathsup by imad last updated on 28/Jun/18

let n from N and A_n = ∫_(−∞) ^(+∞) ((cos(ax))/((x^2 +x+1)^n ))dx and B_n = ∫_(−∞) ^(+∞) ((sin(ax))/((x^2 +x+1)^n ))dx find the value of A_(n ) and B_n .

$${let}\:{n}\:{from}\:{N}\:\:{and}\:\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left({ax}\right)}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }{dx}\:\:{and} \\ $$$${B}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({ax}\right)}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }{dx} \\ $$$${find}\:{the}\:{value}\:{of}\:{A}_{{n}\:} \:\:{and}\:{B}_{{n}} \:\:. \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 01/Jul/18

we have A_n +iB_n = ∫_(−∞) ^(+∞) (e^(iax) /((x^2 +x+1)^n ))dx =W_n W_n = ∫_(−∞) ^(+∞) (e^(iax) /(((x+(1/2))^2 +(3/4))^n ))dx changement x+(1/2)=((√3)/2) t give W_n = ((4/3))^n ∫_(−∞) ^(+∞) (e^(ia(((√3)/2)t−(1/2))) /((t^2 +1)^n )) ((√3)/2)dt =((√3)/2)((4/3))^n e^(−((ia)/2)) ∫_(−∞) ^(+∞) (e^(i((√3)/2)at) /((t^2 +1)^n ))dt let λ =((√3)/2)a and find ∫_(−∞) ^(+∞) (e^(iλt) /((t^2 +1)^n )) =I_λ let ϕ(z)= (e^(iλz) /((z^2 +1)^n )) ϕ(z) = (e^(iλz) /((z−i)^n (z+i)^n )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((n−1)!)) { (z−i)^n ϕ(z)}^((n−1)) lim_(z→i) (1/((n−1)!)) { e^(iλz) (z+i)^(−n) }^((n−1)) but {e^(iλz) (z+i)^((−n)) }^((p)) =Σ_(k=0) ^p C_p ^k {(z+i)^((−n)) }^((k)) (e^(iλz) )^((p−k)) ...be continued...

$${we}\:{have}\:{A}_{{n}} \:+{iB}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{iax}} }{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }{dx}\:={W}_{{n}} \\ $$$${W}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{e}^{{iax}} }{\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} }{dx}\:{changement} \\ $$$${x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{t}\:\:{give}\: \\ $$$${W}_{{n}} =\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ia}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:{e}^{−\frac{{ia}}{\mathrm{2}}} \:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{e}^{{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{at}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }{dt}\:{let}\:\lambda\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{a}\: \\ $$$${and}\:{find}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\lambda{t}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:={I}_{\lambda} \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{{e}^{{i}\lambda{z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\lambda{z}} }{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{\:\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$${lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\:\left\{\:{e}^{{i}\lambda{z}} \:\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:\:\:\:{but} \\ $$$$\left\{{e}^{{i}\lambda{z}} \:\left({z}+{i}\right)^{\left(−{n}\right)} \right\}^{\left({p}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}} ^{{k}} \:\left\{\left({z}+{i}\right)^{\left(−{n}\right)} \right\}^{\left({k}\right)} \:\left({e}^{{i}\lambda{z}} \right)^{\left({p}−{k}\right)} \\ $$$$...{be}\:{continued}... \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 02/Jul/18

let find ((z +i)^(−n) )^((k)) (z+i)^(−n) }^((1)) =−n(z+i)^(−n−1) {(z+i)^(−n) }^((2)) =(−1)^2 n(n+1) (z+i)^(−n−2) ⇒ {(z+i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) also we have (e^(iλz) )^((1)) =iλ e^(iλz) (e^(iλz) )^((2)) =(iλ)^2 e^(iλz) ⇒(e^(iλz) )^((p−k)) =(iλ)^(p−k) e^(iλz) ⇒ {e^(iλz) (z+i)^(−n) }^((p)) =Σ_(k=0) ^p C_p ^k (−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) (iλ)^(p−k) e^(iλz) p=n−1 ⇒ {e^(iλz) (z+i)^(−n) }^((n−1)) =Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(2n−1)(z+i)^(−2n+1) (iλ)^(n−1−k) e^(iλz) Res(ϕ,i) = (1/((n−1)!)) Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(2n−1)(2i)^(−2n+1) (iλ)^(n−1−k) e^(−λ)

$${let}\:{find}\:\left(\left({z}\:+{i}\right)^{−{n}} \right)^{\left({k}\right)} \\ $$$$\left.\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({z}+{i}\right)^{−{n}−\mathrm{2}} \:\Rightarrow \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({k}\right)} =\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)...\left({n}+{k}−\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{k}} \\ $$$${also}\:{we}\:{have}\:\left({e}^{{i}\lambda{z}} \right)^{\left(\mathrm{1}\right)} ={i}\lambda\:{e}^{{i}\lambda{z}} \\ $$$$\left({e}^{{i}\lambda{z}} \right)^{\left(\mathrm{2}\right)} =\left({i}\lambda\right)^{\mathrm{2}} \:{e}^{{i}\lambda{z}} \:\Rightarrow\left({e}^{{i}\lambda{z}} \right)^{\left({p}−{k}\right)} =\left({i}\lambda\right)^{{p}−{k}} \:{e}^{{i}\lambda{z}} \:\Rightarrow \\ $$$$\left\{{e}^{{i}\lambda{z}} \left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} =\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)...\left({n}+{k}−\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{k}} \left({i}\lambda\right)^{{p}−{k}} \:{e}^{{i}\lambda{z}} \\ $$$${p}={n}−\mathrm{1}\:\Rightarrow \\ $$$$\left\{{e}^{{i}\lambda{z}} \left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{n}\left({n}+\mathrm{1}\right)...\left(\mathrm{2}{n}−\mathrm{1}\right)\left({z}+{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \left({i}\lambda\right)^{{n}−\mathrm{1}−{k}} \:{e}^{{i}\lambda{z}} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)...\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \left({i}\lambda\right)^{{n}−\mathrm{1}−{k}} \:{e}^{−\lambda} \\ $$$$ \\ $$$$ \\ $$

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