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Question Number 42305 by maxmathsup by imad last updated on 22/Aug/18

let f(x) = ∫_(−∞) ^(+∞) ((cos(xt))/((t−i)^2 )) dt 1) let R =Re(f(x)) and I =Im(f(x)) extract R and I 2) calculate R and I 3) conclude the value of f(x) 4) calculate ∫_(−∞) ^(+∞) ((cos(2t))/((t−i)^2 ))dt 5) let u_n = ∫_(−∞) ^(+∞) ((cos((t/n)))/((t−i)^2 ))dt (n natral integer not o) find lim_(n→+∞) u_n and study the convergence of Σu_n

$${let}\:{f}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({xt}\right)}{\left({t}−{i}\right)^{\mathrm{2}} }\:{dt} \\ $$$$\left.\mathrm{1}\right)\:{let}\:{R}\:={Re}\left({f}\left({x}\right)\right)\:{and}\:{I}\:={Im}\left({f}\left({x}\right)\right)\:{extract}\:{R}\:{and}\:{I} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{R}\:{and}\:{I} \\ $$$$\left.\mathrm{3}\right)\:{conclude}\:{the}\:{value}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{\left({t}−{i}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{5}\right)\:{let}\:{u}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\frac{{t}}{{n}}\right)}{\left({t}−{i}\right)^{\mathrm{2}} }{dt}\:\:\:\:\left({n}\:{natral}\:{integer}\:{not}\:{o}\right) \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} \:\:\:\:{and}\:\:{study}\:{the}\:{convergence}\:{of}\:\Sigma{u}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 22/Aug/18

i^2 =−1

$${i}^{\mathrm{2}} \:=−\mathrm{1} \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

1) we have f(x) = ∫_(−∞) ^(+∞) ((cos(xt)(t+i)^2 )/((t−i)^2 (t+i)^2 ))dt =∫_(−∞) ^(+∞) (((t^2 +2it −1)cos(xt))/((t^2 +1)^2 ))dt = ∫_(−∞) ^(+∞) (((t^2 −1)cos(xt))/((t^2 +1)^2 )) dt +i ∫_(−∞) ^(+∞) ((2t cos(xt))/((t^2 +1)^2 ))dt ⇒ Re(f(x)) = ∫_(−∞) ^(+∞) (((t^2 −1)cos(xt))/((t^2 +1)^2 ))dt and Im (f(x)) = ∫_(−∞) ^(+∞) ((2t cos(xt))/((t^2 +1)^2 )) dt(=0)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({xt}\right)\left({t}+{i}\right)^{\mathrm{2}} }{\left({t}−{i}\right)^{\mathrm{2}} \left({t}+{i}\right)^{\mathrm{2}} }{dt}\:=\int_{−\infty} ^{+\infty} \:\:\frac{\left({t}^{\mathrm{2}} \:+\mathrm{2}{it}\:−\mathrm{1}\right){cos}\left({xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right){cos}\left({xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:+{i}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{2}{t}\:{cos}\left({xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$${Re}\left({f}\left({x}\right)\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right){cos}\left({xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:{and}\:{Im}\:\left({f}\left({x}\right)\right)\:=\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}{t}\:{cos}\left({xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\left(=\mathrm{0}\right) \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

2) let caoculate R =Re( ∫_(−∞) ^(+∞) (((x^2 −1)e^(ixt) )/((x^2 +1)^2 ))dt) let consider the complex function ϕ(z) =(((z^2 −1)e^(ixz) )/((z^2 +1)^2 )) we have ϕ(z) =(((z^2 −1)e^(ixz) )/((z−i)^2 (z+i)^2 ))( iand −i double poles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (((z^2 −1)e^(ixz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((2z +ix(z^2 −1))e^(ixz) (z+i)^2 −2(z+i)(z^2 −1)e^(ixz) )/((z+i)^4 )) =lim_(z→i) (((2z +ixz^2 −ix)e^(ixz) (z+i)−2(z^2 −1)e^(ixz) )/((z+i)^3 )) = (((2i−ix −ix)e^(−x) (2i) +4 e^(−x) )/((2i)^3 )) =((2i(2i−2ix)e^(−x) +4 e^(−x) )/(−8i)) =((−4(1−x)e^(−x) +4e^(−x) )/(−8i)) = ((4x e^(−x) )/(−8i)) =−(x/(2i)) e^(−x) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(−(x/(2i))e^(−x) ) = −πx e^(−x) ⇒ R =−π x e^(−x) also I =Im(f(x))=0

$$\left.\mathrm{2}\right)\:{let}\:{caoculate}\:{R}\:\:\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixt}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right)\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{we}\:{have}\:\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\left(\:\:\:{iand}\:−{i}\:{double}\:{poles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but}\:{Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\frac{\left(\mathrm{2}{z}\:\:+{ix}\left({z}^{\mathrm{2}} −\mathrm{1}\right)\right){e}^{{ixz}} \left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right)\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left(\mathrm{2}{z}\:+{ixz}^{\mathrm{2}} −{ix}\right){e}^{{ixz}} \left({z}+{i}\right)−\mathrm{2}\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{\left(\mathrm{2}{i}−{ix}\:−{ix}\right){e}^{−{x}} \left(\mathrm{2}{i}\right)\:\:+\mathrm{4}\:{e}^{−{x}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{\mathrm{2}{i}\left(\mathrm{2}{i}−\mathrm{2}{ix}\right){e}^{−{x}} \:+\mathrm{4}\:{e}^{−{x}} }{−\mathrm{8}{i}} \\ $$$$=\frac{−\mathrm{4}\left(\mathrm{1}−{x}\right){e}^{−{x}} \:+\mathrm{4}{e}^{−{x}} }{−\mathrm{8}{i}}\:=\:\frac{\mathrm{4}{x}\:{e}^{−{x}} }{−\mathrm{8}{i}}\:=−\frac{{x}}{\mathrm{2}{i}}\:{e}^{−{x}} \:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(−\frac{{x}}{\mathrm{2}{i}}{e}^{−{x}} \right) \\ $$$$=\:−\pi{x}\:{e}^{−{x}} \:\:\Rightarrow\:{R}\:\:=−\pi\:{x}\:{e}^{−{x}} \:\:\:\:{also}\:\:{I}\:={Im}\left({f}\left({x}\right)\right)=\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

3) we have f(x) =Re(f(x)) +i Im(f(x)) ⇒ f(x)=−πx e^(−x) .

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)\:={Re}\left({f}\left({x}\right)\right)\:+{i}\:{Im}\left({f}\left({x}\right)\right)\:\Rightarrow\:{f}\left({x}\right)=−\pi{x}\:{e}^{−{x}} \:\:. \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

4) ∫_(−∞) ^(+∞) ((cos(2t))/((t−i)^2 )) dt =f(2) = −2π e^(−2) =((−2π)/e^2 ) .

$$\left.\mathrm{4}\right)\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{\left({t}−{i}\right)^{\mathrm{2}} }\:{dt}\:\:={f}\left(\mathrm{2}\right)\:=\:−\mathrm{2}\pi\:{e}^{−\mathrm{2}} \:\:\:=\frac{−\mathrm{2}\pi}{{e}^{\mathrm{2}} }\:\:. \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

5) we have u_n = ∫_(−∞) ^(+∞) ((cos((t/n)))/((t−i)^2 ))dt =f((1/n)) =−(π/n) e^(−(1/n)) ⇒ lim_(n→+∞) u_n =0 and Σ_(n=1) ^∞ u_n =−Σ_(n=1) ^∞ (π/n) e^(−(1/n)) but e^u =1+u +o(u^2 )(u→0) ⇒ e^(−(1/n)) =1−(1/n) +o((1/n^2 )) ⇒(π/n) e^(−(1/n)) =(π/n) −(π/n^2 ) +o((1/n)) but Σ (π/n) diverges Σ (π/n^2 ) converges ⇒ Σ u_n diverges .

$$\left.\mathrm{5}\right)\:{we}\:{have}\:{u}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\frac{{t}}{{n}}\right)}{\left({t}−{i}\right)^{\mathrm{2}} }{dt}\:\:={f}\left(\frac{\mathrm{1}}{{n}}\right)\:=−\frac{\pi}{{n}}\:{e}^{−\frac{\mathrm{1}}{{n}}} \:\:\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} =\mathrm{0} \\ $$$${and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{u}_{{n}} =−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\pi}{{n}}\:{e}^{−\frac{\mathrm{1}}{{n}}} \:\:\:\:\:{but}\:\:\:{e}^{{u}} \:\:=\mathrm{1}+{u}\:+{o}\left({u}^{\mathrm{2}} \right)\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${e}^{−\frac{\mathrm{1}}{{n}}} \:\:=\mathrm{1}−\frac{\mathrm{1}}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:\Rightarrow\frac{\pi}{{n}}\:{e}^{−\frac{\mathrm{1}}{{n}}} \:\:\:=\frac{\pi}{{n}}\:−\frac{\pi}{{n}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:{but}\:\:\Sigma\:\frac{\pi}{{n}}\:{diverges} \\ $$$$\Sigma\:\frac{\pi}{{n}^{\mathrm{2}} }\:{converges}\:\Rightarrow\:\Sigma\:{u}_{{n}} \:\:{diverges}\:. \\ $$$$ \\ $$

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