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Question Number 54830 by maxmathsup by imad last updated on 12/Feb/19
let V_n = ∫_0 ^∞ ((cos(nx))/(n +x^2 ))dx with n integr nstural not 0 . 1) calculate V_n 2)calculate lim_(n→+∞) nV_n 3) calculate the sum Σ_(n=0) ^∞ V_n
$${let}\:{V}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}\:+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:{n}\:{integr}\:{nstural}\:{not}\:\mathrm{0}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{V}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} {nV}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{the}\:{sum}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{V}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 13/Feb/19
1) we have V_n =_(x=(√n)t) ∫_0 ^∞ ((cos(n(√n)t))/(n(1+t^2 ))) (√n)dt =(1/( (√n))) ∫_0 ^∞ ((cos(n(√n)t))/(1+t^2 ))dt ⇒ 2(√n)V_n =∫_(−∞) ^(+∞) ((cos(n(√n)t))/(t^2 +1))dt =Re ( ∫_(−∞) ^(+∞) (e^(in(√n)t) /(t^2 +1))dt) let ϕ(z)=(e^(in(√n)z) /(z^2 +1)) the poles of ϕ are i and −i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but ϕ(z) =(e^(in(√n)z) /((z−i)(z+i))) Res(ϕ,i) =lim_(z→i) (z−i)ϕ(z) = (e^(−n(√n)) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−n(√n)) /(2i)) =π e^(−n(√n)) ⇒2(√n)V_n =π e^(−n(√n)) ⇒ V_n =(π/(2(√n))) e^(−n(√n)) 2) lim_(n→+∞) n V_n =lim_(n→+∞) (π/2) (√n) e^(−n(√n)) =0
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{V}_{{n}} =_{{x}=\sqrt{{n}}{t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({n}\sqrt{{n}}{t}\right)}{{n}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\sqrt{{n}}{dt}\:=\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({n}\sqrt{{n}}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{{n}}{V}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({n}\sqrt{{n}}{t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:={Re}\:\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{in}\sqrt{{n}}{t}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\right)\:\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{in}\sqrt{{n}}{z}} }{{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but}\:\varphi\left({z}\right)\:=\frac{{e}^{{in}\sqrt{{n}}{z}} }{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)\:=\:\frac{{e}^{−{n}\sqrt{{n}}} }{\mathrm{2}{i}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{n}\sqrt{{n}}} }{\mathrm{2}{i}} \\ $$$$=\pi\:{e}^{−{n}\sqrt{{n}}} \:\:\:\Rightarrow\mathrm{2}\sqrt{{n}}{V}_{{n}} =\pi\:{e}^{−{n}\sqrt{{n}}} \:\Rightarrow\:{V}_{{n}} =\frac{\pi}{\mathrm{2}\sqrt{{n}}}\:{e}^{−{n}\sqrt{{n}}} \\ $$$$\left.\mathrm{2}\right)\:{lim}_{{n}\rightarrow+\infty} {n}\:{V}_{{n}} ={lim}_{{n}\rightarrow+\infty} \:\:\frac{\pi}{\mathrm{2}}\:\sqrt{{n}}\:{e}^{−{n}\sqrt{{n}}} \:\:=\mathrm{0} \\ $$

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