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Question Number 64525 by mathmax by abdo last updated on 19/Jul/19
study the convergence of Σ U_n with U_n =∫_0 ^∞ ((cos(nx))/(x^2 +n^2 ))dx (n≥1)
$${study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \:\:\:{with} \\ $$$${U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({nx}\right)}{{x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{dx}\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
we have 2U_n =∫_(−∞) ^(+∞) ((cos(nx))/(x^2 +n^2 )) =_(x=nt) ∫_(−∞) ^(+∞) ((cos(n^2 t))/(n^2 (1+t^2 )))ndt =(1/n) ∫_(−∞) ^(+∞) ((cos(n^2 t))/(t^2 +1))dt =(1/n) Re( ∫_(−∞) ^(+∞) (e^(in^2 t) /(t^2 +1))dt) let ϕ(z)=(e^(in^2 z) /(z^2 +1)) we have ϕ(z) =(e^(inz^2 ) /((z−i)(z+i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (z−i)ϕ(z) =lim_(z→i) (e^(in^2 z) /(z+i)) =(e^(−n^2 ) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−n^2 ) /(2i)) =π e^(−n^2 ) ⇒2U_n =(π/n) e^(−n^2 ) ⇒ U_n =(π/(2n)) e^(−n^2 ) (n≥1) its claear that Σ_(n=1) ^∞ U_n ≤(π/2)Σ_(n=1) ^∞ e^(−n^2 ) ≤(π/2)Σ_(n=1) ^∞ e^(−n) and this serie is convergent ⇒ΣU_n converges
$${we}\:{have}\:\:\mathrm{2}{U}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({nx}\right)}{{x}^{\mathrm{2}} +{n}^{\mathrm{2}} }\:=_{{x}={nt}} \:\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({n}^{\mathrm{2}} {t}\right)}{{n}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{ndt} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({n}^{\mathrm{2}} {t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\frac{\mathrm{1}}{{n}}\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{in}^{\mathrm{2}} {t}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\right)\:{let}\:\varphi\left({z}\right)=\frac{{e}^{{in}^{\mathrm{2}} {z}} }{{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\frac{{e}^{{inz}^{\mathrm{2}} } }{\left({z}−{i}\right)\left({z}+{i}\right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{{e}^{{in}^{\mathrm{2}} {z}} }{{z}+{i}}\:=\frac{{e}^{−{n}^{\mathrm{2}} } }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{n}^{\mathrm{2}} } }{\mathrm{2}{i}}\:=\pi\:{e}^{−{n}^{\mathrm{2}} \:} \:\Rightarrow\mathrm{2}{U}_{{n}} =\frac{\pi}{{n}}\:{e}^{−{n}^{\mathrm{2}} } \:\Rightarrow \\ $$$${U}_{{n}} =\frac{\pi}{\mathrm{2}{n}}\:{e}^{−{n}^{\mathrm{2}} } \:\:\:\:\:\left({n}\geqslant\mathrm{1}\right)\:\:{its}\:{claear}\:{that}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{U}_{{n}} \leqslant\frac{\pi}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}^{\mathrm{2}} } \\ $$$$\leqslant\frac{\pi}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} \:\:\:\:{and}\:{this}\:{serie}\:{is}\:{convergent}\:\Rightarrow\Sigma{U}_{{n}} \:{converges} \\ $$

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