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Question Number 56311 by maxmathsup by imad last updated on 13/Mar/19

let f(x) =∫_0 ^∞ ((cos(xt))/(x^2 +t^2 )) dt with x>0 1) find f(x) 2) find the values of ∫_0 ^∞ ((cos(t))/(1+t^2 ))dt and ∫_0 ^∞ ((cos(2t))/(4+t^2 ))dt 3) let U_n =∫_0 ^∞ ((cos(nt))/(n^2 +t^2 ))dt find lim_(n→+∞) U_n and study the convergenge of Σ U_n and Σ U_n ^2

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({xt}\right)}{{x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }\:{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{find}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{t}\right)}{\mathrm{4}+{t}^{\mathrm{2}} }{dt} \\ $$ $$\left.\mathrm{3}\right)\:{let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nt}\right)}{{n}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:\:\:{find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \:\:\:\:{and}\:{study}\:{the}\:{convergenge}\:{of} \\ $$ $$\Sigma\:{U}_{{n}} \:\:\:{and}\:\Sigma\:{U}_{{n}} ^{\mathrm{2}} \\ $$ $$ \\ $$

Commented bymaxmathsup by imad last updated on 15/Mar/19

1) we have 2f(x) =∫_(−∞) ^(+∞) ((cos(xt))/(t^2 +x^2 ))dt =Re(∫_(−∞) ^(+∞) (e^(ixt) /(t^2 +x^2 ))dt) let consider the complex function ϕ(z) =(e^(ixz) /(z^2 +x^2 )) ⇒ϕ(z)=(e^(ixz) /((z−ix)(z+ix))) so the poles of ϕ are +^− ix residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ix) Res(ϕ,ix) =lim_(z→ix) (z−ix)ϕ(z) =(e^(ix(ix)) /(2ix)) =(e^(−x^2 ) /(2ix)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−x^2 ) /(2ix)) =(π/x) e^(−x^2 ) ⇒★f(x) =(π/(2x))e^(−x^2 ) ★ 2) ∫_0 ^∞ ((cos(t))/(1+t^2 ))dt =f(1) =(π/(2e)) ∫_0 ^∞ ((cos(2t))/(4+t^2 )) =f(2) =(π/4) e^(−4) 3) we have U_n =f(n) =(π/(2n)) e^(−n^2 ) ⇒lim_(n→+∞) U_n =0 we have U_n >0 and U_n ≤(π/2) e^(−n^2 ) ≤ (π/2) e^(−n) (n>0) ⇒Σ_(n=1) ^∞ U_n ≤ (π/2)Σ_(n=1) ^∞ e^(−n) and this serie converges ⇒Σ U_n is convergente we have U_n ^2 =(π^2 /(4n^2 )) e^(−2n^2 ) ⇒U_n ^2 ≤ (π^2 /4) e^(−2n) ⇒Σ_(n=1) ^∞ U_n ^2 ≤(π^2 /4) Σ_(n=1) ^∞ e^(−2n) and this serie converges ⇒Σ U_n ^2 converges.

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{f}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({xt}\right)}{{t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dt}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ixt}} }{{t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dt}\right)\:{let}\:{consider}\:{the} \\ $$ $${complex}\:{function}\:\varphi\left({z}\right)\:=\frac{{e}^{{ixz}} }{{z}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{ixz}} }{\left({z}−{ix}\right)\left({z}+{ix}\right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$ $$\overset{−} {+}{ix}\:\:\:\:{residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ix}\right) \\ $$ $${Res}\left(\varphi,{ix}\right)\:={lim}_{{z}\rightarrow{ix}} \left({z}−{ix}\right)\varphi\left({z}\right)\:=\frac{{e}^{{ix}\left({ix}\right)} }{\mathrm{2}{ix}}\:=\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{ix}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{ix}} \\ $$ $$=\frac{\pi}{{x}}\:{e}^{−{x}^{\mathrm{2}} } \:\:\:\:\Rightarrow\bigstar{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}{x}}{e}^{−{x}^{\mathrm{2}} } \:\bigstar \\ $$ $$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}{e}} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{\mathrm{4}+{t}^{\mathrm{2}} }\:={f}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{4}}\:{e}^{−\mathrm{4}} \\ $$ $$\left.\mathrm{3}\right)\:{we}\:{have}\:{U}_{{n}} ={f}\left({n}\right)\:=\frac{\pi}{\mathrm{2}{n}}\:{e}^{−{n}^{\mathrm{2}} } \:\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} =\mathrm{0} \\ $$ $${we}\:{have}\:\:{U}_{{n}} >\mathrm{0}\:\:{and}\:\:\:{U}_{{n}} \leqslant\frac{\pi}{\mathrm{2}}\:{e}^{−{n}^{\mathrm{2}} } \leqslant\:\frac{\pi}{\mathrm{2}}\:{e}^{−{n}} \:\:\:\:\:\:\:\left({n}>\mathrm{0}\right)\:\: \\ $$ $$\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{U}_{{n}} \:\leqslant\:\frac{\pi}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} \:\:\:{and}\:{this}\:{serie}\:{converges}\:\Rightarrow\Sigma\:{U}_{{n}} \:{is}\:{convergente} \\ $$ $${we}\:{have}\:{U}_{{n}} ^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} }\:{e}^{−\mathrm{2}{n}^{\mathrm{2}} } \:\Rightarrow{U}_{{n}} ^{\mathrm{2}} \:\leqslant\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:{e}^{−\mathrm{2}{n}} \:\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{U}_{{n}} ^{\mathrm{2}} \:\leqslant\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−\mathrm{2}{n}} \:{and}\:{this} \\ $$ $${serie}\:{converges}\:\Rightarrow\Sigma\:{U}_{{n}} ^{\mathrm{2}} \:{converges}. \\ $$

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