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Question Number 55760 by maxmathsup by imad last updated on 03/Mar/19
let f(x) =∫_0 ^∞ ((cos(xt))/((xt^2 +i)^2 ))dx with x from R and x≠0 1) find a explicit form of f(x) 2) extract A =Re(f(x)) and B =Im(f(x)) and find its values . 3) calculate ∫_0 ^∞ ((cos(2t))/((2t^2 +i)^2 ))dt 4) let U_n =∫_0 ^∞ ((cos(nt))/((nt^2 +i)^2 ))dt .calculate lim_(n→+∞) u_n and study the convergence of Σu_n
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({xt}\right)}{\left({xt}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }{dx}\:\:\:{with}\:{x}\:{from}\:{R}\:\:{and}\:{x}\neq\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{extract}\:\:{A}\:={Re}\left({f}\left({x}\right)\right)\:{and}\:\:{B}\:={Im}\left({f}\left({x}\right)\right)\:{and}\:{find}\:{its}\:{values}\:. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{\left(\mathrm{2}{t}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{4}\right)\:{let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nt}\right)}{\left({nt}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} }{dt}\:\:\:.{calculate}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} \\ $$$${and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma{u}_{{n}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 04/Mar/19
i^2 =−1
$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 04/Mar/19
f(x)=∫_0 ^∞ ((cos(xt))/((xt^2 +i)^2 ))dt
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({xt}\right)}{\left({xt}^{\mathrm{2}} +{i}\right)^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 10/Mar/19
case 1 x>0 changement xt =u give f(x)=∫_0 ^∞ ((cosu)/((x(u^2 /x^2 )+i)^2 ))(du/x) =(1/x)∫_0 ^∞ ((cosu)/(((u^2 /x)+i)^2 )) =x∫_0 ^∞ ((cosu)/((u^2 +ix)^2 )) du=x ∫_0 ^∞ (((u^2 −ix)^2 cosu)/((u^4 +x^2 )^2 ))du =x ∫_0 ^∞ (((u^2 −2ix u^2 −x^2 )cosu)/((u^4 +x^2 )^2 )) du =x ∫_0 ^∞ (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 ))du −2ix^2 ∫_0 ^∞ ((u^2 cosu)/((u^4 +x^2 )^2 )) ⇒Re(f(x))=x∫_0 ^∞ (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 ))du =I and Im(f(x)) =−2x^2 ∫_0 ^∞ ((u^2 cosu)/((u^4 +x^2 ))) du let find I we have 2I =x ∫_(−∞) ^(+∞) (((u^2 −x^2 )cosu)/((u^4 +x^2 )^2 )) du =x Re(∫_(−∞) ^(+∞) (((u^2 −x^2 )e^(iu) )/((u^4 +x^2 )^2 )) du) let ϕ(z) =(((z^2 −x^2 )e^(iz) )/((z^4 +x^2 )^2 )) ⇒ϕ(z) =(((z^2 −x^2 )e^(iz) )/((z^2 −ix)^2 (z^2 +ix)^2 )) =(((z^2 −x^2 )e^(iz) )/((z−(√x)e^(i(π/4)) )^2 (z+(√x)e^((iπ)/4) )^2 (z−(√x)e^(−((iπ)/4)) )^2 (z+(√x)e^(−((iπ)/4)) )^2 )) so the poles of ϕ are +^− (√x)e^((iπ)/4) and +^− (√x)e^(−((iπ)/4)) (double poles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√x)e^((iπ)/4) ) +Res(ϕ,−(√x)e^(−((iπ)/4)) )} Res(ϕ, (√x)e^((iπ)/4) ) =lim_(z→(√x)e^((iπ)/4) ) (1/((2−1)!)) {(z−(√x)e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→(√x)e^((iπ)/4) ) { (((z^2 −x^2 )e^(iz) )/((z +(√x)e^((iπ)/4) )^2 (z^2 +ix)^2 ))}^((1)) ....be continued... mim
$${case}\:\mathrm{1}\:\:\:{x}>\mathrm{0}\:\:{changement}\:{xt}\:={u}\:\:{give}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosu}}{\left({x}\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+{i}\right)^{\mathrm{2}} }\frac{{du}}{{x}} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosu}}{\left(\frac{{u}^{\mathrm{2}} }{{x}}+{i}\right)^{\mathrm{2}} }\:={x}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosu}}{\left({u}^{\mathrm{2}} \:+{ix}\right)^{\mathrm{2}} }\:{du}={x}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left({u}^{\mathrm{2}} −{ix}\right)^{\mathrm{2}} {cosu}}{\left({u}^{\mathrm{4}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{du} \\ $$$$={x}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left({u}^{\mathrm{2}} −\mathrm{2}{ix}\:{u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du} \\ $$$$={x}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:−\mathrm{2}{ix}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\mathrm{2}} {cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow{Re}\left({f}\left({x}\right)\right)={x}\int_{\mathrm{0}} ^{\infty} \:\frac{\left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:={I} \\ $$$${and}\:{Im}\left({f}\left({x}\right)\right)\:=−\mathrm{2}{x}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\mathrm{2}} {cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)}\:{du}\:\:{let}\:{find}\:\:{I}\:\:\:{we}\:{have}\: \\ $$$$\mathrm{2}{I}\:={x}\:\int_{−\infty} ^{+\infty} \:\:\frac{\left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){cosu}}{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du}\:={x}\:{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){e}^{{iu}} }{\left({u}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){e}^{{iz}} }{\left({z}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\Rightarrow\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){e}^{{iz}} }{\left({z}^{\mathrm{2}} −{ix}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} +{ix}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({z}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){e}^{{iz}} }{\left({z}−\sqrt{{x}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}−\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\overset{−} {+}\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:{and}\:\overset{−} {+}\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\left({double}\:{poles}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,\:\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\:\:\left\{\left({z}−\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left\{\:\:\frac{\left({z}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){e}^{{iz}} }{\left({z}\:+\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{ix}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:….{be}\:{continued}… \\ $$$${mim} \\ $$

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