Menu Close

let-f-x-0-cos-pixt-t-2-3x-2-2-dt-with-x-gt-0-1-find-a-explicit-form-for-f-x-2-find-the-value-of-0-cos-pit-t-2-3-2-dt-3-let-U-n-f-n-find-nature-of-U-n-

Tinku Tara 0 Comments
Pin




Question Number 57900 by maxmathsup by imad last updated on 13/Apr/19
let f(x) =∫_0 ^∞ ((cos(πxt))/((t^2 +3x^2 )^2 )) dt with x>0 1) find a explicit form for f(x) 2) find the value of ∫_0 ^∞ ((cos(πt))/((t^2 +3)^2 ))dt 3) let U_n =f(n) find nature of Σ U_n
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi{xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi{t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{U}_{{n}} ={f}\left({n}\right)\:\:{find}\:{nature}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 17/Apr/19
1) we have 2f(x) =∫_(−∞) ^(+∞) ((cos(πxt))/((t^2 +3x^2 )^2 ))dt =Re(∫_(−∞) ^(+∞) (e^(iπxt) /((t^2 +3x^2 )^2 ))dt) let ϕ(z) =(e^(iπxz) /((z^2 +3x^2 )^2 )) ⇒ ϕ(z) =(e^(iπxz) /((z−ix(√3))^2 (z+ix(√3))^2 )) so the poles of ϕ are +^− ix(√3) (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ix(√3)) Res(ϕ,ix(√3)) =lim_(z→ix(√3)) (1/((2−1)!)) {(z−ix(√3))^2 ϕ(z)}^((1)) =lim_(z→ix(√x)) { (e^(iπxz) /((z+ix(√3))^2 ))}^((1)) =lim_(z→ix(√3)) ((iπxe^(iπxz) (z+ix(√3))^2 −2(z+ix(√3))e^(iπxz) )/((z+ix(√3))^4 )) =lim_(z→ix(√3)) (({iπx(z+ix(√3))−2}e^(iπxz) )/((z+ix(√3))^3 )) (({iπx(2ix(√3))−2)e^(iπx(ix(√3))) )/((2ix(√3))^3 )) =(({−2π(√3)x^2 −2) e^(−π(√3)x^2 ) )/(−8i (3(√3)))) =(((π(√3)x^2 +1)e^(−π(√3)x^2 ) )/(12i(√3))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((π(√3)x^2 +1)e^(−π(√3)x^2 ) )/(12i(√3))) =(π/(6(√3))) (π(√3)x^2 +1)e^(−π(√3)x^2 ) ⇒ f(x) =(π/(12(√3)))(π(√3)x^2 +1)e^(−π(√3)x^2 ) . 2) ∫_0 ^∞ ((cos(πt))/((t^2 +3)^2 )) dt =f(1) =(π/(12(√3)))(π(√3) +1)e^(−π(√3))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{f}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\pi{xt}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\pi{xt}} }{\left({t}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\right)\:{let} \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\pi{xz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\Rightarrow\:\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\pi{xz}} }{\left({z}−{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left({z}+{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:\varphi \\ $$$${are}\:\overset{−} {+}{ix}\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\left({doubles}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ix}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left(\varphi,{ix}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{ix}\sqrt{\mathrm{3}}} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left({z}−{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{ix}\sqrt{{x}}} \:\:\:\:\left\{\:\:\frac{{e}^{{i}\pi{xz}} }{\left({z}+{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{ix}\sqrt{\mathrm{3}}} \:\:\:\:\:\frac{{i}\pi{xe}^{{i}\pi{xz}} \left({z}+{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{ix}\sqrt{\mathrm{3}}\right){e}^{{i}\pi{xz}} }{\left({z}+{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{ix}\sqrt{\mathrm{3}}} \:\:\:\:\:\:\frac{\left\{{i}\pi{x}\left({z}+{ix}\sqrt{\mathrm{3}}\right)−\mathrm{2}\right\}{e}^{{i}\pi{xz}} }{\left({z}+{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$\frac{\left\{{i}\pi{x}\left(\mathrm{2}{ix}\sqrt{\mathrm{3}}\right)−\mathrm{2}\right){e}^{{i}\pi{x}\left({ix}\sqrt{\mathrm{3}}\right)} }{\left(\mathrm{2}{ix}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:=\frac{\left\{−\mathrm{2}\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} −\mathrm{2}\right)\:{e}^{−\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} } }{−\mathrm{8}{i}\:\left(\mathrm{3}\sqrt{\mathrm{3}}\right)}\:=\frac{\left(\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} } }{\mathrm{12}{i}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\left(\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \:+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} } }{\mathrm{12}{i}\sqrt{\mathrm{3}}}\:=\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}}\:\left(\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \:+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \:} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\pi}{\mathrm{12}\sqrt{\mathrm{3}}}\left(\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \:+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}{x}^{\mathrm{2}} } \:\:. \\ $$$$\left.\mathrm{2}\right)\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\pi{t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:{dt}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{12}\sqrt{\mathrm{3}}}\left(\pi\sqrt{\mathrm{3}}\:+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *