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Question Number 38470 by maxmathsup by imad last updated on 25/Jun/18
calculate f(t)=∫_0 ^∞ ((cos(tx))/((1+tx^2 )^2 )) dx with t≥0 2) find the values of ∫_0 ^∞ ((cos(2x))/((1+2x^2 )^2 ))dx and ∫_0 ^∞ ((cosx)/((2+x^2 )^2 ))dx
$${calculate}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
t>0
$${t}>\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 27/Jun/18
we have 2f(x)=∫_(−∞) ^(+∞) ((cos(tx))/((1+tx^2 )^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(itx) /((1+tx^2 )^2 ))dx) let consider ϕ(z) = (e^(itz) /((tz^2 +1)^2 )) we have ϕ(z)= (e^(itz) /(t^2 (z^2 +(1/t))^2 )) = (e^(itz) /(t^2 (z−(i/( (√t))))^2 (z+(i/( (√t))))^2 )) the poles of ϕ are (i/( (√t))) and ((−i)/( (√t))) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/( (√t)))) but Res(ϕ,(i/( (√t))))=lim_(z→(i/( (√t)))) (1/((2−1)!)){ (z−(i/( (√t))))^2 ϕ(z)}^((1)) =lim_(z→(i/( (√t)))) (1/t^2 ){ (e^(itz) /((z+(i/( (√t))))^2 ))}^((1)) =lim_(z→(i/( (√t)))) (1/t^2 ){ ((it e^(itz) (z+(i/( (√t))))^2 −2(z+(i/( (√t))))e^(itz) )/((z+(i/( (√t))))^4 ))} =lim_(z→(i/( (√t)))) (1/t^2 ){ ((it e^(itz) (z+(i/( (√t)))) −2 e^(itz) )/((z+(i/( (√t))))^3 ))} =(1/t^2 )((it e^(it(i/( (√t)))) (((2i)/( (√t))))−2 e^(it((i/( (√t))))) )/((((2i)/( (√t))))^3 ))=(1/t^2 ) ((−2(√t)e^(−(√t)) −2 e^(−(√t)) )/((−8i)/(t(√t)))) =((t(√t))/t^2 ) (((1+(√t))e^(−(√t)) )/(4i)) =((√t)/(4ti))(1+(√t))e^(−(√t)) =(((t+(√t))e^(−(√t)) )/(4ti)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((t+(√t))e^(−(√t)) )/(4it)) =(π/(2t)) (t+(√t))e^(−(√t)) 2f(x)= Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ f(x)= (π/(4t))(t+(√t))e^(−(√t)) (t>0) 2) ∫_0 ^∞ ((cos(2x))/((1+2x^2 )^2 ))dx=f(2) =(π/8)(2+(√2))e^(−(√2))
$${we}\:{have}\:\mathrm{2}{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{itx}} }{\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\left({tz}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)=\:\:\frac{{e}^{{itz}} }{{t}^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} }\:=\:\frac{{e}^{{itz}} }{{t}^{\mathrm{2}} \left({z}−\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} \left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\frac{{i}}{\:\sqrt{{t}}}\:{and}\:\frac{−{i}}{\:\sqrt{{t}}}\:\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{{t}}}\right)\:\:{but} \\ $$$${Res}\left(\varphi,\frac{{i}}{\:\sqrt{{t}}}\right)={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{{t}}}} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{{t}}}} \:\:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\:\:\:\frac{{e}^{{itz}} }{\left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{{t}}}} \:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\:\:\frac{{it}\:{e}^{{itz}} \left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+\frac{{i}}{\:\sqrt{{t}}}\right){e}^{{itz}} }{\left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\:\sqrt{{t}}}} \:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\:\:\frac{{it}\:{e}^{{itz}} \left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)\:−\mathrm{2}\:{e}^{{itz}} }{\left({z}+\frac{{i}}{\:\sqrt{{t}}}\right)^{\mathrm{3}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\frac{{it}\:{e}^{{it}\frac{{i}}{\:\sqrt{{t}}}} \left(\frac{\mathrm{2}{i}}{\:\sqrt{{t}}}\right)−\mathrm{2}\:{e}^{{it}\left(\frac{{i}}{\:\sqrt{{t}}}\right)} }{\left(\frac{\mathrm{2}{i}}{\:\sqrt{{t}}}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\frac{−\mathrm{2}\sqrt{{t}}{e}^{−\sqrt{{t}}} \:−\mathrm{2}\:{e}^{−\sqrt{{t}}} }{\frac{−\mathrm{8}{i}}{{t}\sqrt{{t}}}} \\ $$$$=\frac{{t}\sqrt{{t}}}{{t}^{\mathrm{2}} }\:\frac{\left(\mathrm{1}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} }{\mathrm{4}{i}}\:=\frac{\sqrt{{t}}}{\mathrm{4}{ti}}\left(\mathrm{1}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} \\ $$$$=\frac{\left({t}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} }{\mathrm{4}{ti}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left({t}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} }{\mathrm{4}{it}}\:=\frac{\pi}{\mathrm{2}{t}}\:\left({t}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} \\ $$$$\mathrm{2}{f}\left({x}\right)=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\pi}{\mathrm{4}{t}}\left({t}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} \:\:\:\:\:\:\:\:\left({t}>\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}={f}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{8}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){e}^{−\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 27/Jun/18
let calculate I =∫_0 ^∞ ((cosx)/((2+x^2 )^2 ))dx 2I =∫_(−∞) ^(+∞) ((cosx)/((2+x^2 )^2 ))dx=Re( ∫_(−∞) ^(+∞) (e^(ix) /((x^2 +2)^2 ))dx) let ϕ(z)= (e^(iz) /((z^2 +2)^2 )) ϕ(z)= (e^(iz) /((z−i(√2))^2 (z+i(√2))^2 )) so the polesof ϕ are i(√2) and −i(√2) (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i(√2)) Res(ϕ,i(√2))=lim_(z→i(√2)) (1/((2−1)!)){ (z−i(√2))^2 ϕ(z)}^((1)) =lim_(z→i(√2)) { (e^(iz) /((z+i(√2))^2 ))}^((1)) =lim_(z→i(√2)) ((i e^(iz) (z+i(√2))^2 −2(z+i(√2))e^(iz) )/((z+i(√2))^4 )) =lim_(z→i(√2)) ((i e^(iz) (z+i(√2)) −2 e^(iz) )/((z+i(√2))^3 )) =((i e^(i(i(√2))) (2i(√2)) −2 e^(i(i(√2))) )/((2i(√2))^3 )) = ((−2(√2) e^(−(√2)) −2 e^(−(√2)) )/(−8i(2(√2)))) =(((1+(√2))e^(−(√2)) )/(8i(√2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((1+(√2))e^(−(√2)) )/(8i(√2))) = (π/(4(√2)))(1+(√2))e^(−(√2)) 2I =Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ I = (π/(8(√2)))(1+(√2))e^(−(√2)) .
$${let}\:{calculate}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cosx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ix}} }{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} }{dx}\right) \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{{e}^{{iz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{iz}} }{\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:{so}\:{the}\:{polesof}\:\varphi\:{are} \\ $$$${i}\sqrt{\mathrm{2}}\:{and}\:−{i}\sqrt{\mathrm{2}}\:\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\:\left\{\:\:\frac{{e}^{{iz}} }{\left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\frac{{i}\:{e}^{{iz}} \left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\sqrt{\mathrm{2}}\right){e}^{{iz}} }{\left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\:\:\frac{{i}\:{e}^{{iz}} \left({z}+{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}\:{e}^{{iz}} }{\left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\frac{{i}\:{e}^{{i}\left({i}\sqrt{\mathrm{2}}\right)} \left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}\:{e}^{{i}\left({i}\sqrt{\mathrm{2}}\right)} }{\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{−\mathrm{2}\sqrt{\mathrm{2}}\:{e}^{−\sqrt{\mathrm{2}}} \:\:−\mathrm{2}\:{e}^{−\sqrt{\mathrm{2}}} }{−\mathrm{8}{i}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){e}^{−\sqrt{\mathrm{2}}} }{\mathrm{8}{i}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){e}^{−\sqrt{\mathrm{2}}} }{\mathrm{8}{i}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){e}^{−\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}{I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){e}^{−\sqrt{\mathrm{2}}} . \\ $$
Commented by abdo.msup.com last updated on 27/Jun/18
f(t)=(π/(4t))(t+(√t))e^(−(√t))
$${f}\left({t}\right)=\frac{\pi}{\mathrm{4}{t}}\left({t}+\sqrt{{t}}\right){e}^{−\sqrt{{t}}} \\ $$

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