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Question Number 36187 by prof Abdo imad last updated on 30/May/18
let I_n (x)= ∫_0 ^∞ ((t sin(t))/((t^2 +x^2 )^n ))dt 1) find a relation between I_(n+1) and I_n 2) calculate I_2 (x) and I_3 (x) 3) calculate ∫_0 ^∞ ((tsin(t))/((2+t^2 )^2 ))dt
$${let}\:{I}_{{n}} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{sin}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{{n}} }{dt}\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{relation}\:{between}\:{I}_{{n}+\mathrm{1}} \:\:{and}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{\mathrm{2}} \left({x}\right)\:{and}\:{I}_{\mathrm{3}} \left({x}\right)\: \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{tsin}\left({t}\right)}{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
3) let A = ∫_0 ^∞ ((t sint)/((2+t^2 )^2 )) dt changement t =(√2)x give A =∫_0 ^∞ (((√2)xsin((√2)x))/(4(1+x^2 ))) (√2)dx =(1/2) ∫_0 ^∞ ((xsin(x(√2)))/((x^2 +1)^2 ))dx =(1/4) ∫_(−∞) ^(+∞) ((x sin(x(√2)))/((x^2 +1)^2 ))dx ⇒ 4A = Im( ∫_(−∞) ^(+∞) ((x e^(ix(√2)) )/((x^2 +1)^2 ))) let considere the complex function ϕ(z) = ((z e^(iz(√2)) )/((z^2 +1)^2 )) ⇒ϕ(z) = ((z e^(iz(√2)) )/((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and −i (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1)! )) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(iz(√2)) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iz(√2)) +i(√2)z e^(iz(√2)) )(z+i)^2 −2(z+i)ze^(iz(√2)) )/((z+i)^4 )) =lim_(z→i) (((z+i) (e^(iz(√2)) +i(√2)z e^(iz(√2)) )−2 z e^(iz(√2)) )/((z+i)^3 )) =(((2i)( e^(−(√2)) −(√2)e^(−(√2)) )−2i e^(−(√2)) )/((2i)^3 )) ((−2(√2)i e^(−(√2)) )/(−8i)) = ((√2)/4) e^(−(√2)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((√2)/4) e^(−(√2)) =((iπ(√2))/2) e^(−(√2)) 4A =Im( ∫_(−∞) ^(+∞) ϕ(z)dz ) =((π(√2))/2) e^(−(√2)) ⇒ A =((π(√2))/8) e^(−(√2)) .
$$\left.\mathrm{3}\right)\:\:{let}\:\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}\:{sint}}{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:\:{changement}\:{t}\:=\sqrt{\mathrm{2}}{x}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\sqrt{\mathrm{2}}{xsin}\left(\sqrt{\mathrm{2}}{x}\right)}{\mathrm{4}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\sqrt{\mathrm{2}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xsin}\left({x}\sqrt{\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}\:{sin}\left({x}\sqrt{\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\mathrm{4}{A}\:=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}\:{e}^{{ix}\sqrt{\mathrm{2}}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\right)\:{let}\:{considere}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)\:=\:\:\frac{{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({doubles}\right)\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:\:{but} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!\:}\:\:\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\:\:\frac{{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\frac{\left({e}^{{iz}\sqrt{\mathrm{2}}} \:+{i}\sqrt{\mathrm{2}}{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} \right)\left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){ze}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left({z}+{i}\right)\:\left({e}^{{iz}\sqrt{\mathrm{2}}} +{i}\sqrt{\mathrm{2}}{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} \right)−\mathrm{2}\:{z}\:{e}^{{iz}\sqrt{\mathrm{2}}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2}{i}\right)\left(\:{e}^{−\sqrt{\mathrm{2}}} \:−\sqrt{\mathrm{2}}{e}^{−\sqrt{\mathrm{2}}} \right)−\mathrm{2}{i}\:{e}^{−\sqrt{\mathrm{2}}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$\frac{−\mathrm{2}\sqrt{\mathrm{2}}{i}\:{e}^{−\sqrt{\mathrm{2}}} }{−\mathrm{8}{i}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{−\sqrt{\mathrm{2}}} \:\:\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{−\sqrt{\mathrm{2}}} \:\:=\frac{{i}\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−\sqrt{\mathrm{2}}} \\ $$$$\mathrm{4}{A}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:\right)\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−\sqrt{\mathrm{2}}} \:\Rightarrow\:{A}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:{e}^{−\sqrt{\mathrm{2}}} \:\:. \\ $$
Commented by maxmathsup by imad last updated on 19/Aug/18
2) we have I_2 (x)= ∫_0 ^∞ ((t sint)/((t^2 +x^2 )^2 )) dt =(1/2) ∫_(−∞) ^(+∞) ((tsint)/((t^2 +x^2 )^2 ))dt ⇒ 2I_2 (x) =Im( ∫_(−∞) ^(+∞) ((t e^(it) )/((t^2 +x^2 )^2 ))dt)=Im(A(x)) changement t=xu give A(x) = ∫_(−∞) ^(+∞) ((xu e^(ixu) )/(x^4 (1+u^2 )^2 )) x du = (1/x^2 ) ∫_(−∞) ^(+∞) ((u e^(ixu) )/((u^2 +1)^2 )) du (we suppose x>0) let consider the complex function ϕ(z) =((z e^(ixz) )/((z^2 +1)^2 )) ϕ(z) =((z e^(ixz) )/((z−i)^2 (z+i)^2 )) the poles of ϕ are i and −i(doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) (1/((2−1!)) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(ixz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(ixz) +ixz e^(ixz) )(z+i)^2 −2(z+i)z e^(ixz) )/((z+i)^4 )) =lim_(z→i) (((1+ixz)e^(ixz) (z+i)−2z e^(ixz) )/((z+i)^3 )) =(((1−x)e^(−x) (2i)−2i e^(−x) )/((2i)^3 )) = ((2i(1−x)e^(−x) −2i e^(−x) )/(−8i)) =(((2−2x−2)e^(−x) )/(−8)) = ((x e^(−x) )/4) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((x e^(−x) )/4) =((iπ)/2) x e^(−x) ⇒ 2I_2 (x) = ((πx)/2) e^(−x) ⇒ I_2 (x) =((πx)/4) e^(−x) . if x<0 x =−α with α>0 ⇒I_2 (x)=I_2 (−α) =∫_0 ^∞ ((tsint)/((t^2 +α^2 )^2 )) =I_2 (α) =((πα)/4) e^(−α) =−((πx)/4) e^x .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{I}_{\mathrm{2}} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}\:{sint}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{tsint}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\mathrm{2}{I}_{\mathrm{2}} \left({x}\right)\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}\:{e}^{{it}} }{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\right)={Im}\left({A}\left({x}\right)\right)\:\:{changement}\:{t}={xu}\:{give} \\ $$$${A}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{xu}\:{e}^{{ixu}} }{{x}^{\mathrm{4}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{x}\:{du}\:\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{u}\:{e}^{{ixu}} }{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{du}\:\left({we}\:{suppose}\:{x}>\mathrm{0}\right) \\ $$$${let}\:{consider}\:{the}\:{complex}\:{function}\:\varphi\left({z}\right)\:=\frac{{z}\:{e}^{{ixz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\frac{{z}\:{e}^{{ixz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\left({doubles}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:\:{but} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}!\right.}\:\left\{\:\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\:\frac{{z}\:{e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \:\:\frac{\left({e}^{{ixz}} \:\:+{ixz}\:{e}^{{ixz}} \right)\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){z}\:{e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left(\mathrm{1}+{ixz}\right){e}^{{ixz}} \left({z}+{i}\right)−\mathrm{2}{z}\:{e}^{{ixz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{1}−{x}\right){e}^{−{x}} \left(\mathrm{2}{i}\right)−\mathrm{2}{i}\:{e}^{−{x}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:\:=\:\frac{\mathrm{2}{i}\left(\mathrm{1}−{x}\right){e}^{−{x}} \:−\mathrm{2}{i}\:{e}^{−{x}} }{−\mathrm{8}{i}} \\ $$$$=\frac{\left(\mathrm{2}−\mathrm{2}{x}−\mathrm{2}\right){e}^{−{x}} }{−\mathrm{8}}\:=\:\frac{{x}\:{e}^{−{x}} }{\mathrm{4}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{x}\:{e}^{−{x}} }{\mathrm{4}}\:=\frac{{i}\pi}{\mathrm{2}}\:{x}\:{e}^{−{x}} \:\Rightarrow \\ $$$$\mathrm{2}{I}_{\mathrm{2}} \left({x}\right)\:=\:\frac{\pi{x}}{\mathrm{2}}\:{e}^{−{x}} \:\Rightarrow\:{I}_{\mathrm{2}} \left({x}\right)\:=\frac{\pi{x}}{\mathrm{4}}\:{e}^{−{x}} \:\:. \\ $$$${if}\:\:{x}<\mathrm{0}\:\:{x}\:=−\alpha\:\:{with}\:\alpha>\mathrm{0}\:\Rightarrow{I}_{\mathrm{2}} \left({x}\right)={I}_{\mathrm{2}} \left(−\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tsint}}{\left({t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:={I}_{\mathrm{2}} \left(\alpha\right) \\ $$$$=\frac{\pi\alpha}{\mathrm{4}}\:{e}^{−\alpha} \:\:=−\frac{\pi{x}}{\mathrm{4}}\:{e}^{{x}} \:. \\ $$
Commented by maxmathsup by imad last updated on 19/Aug/18
error at the final line we have A(x)=(1/x^2 ) ∫_(−∞) ^(+∞) ϕ(z)dz =(1/x^2 ) ((iπ)/2) x e^(−x) =((iπ)/(2x)) e^(−x) ⇒I_2 (x)=(1/2) Im(A(x))=(1/2) (π/(2x)) e^(−x) ⇒ I_2 (x) =(π/(4x)) e^(−x) with x>0 and if x<0 I_2 (x)=−(π/(4x)) e^x .
$${error}\:{at}\:{the}\:{final}\:{line}\:\:{we}\:{have}\:{A}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\: \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\frac{{i}\pi}{\mathrm{2}}\:{x}\:{e}^{−{x}} \:=\frac{{i}\pi}{\mathrm{2}{x}}\:{e}^{−{x}} \:\:\Rightarrow{I}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{Im}\left({A}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi}{\mathrm{2}{x}}\:{e}^{−{x}} \:\Rightarrow \\ $$$${I}_{\mathrm{2}} \left({x}\right)\:=\frac{\pi}{\mathrm{4}{x}}\:{e}^{−{x}} \:\:\:{with}\:{x}>\mathrm{0}\:\:{and}\:{if}\:{x}<\mathrm{0} \\ $$$${I}_{\mathrm{2}} \left({x}\right)=−\frac{\pi}{\mathrm{4}{x}}\:{e}^{{x}} \:\:\:. \\ $$

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