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Question Number 133119 by abdomsup last updated on 19/Feb/21
find ∫_0 ^∞ ((xsin(2x))/((x^2 +4)^3 ))dx
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xsin}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{3}} }{dx} \\ $$
Answered by mindispower last updated on 19/Feb/21
let f(a)=∫_0 ^∞ ((xsin(2x))/(x^2 +a))dx,∀a∈R_+ −{0} f(a)=(1/2)Im∫_(−∞) ^∞ ((xe^(2ix) )/(x^2 +a))dx.....E ∫_(−∞) ^∞ ((xe^(2ix) )/(x^2 +a))=2iπRes(((xe^(2ix) )/(x^2 +a)),x=i(√a)) =2iπ.((i(√a).e^(−2(√a)) )/(2i(√a)))=iπe^(−2(√a)) E⇒f(a)=(π/2)e^(−2(√a)) f′(a)=∫_0 ^∞ ((−xsin(2x))/((x^2 +a)^2 ))dx f′′(a)=2∫((xsin(2x)dx)/((x^2 +a)^3 )) (1/2)f′′(4)=∫_0 ^∞ ((xsin(2x))/((x^2 +4)^2 ))dx...our integral f′(a)=−(π/(2(√a)))e^(−2(√a)) ,f′′(a)=(π/(2a))e^(−2(√a)) +(π/(4a(√a)))e^(−2(√a)) f′′(4)=((5π)/(32))e^(−4)
$${let} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{xsin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} +{a}}{dx},\forall{a}\in\mathbb{R}_{+} −\left\{\mathrm{0}\right\} \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}{Im}\int_{−\infty} ^{\infty} \frac{{xe}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} +{a}}{dx}…..{E} \\ $$$$\int_{−\infty} ^{\infty} \frac{{xe}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} +{a}}=\mathrm{2}{i}\pi{Res}\left(\frac{{xe}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} +{a}},{x}={i}\sqrt{{a}}\right) \\ $$$$=\mathrm{2}{i}\pi.\frac{{i}\sqrt{{a}}.{e}^{−\mathrm{2}\sqrt{{a}}} }{\mathrm{2}{i}\sqrt{{a}}}={i}\pi{e}^{−\mathrm{2}\sqrt{{a}}} \\ $$$${E}\Rightarrow{f}\left({a}\right)=\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{2}\sqrt{{a}}} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{−{xsin}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} }{dx} \\ $$$${f}''\left({a}\right)=\mathrm{2}\int\frac{{xsin}\left(\mathrm{2}{x}\right){dx}}{\left({x}^{\mathrm{2}} +{a}\right)^{\mathrm{3}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{f}''\left(\mathrm{4}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{xsin}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}…{our}\:{integral} \\ $$$${f}'\left({a}\right)=−\frac{\pi}{\mathrm{2}\sqrt{{a}}}{e}^{−\mathrm{2}\sqrt{{a}}} ,{f}''\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}{e}^{−\mathrm{2}\sqrt{{a}}} +\frac{\pi}{\mathrm{4}{a}\sqrt{{a}}}{e}^{−\mathrm{2}\sqrt{{a}}} \\ $$$${f}''\left(\mathrm{4}\right)=\frac{\mathrm{5}\pi}{\mathrm{32}}{e}^{−\mathrm{4}} \\ $$
Answered by mathmax by abdo last updated on 20/Feb/21
Φ=∫_0 ^∞ ((xsin(2x))/((x^2 +4)^3 ))dx ⇒Φ=_(x=2t) ∫_0 ^∞ ((2tsin(4t))/(64(t^2 +1)^3 ))(2dt) =(1/(16))∫_0 ^∞ ((tsin(4t))/((t^2 +1)^3 ))dt =(1/(32))∫_(−∞) ^(+∞) ((tsin(4t))/((t^2 +1)^3 ))dt =(1/(32))Im(∫_(−∞) ^(+∞) ((te^(4it) )/((t^2 +1)^3 ))dt) ϕ(z)=((ze^(4iz) )/((z^2 +1)^3 )) ⇒ϕ(z)=((ze^(4iz) )/((z−i)^3 (z+i)^3 )) ∫_R ϕ(z)dz =2iπ Res(ϕ,i) and Res(ϕ,i)=lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) {((ze^(4iz) )/((z+i)^3 ))}^((2)) ⇒2Res(ϕ,i)=lim_(z→i) { (((e^(4iz) +4iz e^(4iz) )(z+i)^3 −3(z+i)^2 ze^(4iz) )/((z+i)^6 ))}^((1)) lim_(z→i) {(((1+4iz)e^(4iz) (z+i)−3ze^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) {(((1+4iz)(z+i)−3z)e^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) {(((i+4iz^2 −3z)e^(4iz) )/((z+i)^4 ))}^((1)) =lim_(z→i) (((8iz−3)e^(4iz) +4i (4iz^2 −3z+1)e^(4iz) )(z+i)^4 −4(z+i)^3 (4iz^2 −3z+i)e^(4iz) )/((z+i)^8 )) ....rest to finish the calculus ...
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{xsin}\left(\mathrm{2x}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} }\mathrm{dx}\:\Rightarrow\Phi=_{\mathrm{x}=\mathrm{2t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2tsin}\left(\mathrm{4t}\right)}{\mathrm{64}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\left(\mathrm{2dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{tsin}\left(\mathrm{4t}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{32}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{tsin}\left(\mathrm{4t}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{32}}\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{te}^{\mathrm{4it}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dt}\right) \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{4iz}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{4iz}} }{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} } \\ $$$$\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\:\mathrm{and}\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{ze}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:\Rightarrow\mathrm{2Res}\left(\varphi,\mathrm{i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \left\{\:\:\frac{\left(\mathrm{e}^{\mathrm{4iz}} +\mathrm{4iz}\:\mathrm{e}^{\mathrm{4iz}} \right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} \mathrm{ze}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\left(\mathrm{1}+\mathrm{4iz}\right)\mathrm{e}^{\mathrm{4iz}} \left(\mathrm{z}+\mathrm{i}\right)−\mathrm{3ze}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\left.\left(\mathrm{1}+\mathrm{4iz}\right)\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{3z}\right)\mathrm{e}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\left(\mathrm{i}+\mathrm{4iz}^{\mathrm{2}} −\mathrm{3z}\right)\mathrm{e}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\frac{\left.\left(\mathrm{8iz}−\mathrm{3}\right)\mathrm{e}^{\mathrm{4iz}} \:+\mathrm{4i}\:\left(\mathrm{4iz}^{\mathrm{2}} −\mathrm{3z}+\mathrm{1}\right)\mathrm{e}^{\mathrm{4iz}} \right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{4iz}^{\mathrm{2}} −\mathrm{3z}+\mathrm{i}\right)\mathrm{e}^{\mathrm{4iz}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{8}} } \\ $$$$….\mathrm{rest}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{the}\:\mathrm{calculus}\:… \\ $$

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