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En-utilisant-la-transforme-de-laplace-Calculer-0-tsin-xt-a-2-t-2-dt-a-x-R-

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Question Number 144057 by lapache last updated on 21/Jun/21
En utilisant la transforme de laplace Calculer ∫_0 ^(+∞) ((tsin(xt))/(a^2 +t^2 ))dt ∀a,x∈R^∗
$${En}\:{utilisant}\:{la}\:{transforme}\:{de}\:{laplace} \\ $$$${Calculer} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{tsin}\left({xt}\right)}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:\:\:\forall{a},{x}\in\mathbb{R}^{\ast} \\ $$
Answered by qaz last updated on 21/Jun/21
I(x)=∫_0 ^∞ ((tsin (xt))/(a^2 +t^2 ))dt L(I(x))(s)=∫_0 ^∞ ∫_0 ^∞ ((tsin (xt))/(a^2 +t^2 ))e^(−sx) dtdx =∫_0 ^∞ (t/(a^2 +t^2 ))∙L(sin (xt))(s)dt =∫_0 ^∞ (t/(a^2 +t^2 ))∙(t/(s^2 +t^2 ))dt =(1/(s^2 −a^2 ))∫_0 ^∞ ((s^2 /(s^2 +t^2 ))−(a^2 /(a^2 +t^2 )))dt =(1/(s^2 −a^2 ))(stan^(−1) (t/s)−atan^(−1) (t/a))∣_0 ^∞ =(1/(s^2 −a^2 ))∙(π/2)(s−a) =(π/(2(s+a))) I(x)=L^(−1) ((π/(2(s+a))))=(π/2)e^(−ax)
$$\mathrm{I}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tsin}\:\left(\mathrm{xt}\right)}{\mathrm{a}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathscr{L}\left(\mathrm{I}\left(\mathrm{x}\right)\right)\left(\mathrm{s}\right)=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tsin}\:\left(\mathrm{xt}\right)}{\mathrm{a}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\mathrm{e}^{−\mathrm{sx}} \mathrm{dtdx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}}{\mathrm{a}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\centerdot\mathscr{L}\left(\mathrm{sin}\:\left(\mathrm{xt}\right)\right)\left(\mathrm{s}\right)\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}}{\mathrm{a}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\centerdot\frac{\mathrm{t}}{\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{s}^{\mathrm{2}} }{\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\left(\mathrm{stan}^{−\mathrm{1}} \frac{\mathrm{t}}{\mathrm{s}}−\mathrm{atan}^{−\mathrm{1}} \frac{\mathrm{t}}{\mathrm{a}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\centerdot\frac{\pi}{\mathrm{2}}\left(\mathrm{s}−\mathrm{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{2}\left(\mathrm{s}+\mathrm{a}\right)} \\ $$$$\mathrm{I}\left(\mathrm{x}\right)=\mathscr{L}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{2}\left(\mathrm{s}+\mathrm{a}\right)}\right)=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{ax}} \\ $$
Answered by mathmax by abdo last updated on 21/Jun/21
residus method Φ=∫_0 ^∞ ((tsin(xt))/(a^2 +t^2 ))dt ⇒ Φ=_(t=au) ∫_0 ^∞ ((ausin(xau))/(a^2 (1+u^2 )))(adu) =∫_0 ^∞ ((usin(xau))/(u^2 +1))du (we suppose a>0) Φ=(1/2)∫_(−∞) ^(+∞) ((ue^(ixau) )/(u^2 +1))du let ϕ(z)=((ze^(ixaz) )/(z^2 +1)) ⇒ϕ(z)=((ze^(ixaz) )/((z−i)(z+i))) ∫_R ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×((ie^(−xa) )/(2i))=πi e^(−xa) ⇒ Φ=(π/2)e^(−xa) if a<0 we do the changement t=−au generally we get Φ=(π/2)e^(−x∣a∣)
$$\mathrm{residus}\:\mathrm{method}\:\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\Rightarrow \\ $$$$\Phi=_{\mathrm{t}=\mathrm{au}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ausin}\left(\mathrm{xau}\right)}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\left(\mathrm{adu}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{usin}\left(\mathrm{xau}\right)}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{du}\:\:\:\:\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{a}>\mathrm{0}\right) \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{ue}^{\mathrm{ixau}} }{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{du}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{ixaz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{ixaz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{ie}^{−\mathrm{xa}} }{\mathrm{2i}}=\pi\mathrm{i}\:\mathrm{e}^{−\mathrm{xa}} \:\Rightarrow \\ $$$$\Phi=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{xa}} \:\:\:\:\mathrm{if}\:\mathrm{a}<\mathrm{0}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{t}=−\mathrm{au} \\ $$$$\mathrm{generally}\:\mathrm{we}\:\mathrm{get}\:\Phi=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{x}\mid\mathrm{a}\mid} \\ $$

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