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Question Number 144057 by lapache last updated on 21/Jun/21

$$Enutilisantlatransformedelaplace$$$$Calculer$$$${\int }_0^{+\mathrm{\infty }}\frac{tsin\left(xt\right)}{a^2+t^2}dt\mathrm{\forall }a,x\in {\mathbb{R}}^{\ast }$$

Answered by qaz last updated on 21/Jun/21

$$\mathrm{I}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{{\mathrm{a}}^2+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathcal{L}\left(\mathrm{I}\left(\mathrm{x}\right)\right)\left(\mathrm{s}\right)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{{\mathrm{a}}^2+{\mathrm{t}}^2}{\mathrm{e}}^{-\mathrm{sx}}\mathrm{dtdx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{{\mathrm{a}}^2+{\mathrm{t}}^2}\cdot \mathcal{L}\left(\sin \left(\mathrm{xt}\right)\right)\left(\mathrm{s}\right)\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{{\mathrm{a}}^2+{\mathrm{t}}^2}\cdot \frac{\mathrm{t}}{{\mathrm{s}}^2+{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{{\mathrm{s}}^2-{\mathrm{a}}^2}{\int }_0^{\mathrm{\infty }}(\frac{{\mathrm{s}}^2}{{\mathrm{s}}^2+{\mathrm{t}}^2}-\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+{\mathrm{t}}^2})\mathrm{dt}$$$$=\frac{1}{{\mathrm{s}}^2-{\mathrm{a}}^2}({\mathrm{stan}}^{-1}\frac{\mathrm{t}}{\mathrm{s}}-{\mathrm{atan}}^{-1}\frac{\mathrm{t}}{\mathrm{a}}){\mid }_0^{\mathrm{\infty }}$$$$=\frac{1}{{\mathrm{s}}^2-{\mathrm{a}}^2}\cdot \frac{\pi }{2}(\mathrm{s}-\mathrm{a})$$$$=\frac{\pi }{2(\mathrm{s}+\mathrm{a})}$$$$\mathrm{I}\left(\mathrm{x}\right)={\mathcal{L}}^{-1}\left(\frac{\pi }{2(\mathrm{s}+\mathrm{a})}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{ax}}$$

Answered by mathmax by abdo last updated on 21/Jun/21

$$\mathrm{residus}\mathrm{method}\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{{\mathrm{a}}^2+{\mathrm{t}}^2}\mathrm{dt}\Rightarrow$$$$\mathrm{\Phi }{=}_{\mathrm{t}=\mathrm{au}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{ausin}\left(\mathrm{xau}\right)}{{\mathrm{a}}^2(1+{\mathrm{u}}^2)}\left(\mathrm{adu}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{usin}\left(\mathrm{xau}\right)}{{\mathrm{u}}^2+1}\mathrm{du}(\mathrm{we}\mathrm{suppose}\mathrm{a}>0)$$$$\mathrm{\Phi }=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{ue}}^{\mathrm{ixau}}}{{\mathrm{u}}^2+1}\mathrm{du}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{ze}}^{\mathrm{ixaz}}}{{\mathrm{z}}^2+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{ze}}^{\mathrm{ixaz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})=2\mathrm{i}\pi \times \frac{{\mathrm{ie}}^{-\mathrm{xa}}}{2\mathrm{i}}=\pi \mathrm{i}{\mathrm{e}}^{-\mathrm{xa}}\Rightarrow$$$$\mathrm{\Phi }=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{xa}}\mathrm{if}\mathrm{a}<0\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{t}=-\mathrm{au}$$$$\mathrm{generally}\mathrm{we}\mathrm{get}\mathrm{\Phi }=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{x}\mid \mathrm{a}\mid }$$

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