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Question Number 130417 by Lordose last updated on 25/Jan/21

$${\int }_0^{\mathrm{\infty }}\frac{\sin \left(\alpha \mathrm{x}\right)\sin \left(\beta \mathrm{x}\right)}{{\mathrm{x}}^2}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\alpha \mathrm{x}\right)\sin \left(\beta \mathrm{x}\right)}{{\mathrm{x}}^2}\mathrm{dx}\Rightarrow \mathrm{I}{=}_{\alpha \mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{sintsin}\left(\frac{\beta }{\alpha }\mathrm{t}\right)}{\frac{{\mathrm{t}}^2}{{\alpha }^2}}\times \frac{\mathrm{dt}}{\alpha }$$$$={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{t}\right)\sin \left(\frac{\beta }{\alpha }\mathrm{t}\right)}{{\mathrm{t}}^2}\mathrm{dt}(\mathrm{we}\mathrm{suppose}\alpha >0\mathrm{and}\beta >0)\mathrm{let}\lambda =\frac{\beta }{\alpha }\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sintsin}\left(\lambda \mathrm{t}\right)}{{\mathrm{t}}^2}\mathrm{dt}\mathrm{by}\mathrm{parts}\mathrm{we}\mathrm{get}$$$$\mathrm{I}={[-\frac{1}{\mathrm{t}}\mathrm{sint}\sin \left(\lambda \mathrm{t}\right)]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{t}}(\mathrm{costsin}\left(\lambda \mathrm{t}\right)+\lambda \mathrm{sint}\cos \left(\lambda \mathrm{t}\right))\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{costsin}\left(\lambda \mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}+\lambda {\int }_0^{\mathrm{\infty }}\frac{\mathrm{sintcos}\left(\lambda \mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}=\mathrm{u}\left(\lambda \right)+\lambda \mathrm{v}\left(\lambda \right)$$$$\mathrm{we}\mathrm{have}\sin (\mathrm{x}+\mathrm{y})=\mathrm{sinx}\mathrm{cosy}+\mathrm{cosx}\mathrm{siny}$$$$\sin (\mathrm{x}-\mathrm{y})=\mathrm{sinxcosy}-\mathrm{cosxsiny}\Rightarrow \mathrm{sinx}\mathrm{cosy}=\frac{1}{2}(\sin (\mathrm{x}+\mathrm{h})+\sin (\mathrm{x}-\mathrm{y}))\Rightarrow$$$$\mathrm{u}\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{t}}(\sin (\lambda +1)\mathrm{t}+\sin (\lambda -1)\mathrm{t})\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{\sin (\lambda +1)\mathrm{t}}{\mathrm{t}}+{\int }_0^{\mathrm{\infty }}\frac{\sin (\lambda -1)\mathrm{t}}{\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{know}\lambda >0\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\sin (\lambda +1)\mathrm{t}}{\mathrm{t}}\mathrm{dt}{=}_{(\lambda +1)\mathrm{t}=\mathrm{y}}(\lambda +1){\int }_0^{\mathrm{\infty }}\frac{\mathrm{siny}}{\mathrm{y}}\frac{\mathrm{dy}}{\lambda +1}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{siny}}{\mathrm{y}}\mathrm{dy}=\frac{\pi }{2}$$$$\mathrm{if}\lambda >1{\int }_0^{\mathrm{\infty }}\frac{\sin (\lambda -1)\mathrm{t}}{\mathrm{t}}\mathrm{dt}=\frac{\pi }{2}\Rightarrow \mathrm{I}=\frac{\pi }{2}+\lambda \frac{\pi }{2}=\frac{\pi }{2}(\lambda +1)=\frac{\pi }{2}(\frac{\beta }{\alpha }+1)$$$$\mathrm{if}0<\lambda <1\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\sin (\lambda -1)\mathrm{t}}{\mathrm{t}}\mathrm{dt}=-{\int }_0^{\mathrm{\infty }}\frac{\sin (1-\lambda )\mathrm{t}}{\mathrm{t}}\mathrm{dt}=-\frac{\pi }{2}\Rightarrow$$$$\mathrm{I}=\frac{\pi }{2}-\lambda \frac{\pi }{2}=\frac{\pi }{2}(1-\lambda )=\frac{\pi }{2}(1-\frac{\beta }{\alpha })$$

Commented by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{sorry}\mathrm{I}=\alpha {\int }_0^{\mathrm{\infty }}\frac{\mathrm{sintsin}\left(\lambda \mathrm{t}\right)}{{\mathrm{t}}^2}\Rightarrow$$$$\mathrm{I}=\frac{\pi }{2}(\alpha +\beta )\mathrm{or}\mathrm{I}=\frac{\pi }{2}(\alpha -\beta )....$$

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