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Question Number 130326 by rs4089 last updated on 24/Jan/21

Answered by Lordose last updated on 24/Jan/21

$$$$$$\mathrm{\Omega }\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{px}\right)}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}$$$${\mathrm{\Omega }}^{\prime }\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{px}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}$$$$\mathrm{\Omega }\left(\mathrm{p}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}+\mathrm{C}$$$$\mathrm{\Omega }\left(0\right)=\frac{\pi }{2}=\mathrm{C}$$$$\mathrm{\Omega }\left(\mathrm{p}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}+\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 24/Jan/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{px}\right)}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{px}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{px}\right)}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{1}{2}\mathrm{Re}\left({\int }_{\mathrm{R}}\frac{{\mathrm{e}}^{\mathrm{ipx}}}{{\mathrm{x}}^2+1}\mathrm{dx}\right)$$$$=\frac{1}{2}\mathrm{Re}\left(2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{-\mathrm{p}}}{2\mathrm{i}}\right)=\frac{1}{2}\mathrm{Re}\left(\pi {\mathrm{e}}^{-\mathrm{p}}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{p}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}+\mathrm{C}$$$$\mathrm{f}\left(0\right)=0=\frac{\pi }{2}+\mathrm{C}\Rightarrow \mathrm{C}=-\frac{\pi }{2}\Rightarrow \mathrm{f}\left(\mathrm{p}\right)=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{p}}-\frac{\pi }{2}$$

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