Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 84693 by Power last updated on 15/Mar/20

Commented by abdomathmax last updated on 15/Mar/20

$$let\phi \left(t\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(tx\right)}{x(x^2+1)}dxwehave$$$${\phi }^{{}^{\prime }}\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{cos\left(tx\right)}{x^2+1}dx\Rightarrow 2{\phi }^{{}^{\prime }}\left(t\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(tx\right)}{x^2+1}dx$$$$=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx\right)residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx=2i\pi Res(f,i)=2i\pi \times \frac{e^{-t}}{2i}$$$$=\pi e^{-t}\Rightarrow 2{\phi }^{{}^{\prime }}\left(t\right)=\pi e^{-t}\Rightarrow {\phi }^{{}^{\prime }}\left(t\right)=\frac{\pi }{2}{e}^{-t}\Rightarrow$$$$\phi \left(t\right)=k-\frac{\pi }{2}{e}^{-t}$$$$\phi \left(0\right)=0=k-\frac{\pi }{2}\Rightarrow k=\frac{\pi }{2}\Rightarrow \phi \left(t\right)=\frac{\pi }{2}-\frac{\pi }{2}{e}^{-t}$$$$I={\int }_0^{\mathrm{\infty }}\frac{sinx}{x(1+x^2)}dx=\phi \left(1\right)=\frac{\pi }{2}-\frac{\pi }{2e}$$

Answered by Joel578 last updated on 15/Mar/20

$$I\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(tx\right)}{x(1+x^2)}dx$$$$\Rightarrow I{}^{\prime }\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left(tx\right)}{1+x^2}dx$$$$\Rightarrow I{}^{″}\left(t\right)=-{\int }_0^{\mathrm{\infty }}\frac{x\sin \left(tx\right)}{1+x^2}dx=-{\int }_0^{\mathrm{\infty }}\frac{(x^2+1-1)\sin \left(tx\right)}{x(1+x^2)}dx$$$$=-{\int }_0^{\mathrm{\infty }}\frac{\sin \left(tx\right)}{x}dx+{\int }_0^{\mathrm{\infty }}\frac{\sin \left(tx\right)}{x(1+x^2)}dx$$$$\Rightarrow I{}^{″}\left(t\right)=-\frac{\pi }{2}+I\left(t\right)$$$$\Rightarrow I{}^{″}-I=-\frac{\pi }{2}$$$$\mathrm{Solving}\mathrm{non}-\mathrm{homogeneous}\mathrm{ODE}\mathrm{gives}$$$$\Rightarrow I\left(t\right)=C_1{e}^t+C_2{e}^{-t}+\frac{\pi }{2}$$$$\Rightarrow I{}^{\prime }\left(t\right)=C_1{e}^t-C_2{e}^{-t}$$$$\mathrm{Note}\mathrm{that}I\left(0\right)={\int }_0^{\mathrm{\infty }}0dx=C_1+C_2+\frac{\pi }{2}=0$$$$\mathrm{and}I{}^{\prime }\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}dx=C_1-C_2=\frac{\pi }{2}$$$$\Rightarrow C_1=0,C_2=-\frac{\pi }{2}$$$$\Rightarrow I\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(tx\right)}{x(1+x^2)}dx=-\frac{\pi }{2}{e}^{-t}+\frac{\pi }{2}$$$$\Rightarrow I\left(1\right)={\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x(1+x^2)}dx=-\frac{\pi }{2e}+\frac{\pi }{2}$$

Commented by Power last updated on 15/Mar/20

$$\mathrm{thank}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com