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 101970 by Rohit@Thakur last updated on 05/Jul/20

$${\int }_0^{\mathrm{\infty }}\frac{Cos\left(ax\right)}{x^2+b^2}dx$$

Answered by mathmax by abdo last updated on 06/Jul/20

$$\mathrm{let}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{ax}\right)}{{\mathrm{x}}^2+{\mathrm{b}}^2}\mathrm{dx}\mathrm{for}\mathrm{b}>0\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{bt}\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{abt}\right)}{{\mathrm{b}}^2(1+{\mathrm{t}}^2)}\mathrm{bdt}=\frac{1}{\mathrm{b}}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{abt}\right)}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{1}{2\mathrm{b}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{abt}\right)}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\frac{1}{2\mathrm{b}}\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iabt}}}{{\mathrm{t}}^2+1}\mathrm{dt}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iabz}}}{{\mathrm{z}}^2+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iabz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$\mathrm{residus}\mathrm{tbeorem}\mathrm{give}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{\mathrm{iab}\left(\mathrm{i}\right)}}{2\mathrm{i}}=\pi {\mathrm{e}}^{-\mathrm{ab}}\Rightarrow \mathrm{I}=\frac{\pi }{2\mathrm{b}}{\mathrm{e}}^{-\mathrm{ab}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com