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 100511 by mathmax by abdo last updated on 27/Jun/20

$$\mathrm{calculate}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan (\mathrm{cosx}+\mathrm{sinx})}{{\mathrm{x}}^2+4}\mathrm{dx}$$

Answered by abdomathmax last updated on 28/Jun/20

$$\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan (\mathrm{cosx}+\mathrm{sinx})}{{\mathrm{x}}^2+4}\mathrm{dx}\Rightarrow$$$$\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan \left(\sqrt{2}\cos (\mathrm{x}-\frac{\pi }{4})\right)}{{\mathrm{x}}^2+4}\mathrm{dx}\mathrm{let}$$$$\phi \left(\mathrm{z}\right)=\frac{\arctan \left(\sqrt{2}\cos (\mathrm{z}-\frac{\pi }{4})\right)}{{\mathrm{z}}^2+4}\mathrm{the}\mathrm{poles}\mathrm{of}\phi \mathrm{are}$$$$2\mathrm{i}\mathrm{and}-2\mathrm{i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,2\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{\arctan \left(\sqrt{2}\cos (2\mathrm{i}-\frac{\pi }{4})\right)}{4\mathrm{i}}$$$$=\frac{\pi }{2}\arctan \left(\sqrt{2}\cos (2\mathrm{i}-\frac{\pi }{4})\right)$$$$\cos (2\mathrm{i}-\frac{\pi }{4})=\frac{{\mathrm{e}}^{\mathrm{i}(2\mathrm{i}-\frac{\pi }{4})}+{\mathrm{e}}^{-\mathrm{i}(2\mathrm{i}-\frac{\pi }{4})}}{2}$$$$=\frac{{\mathrm{e}}^{-2}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}+{\mathrm{e}}^2{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}}{2}$$$$=\frac{{\mathrm{e}}^{-2}(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})+{\mathrm{e}}^2(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}{2}$$$$=\frac{1}{2\sqrt{2}}\{{\mathrm{e}}^{-2}(1-\mathrm{i})+{\mathrm{e}}^2(1+\mathrm{i})\}=....\mathrm{after}\mathrm{we}\mathrm{use}$$$$\mathrm{arctanz}=\frac{1}{2\mathrm{i}}\ln \left(\frac{1+\mathrm{iz}}{1-\mathrm{iz}}\right)...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com