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 59991 by selimatas01 last updated on 16/May/19

Commented by Mr X pcx last updated on 16/May/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{cosx}{{(x^2+1)}^2}dx\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cosx}{{(x^2+1)}^2}dx=$$$$Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{{(x^2+1)}^2}dx\right)letconsider$$$$thecomplexefunction$$$$\phi \left(z\right)=\frac{e^{iz}}{{(z^2+1)}^2}\Rightarrow \phi \left(z\right)=\frac{e^{iz}}{{(z-i)}^2{(z+i)}^2}$$$$sothepolesof\phi areiand-i\left(doubles\right)$$$$redidustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}{\left\{{(z-i)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}{\left\{\frac{e^{iz}}{{(z+i)}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}\frac{i{e}^{iz}{(z+i)}^2-2(z+i)e^{iz}}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{{ie}^{iz}(z+i)-2e^{iz}}{{(z+i)}^3}$$$$=\frac{-2e^{-1}-2e^{-1}}{{\left(2i\right)}^3}=\frac{-4e^{-1}}{-8i}=\frac{e^{-1}}{2i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-1}}{2i}=\pi e^{-1}\Rightarrow$$$$2I=\frac{\pi }{e}\Rightarrow ★I=\frac{\pi }{2e}★$$

Commented by selimatas01 last updated on 17/May/19

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com