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 82816 by M±th+et£s last updated on 24/Feb/20

$$calculatetheexactvalue$$$${\int }_0^{\mathrm{\infty }}\frac{cos\left(x\right)}{x^2+1}dx$$

Commented by msup trace by abdo last updated on 24/Feb/20

$$I={\int }_0^{\mathrm{\infty }}\frac{cosx}{1+x^2}dx\Rightarrow I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cosx}{1+x^2}dx$$$$\Rightarrow 2I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{x^2+1}dx\right)let$$$$\phi \left(z\right)=\frac{e^{iz}}{z^2+1}\Rightarrow \phi \left(z\right)=\frac{e^{iz}}{(z-i)(z+i)}$$$$redidustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Re(\phi ,i)$$$$=2i\pi \times \frac{e^{-1}}{2i}=\pi e^{-1}\Rightarrow$$$$I=\frac{\pi }{2}{e}^{-1}\Rightarrow I=\frac{\pi }{2e}$$

Commented by M±th+et£s last updated on 24/Feb/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 24/Feb/20

$$youare[welcome$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com