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 32341 by abdo imad last updated on 23/Mar/18

$$letgive\lambda fromRand{\lambda }^2\ne 1and$$$$I_n\left(\lambda \right)={\int }_0^{\pi }\frac{cos\left(nt\right)}{1-2\lambda cost+{\lambda }^2}dt.calculate{I}_n\left(\lambda \right).$$

Commented by abdo imad last updated on 01/Apr/18

$$ch.t=2xgive{I}_n\left(\lambda \right)=2{\int }_0^{2\pi }\frac{cos\left(2nx\right)}{1-2\lambda cos\left(2x\right)+{\lambda }^2}dxafter$$$$weusethech.e^{ix}=z$$$$I_n\left(\lambda \right)=2{\int }_{\mid z\mid =1}\frac{\frac{z^{2n}+z^{-2n}}{2}}{1-2\lambda \frac{z^2+z^{-2}}{2}}\frac{dz}{iz}$$$$=2{\int }_{\mid z\mid =1}\frac{z^{2n}+z^{-2n}}{iz(2-2\lambda (z^2+z^{-2}))}dz$$$$={\int }_{\mid z\mid =1}\frac{z^{2n}+z^{-2n}}{iz(1-\lambda z^2-\lambda z^{-2})}={\int }_{\mid z\mid =1}\frac{-i(z^{2n}+z^{-2n})}{z(1-\lambda z^2-\frac{\lambda }{z^2})}dz$$$$={\int }_{\mid z\mid =1}\frac{-iz(z^{2n}+z^{-2n})}{z^2-\lambda z^4-\lambda }dz$$$$={\int }_{\mid z\mid =1}\frac{i(z^{2n+1}+z^{-2n+1})}{\lambda z^4-z^2+\lambda }dzletput$$$$\phi \left(z\right)=\frac{i(z^{2n+1}+z^{-2n+1})}{\lambda z^4-z^2+\lambda }.polesof\phi ?$$$$\lambda z^4-z^2+\lambda =0\Rightarrow \mathrm{\Delta }=1-4{\lambda }^{2}if\mathrm{\Delta }⩾0therootsarereals$$$$z_1^2=\frac{1+\sqrt{1-4{\lambda }^2}}{2}and{z}_2^2=\frac{1-\sqrt{1-4{\lambda }^2}}{2}\Rightarrow$$$$z_1=\xi \sqrt{\frac{1+\sqrt{1-4{\lambda }^2}}{2}}and{z}_2=\xi \sqrt{\frac{1-\sqrt{1-4{\lambda }^2}}{2}}$$$$with{\xi }^2=1andweget{\int }_R\phi \left(z\right)dx=2i\pi \sum _{i}Res(\phi ,x_i)$$$$if\mathrm{\Delta }<0\mathrm{\Delta }={\left(i\sqrt{4{\lambda }^2-1}\right)}^2\Rightarrow$$$$z_1^2=\frac{1+i\sqrt{4{\lambda }^2-1}}{2}\Rightarrow z_1=\xi \sqrt{\frac{1+i\sqrt{4{\lambda }^2-1}}{2}}$$$$z_2^2=\frac{1-i\sqrt{4{\lambda }^2-1}}{2}\Rightarrow z_2=\xi \sqrt{\frac{1-i\sqrt{4{\lambda }^2-1}}{2}}$$$$andwechoosetherootswichverify\mid z_i\mid ⩽1....becontinued$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com