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Question Number 39037 by maxmathsup by imad last updated on 01/Jul/18

$$calculateF\left(x\right)={\int }_0^{2\pi }\frac{cos\left(4t\right)}{x^2-2xcost+1}dt$$

Commented by math khazana by abdo last updated on 04/Jul/18

$$F\left(x\right)=Re\left({\int }_0^{2\pi }\frac{e^{i4t}}{x^2-2xcost+1}dt\right)changement$$$$e^{it}=zgive$$$${\int }_0^{2\pi }\frac{e^{i4t}}{x^2-2xcost+1}dt={\int }_{\mid z\mid =1}\frac{z^4}{x^2-2x\frac{z+z^{-1}}{2}+1}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{-{iz}^4}{z(x^2-xz-{xz}^{-1}+1)}dz$$$$={\int }_{\mid z\mid =1}\frac{-{iz}^4}{x^{2}z-{xz}^2-x+z}dz$$$$={\int }_{\mid z\mid =1}\frac{-i{z}^4}{-{xz}^2+(x^2+1)z-x}$$$$={\int }_{\mid z\mid =1}\frac{i{z}^4}{x{z}^2-(1+x^2)z+x}letconsider$$$$\phi \left(z\right)=\frac{i{z}^4}{x{z}^2-(1+x^2)z+x}polesof\phi ?$$$$\mathrm{\Delta }={(1+x^2)}^2-4x^2=1+2x^2+x^4-4x^2$$$$={(1-x^2)}^2\Rightarrow \sqrt{\mathrm{\Delta }}=\mid 1-x^2\mid$$$$case1\mid x\mid <1\Rightarrow \sqrt{\mathrm{\Delta }}=1-x^2\Rightarrow$$$$z_1=\frac{1+x^2+1-x^2}{2x}=\frac{1}{x}(x\ne 0)$$$$z_2=\frac{1+x^2-1+x^2}{2x}=x$$$$\mid z_1\mid =\frac{1}{\mid x\mid }>1and\mid z_2\mid =\mid x\mid <1so$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,z_2)but$$$$\phi \left(z\right)=\frac{i{z}^4}{x(z-z_1)(z-z_2)}\Rightarrow$$$$Res(\phi ,z_2)=\frac{i{z}_2^4}{x(z_2-z_1)}=\frac{i{x}^4}{x(x-\frac{1}{x})}=\frac{i{x}^4}{x^2-1}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{{ix}^4}{x^2-1}=\frac{-2\pi x^4}{x^2-1}=\frac{2\pi x^4}{1-x^2}\Rightarrow$$$$F\left(x\right)=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)=\frac{2\pi x^4}{1-x^2}.$$$$$$

Commented by math khazana by abdo last updated on 04/Jul/18

$$case2if\mid x\mid <1wefollowthesameroad...$$

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