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Question Number 74513 by mathmax by abdo last updated on 25/Nov/19

$$findf\left(x\right)={\int }_0^{\pi }\frac{d\theta }{x^2-2xsin\left(2\theta \right)+1}\left(xreal\right)$$

Commented by mathmax by abdo last updated on 26/Nov/19

$$f\left(x\right){=}_{2\theta =t}{\int }_0^{2\pi }\frac{dt}{2(x^2-2xsint+1)}\Rightarrow 2f\left(x\right)={\int }_0^{2\pi }\frac{dt}{x^2-2xsint+1}$$$$changement{e}^{it}=zgive$$$$2f\left(x\right)={\int }_{\mid z\mid =1}\frac{1}{x^2-2x\frac{z-z^{-1}}{2i}+1}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{idz}{iz({ix}^2-xz+{xz}^{-1}+i)}=$$$$={\int }_{\mid z\mid =1}\frac{dz}{{ix}^{2}z-{xz}^2+x+iz}={\int }_{\mid z\mid =1}\frac{dz}{-{xz}^2+i(1+x^2)z+x}$$$$={\int }_{\mid z\mid =1}\frac{dz}{-{xz}^2+i(1+x^2)z+x}=-{\int }_{\mid z\mid =1}\frac{dz}{{xz}^2-i(1+x^2)z-x}$$$$let\phi \left(z\right)=\frac{1}{{xz}^2-i(x^2+1)z-x}poleof\phi ?$$$$\mathrm{\Delta }=-{(x^2+1)}^2+4x^2=4x^2-(x^4+2x^2+1)=-x^4+2x^2-1$$$$=-(x^4-2x^2+1)=-{(x^2-1)}^2=$$$$z_1=\frac{i(x^2+1)+i\mid x^2-1\mid }{2x}and{z}_2=\frac{i(x^2+1)-i\mid x^2-1\mid }{2x}$$$$case1\mid x\mid <1\Rightarrow z_1=\frac{i(x^2+1)+i(1-x^2)}{2x}=\frac{i}{x}$$$$\mid z_1\mid =\frac{1}{\mid x\mid }>1\left(outofcircle\right)$$$$z_2=\frac{i(x^2+1)-i(1-x^2)}{2x}=\frac{2{ix}^2}{2x}=ix\Rightarrow \mid z_2\mid =\mid x\mid <1$$$$\phi \left(z\right)=\frac{1}{x(z-z_1)(z-z_2)}and{\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_2)$$$$=2i\pi \times \frac{1}{x(z_2-z_1)}=\frac{2\pi i}{x(ix-\frac{i}{x})}=\frac{2\pi }{(x^2-1)}\Rightarrow 2f\left(x\right)=\frac{-2\pi }{x^2-1}\Rightarrow$$$$f\left(x\right)=\frac{\pi }{1-x^2}with\mid x\mid <1$$$$case2\mid x\mid >1\Rightarrow z_1=\frac{i(x^2+1)+i(x^2-1)}{2x}=ix\Rightarrow \mid z_1\mid =\mid x\mid >1(outof$$$$circle)z_2=\frac{i(x^2+1)-i(x^2-1)}{2x}=\frac{i}{x}\Rightarrow \mid z_2\mid =\frac{1}{\mid x\mid }<1$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_2)=2i\pi \times \frac{1}{x(z_2-z_1)}$$$$=\frac{2i\pi }{x(\frac{i}{x}-ix)}=\frac{2\pi }{1-x^2}\Rightarrow 2f\left(x\right)=\frac{-2\pi }{1-x^2}\Rightarrow f\left(x\right)=\frac{\pi }{x^2-1}$$$$finallyf\left(x\right)=\frac{\pi }{1-x^2}if\mid x\mid <1andf\left(x\right)=\frac{\pi }{x^2-1}if\mid x\mid >1$$

Answered by mind is power last updated on 25/Nov/19

$$f\left(x\right)={\underset{0}{\int }}^{\pi }\frac{d\theta }{{(x-sin\left(2\theta \right))}^2+{cos}^2\left(2\theta \right)}$$$$x\ne 1$$$$since2-2sin\left(2\theta \right)=0for\theta =\frac{\pi }{2}$$$$f\left(x\right)=f(-x)wesupposex⩾0$$$$={\int }_0^{\pi }\frac{d\theta }{(x-sin\left(2\theta \right)-icos\left(\theta \right))(x-sin\left(2\theta \right)+icos\left(2\theta \right))}$$$$={\int }_0^{\pi }\frac{d\theta }{(x-{ie}^{-i2\theta })(x+{ie}^{i2\theta })}$$$$z=e^{2i\theta }\Rightarrow dz=2{ie}^{2i\theta }d\theta$$$$={\int }_C\frac{dz}{2iz(x-\frac{i}{z})(x+iz)}={\int }_C\frac{dz}{2i(xz-i)(x+iz)}$$$$polesare\frac{i}{x},-ix$$$$ifx=0$$$$f\left(0\right)=\pi$$$$f(-1)={\int }_0^{\pi }\frac{d\theta }{2+2sin\left(2\theta \right)}=\frac{d\theta }{2(1+2sin\left(\theta \right)cos\left(\theta \right))}$$$$={\int }_0^{\frac{\pi }{2}}\frac{1+{tg}^2\left(\theta \right)}{2{(1+tg\left(\theta \right))}^2}d\theta +{\int }_{\frac{\pi }{2}}^{\pi }$$$$={[-\frac{1}{2(1+tg\left(\theta \right)}]}_0^{\frac{\pi }{2}}+{\left[\frac{-1}{2(1+tg\left(\theta \right))}\right]}_{\frac{\pi }{2}}^{\pi }=0$$$$if\mid x\mid >1\Rightarrow \mid \frac{i}{x}\mid <1\in C$$$$f\left(x\right)=\pi .Res(\frac{1}{(xz-i)(x+iz)},z=\frac{i}{x})$$$$=\pi .\frac{x}{x^2-1}x>1$$$$\mid x\mid <1\Rightarrow \mid -ix\mid <1\Rightarrow f\left(x\right)=\pi Res(\frac{1}{(xz-i)(x+iz)},z=ix\}$$$$$$$$=\frac{\pi }{1-x^2}\mid x\mid <1$$$$$$$$$$

Commented by mathmax by abdo last updated on 26/Nov/19

$$thankxsir.$$

Commented by mind is power last updated on 26/Nov/19

$$y^{\prime }rewelcom$$

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