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Question Number 39037 by maxmathsup by imad last updated on 01/Jul/18
calculate F(x)=∫_0 ^(2π) ((cos(4t))/(x^2 −2x cost +1)) dt
$$\:{calculate}\:\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}\left(\mathrm{4}{t}\right)}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cost}\:+\mathrm{1}}\:{dt} \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
F(x)=Re( ∫_0 ^(2π) (e^(i4t) /(x^2 −2x cost +1)) dt) changement e^(it) =z give ∫_0 ^(2π) (e^(i4t) /(x^2 −2x cost +1))dt =∫_(∣z∣=1) (z^4 /(x^2 −2x ((z+z^(−1) )/2)+1)) (dz/(iz)) = ∫_(∣z∣=1) ((−iz^4 )/(z( x^2 −xz−xz^(−1) +1)))dz = ∫_(∣z∣=1) ((−iz^4 )/(x^2 z −xz^2 −x +z))dz = ∫_(∣z∣=1) ((−i z^4 )/(−xz^2 +(x^2 +1)z −x)) = ∫_(∣z∣=1) ((i z^4 )/(x z^2 −(1+x^2 )z +x)) let consider ϕ(z) = ((i z^4 )/(x z^2 −(1+x^2 )z +x)) poles of ϕ? Δ =(1+x^2 )^2 −4x^2 =1+2x^2 +x^4 −4x^2 =(1−x^2 )^2 ⇒(√Δ)= ∣1−x^2 ∣ case1 ∣x∣<1 ⇒(√Δ)=1−x^2 ⇒ z_1 =((1+x^2 +1−x^2 )/(2x)) =(1/x) (x≠0) z_2 = ((1+x^2 −1+x^2 )/(2x)) = x ∣z_1 ∣ = (1/(∣x∣))>1 and ∣z_2 ∣ =∣x∣<1 so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_2 ) but ϕ(z)= ((i z^4 )/(x(z−z_1 )(z−z_2 ))) ⇒ Res(ϕ,z_2 )= ((i z_2 ^4 )/(x(z_2 −z_1 ))) = ((i x^4 )/(x(x−(1/x)))) =((i x^4 )/(x^2 −1)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((ix^4 )/(x^2 −1)) = ((−2π x^4 )/(x^2 −1)) = ((2πx^4 )/(1−x^2 )) ⇒ F(x)=Re( ∫_(−∞) ^(+∞) ϕ(z)dz)=((2πx^4 )/(1−x^2 )) .
$${F}\left({x}\right)={Re}\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{e}^{{i}\mathrm{4}{t}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cost}\:+\mathrm{1}}\:{dt}\right)\:{changement} \\ $$$${e}^{{it}} ={z}\:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{e}^{{i}\mathrm{4}{t}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cost}\:+\mathrm{1}}{dt}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{z}^{\mathrm{4}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+\mathrm{1}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−{iz}^{\mathrm{4}} }{{z}\left(\:{x}^{\mathrm{2}} \:−{xz}−{xz}^{−\mathrm{1}} \:+\mathrm{1}\right)}{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−{iz}^{\mathrm{4}} }{{x}^{\mathrm{2}} {z}\:−{xz}^{\mathrm{2}} \:−{x}\:+{z}}{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{−{i}\:{z}^{\mathrm{4}} }{−{xz}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} +\mathrm{1}\right){z}\:−{x}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{i}\:{z}^{\mathrm{4}} }{{x}\:{z}^{\mathrm{2}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}}\:\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{i}\:{z}^{\mathrm{4}} }{{x}\:{z}^{\mathrm{2}} \:−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}}\:{poles}\:{of}\:\varphi? \\ $$$$\Delta\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:−\mathrm{4}{x}^{\mathrm{2}} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\Rightarrow\sqrt{\Delta}=\:\mid\mathrm{1}−{x}^{\mathrm{2}} \mid \\ $$$${case}\mathrm{1}\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sqrt{\Delta}=\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{1}+{x}^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}}\:=\frac{\mathrm{1}}{{x}}\:\:\:\left({x}\neq\mathrm{0}\right) \\ $$$${z}_{\mathrm{2}} =\:\frac{\mathrm{1}+{x}^{\mathrm{2}} \:−\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}}\:=\:{x} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:=\:\frac{\mathrm{1}}{\mid{x}\mid}>\mathrm{1}\:\:{and}\:\mid{z}_{\mathrm{2}} \mid\:=\mid{x}\mid<\mathrm{1}\:{so} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:{but}\: \\ $$$$\varphi\left({z}\right)=\:\frac{{i}\:{z}^{\mathrm{4}} }{{x}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)=\:\:\frac{{i}\:{z}_{\mathrm{2}} ^{\mathrm{4}} }{{x}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\:=\:\frac{{i}\:{x}^{\mathrm{4}} }{{x}\left({x}−\frac{\mathrm{1}}{{x}}\right)}\:=\frac{{i}\:{x}^{\mathrm{4}} }{{x}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{{ix}^{\mathrm{4}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{−\mathrm{2}\pi\:{x}^{\mathrm{4}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{2}\pi{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({x}\right)={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)=\frac{\mathrm{2}\pi{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 04/Jul/18
case 2 if ∣x∣<1 wefollow the same road...
$${case}\:\mathrm{2}\:{if}\:\mid{x}\mid<\mathrm{1}\:{wefollow}\:{the}\:{same}\:{road}… \\ $$

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