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Question Number 50412 by Abdo msup. last updated on 16/Dec/18
1) calculate U_n =∫_0 ^π (dx/(1+cos^2 (nx))) with n from N 2) f continue from [0,π] to R find lim_(n→+∞) ∫_0 ^π ((f(x))/(1+cos^2 (nx)))dx
$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} \left({nx}\right)}\:{with}\:{n}\:{from}\:{N} \\ $$$$\left.\mathrm{2}\right)\:{f}\:{continue}\:{from}\:\left[\mathrm{0},\pi\right]\:{to}\:{R}\:\:{find} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{f}\left({x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left({nx}\right)}{dx} \\ $$
Commented by Abdo msup. last updated on 25/Dec/18
1)changement nx=t give U_n =(1/n)∫_0 ^(nπ) (1/(1+cos^2 t))dt ⇒nU_n =Σ_(k=0) ^(n−1) ∫_(kπ) ^((k+1)π) (dt/(1+cos^2 t)) but ∫_(kπ) ^((k+1)π) (dt/(1+cos^2 t)) =_(t=kπ +x) ∫_0 ^π (dx/(1+cos^2 x)) = ∫_0 ^π (dx/(1+((1+cos(2x))/2))) =∫_0 ^π ((2dx)/(3+cos(2x))) =_(2x=u) ∫_0 ^(2π) (du/(3+cosu)) du ⇒nU_n =n ∫_0 ^(2π) (du/(3+cosu)) ⇒U_n =∫_0 ^(2π) (du/(3+cos(u))) changement e^(iu) =z give U_n = ∫_(∣z∣=1) (1/(3+((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) ((2dz)/(iz(6+z +z^(−1) ))) =∫_(∣z∣=1) ((−2idz)/(6z +z^2 +1)) let ϕ(z)=((−2i)/(z^2 +6z +1)) poles of ϕ? Δ^′ =3^2 −1=8 ⇒z_1 =−3+2(√2) and z_2 =−3−2(√2) ∣z_1 ∣−1 =3−2(√2)−1 =2−2(√2)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1 =3+2(√2)−1=2+2(√2)( z_2 is out of circle) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) = ((−2i)/(z_1 −z_2 )) =((−2i)/(4(√2))) =((−i)/(2(√2))) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ(((−i)/(2(√2))))=(π/( (√2))) ⇒ ∀n∈N U_n =(π/( (√2))) .
$$\left.\mathrm{1}\right){changement}\:{nx}={t}\:{give} \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{{n}\pi} \:\:\frac{\mathrm{1}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:\Rightarrow{nU}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \:\frac{{dt}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}} \\ $$$${but}\:\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)\pi} \:\:\frac{{dt}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:=_{{t}={k}\pi\:+{x}} \:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\mathrm{1}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{dx}}{\mathrm{3}+{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=_{\mathrm{2}{x}={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{du}}{\mathrm{3}+{cosu}}\:{du}\:\:\:\Rightarrow{nU}_{{n}} ={n}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{du}}{\mathrm{3}+{cosu}} \\ $$$$\Rightarrow{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{du}}{\mathrm{3}+{cos}\left({u}\right)}\:\:{changement}\:{e}^{{iu}} ={z}\:{give} \\ $$$${U}_{{n}} =\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{3}+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{6}+{z}\:+{z}^{−\mathrm{1}} \right)}\:\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{idz}}{\mathrm{6}{z}\:+{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${let}\:\varphi\left({z}\right)=\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} \:+\mathrm{6}{z}\:+\mathrm{1}}\:\:{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} =\mathrm{3}^{\mathrm{2}} −\mathrm{1}=\mathrm{8}\:\Rightarrow{z}_{\mathrm{1}} =−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\:=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\left(\:{z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\: \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\frac{−\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\frac{−\mathrm{2}{i}}{\mathrm{4}\sqrt{\mathrm{2}}}\:=\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\:\forall{n}\in{N}\:\:\:{U}_{{n}} =\frac{\pi}{\:\sqrt{\mathrm{2}}}\:. \\ $$

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