Question Number 39025 by maxmathsup by imad last updated on 01/Jul/18
$${let}\:{f}\left({x}\right)=\:\frac{{cos}\left(\alpha{x}\right)}{{cosx}}\:\:\:\:\left(\mathrm{2}\pi\:{periodic}\:{even}\right) \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 05/Jul/18
![f(x) =(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n = (2/T) ∫_([T]) f(x)cos(nx) dx =(2/(2π)) ∫_(−π) ^π ((cos(αx)cos(nx))/(cosx))dx ⇒ π a_n = (1/2) ∫_(−π) ^π ((cos(n+α)x +cos(n−α)x)/(cosx))dx 2π a_n = ∫_(−π) ^π ((cos(n+α)x)/(cosx))dx +∫_(−π) ^π ((cos(n−α)x)/(cosx))dx let find first I_λ =∫_(−π) ^π ((cos(λx))/(cosx))dx I_λ = Re( ∫_(−π) ^π (e^(iλx) /(cosx))dx) changement e^(ix) =z give ∫_(−π) ^π (e^(iλx) /(cosx)) dx = ∫_(∣z∣=1) ((2z^λ )/(z+z^(−1) )) (dz/(iz)) = ∫_(∣z∣=1) ((−2i z^λ )/(z^2 +1))dz let ϕ(z) = ((−2iz^λ )/(z^(2 ) +1)) ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,−i)} Res(ϕ,i)= ((−2i i^λ )/(2i)) = −i^λ Res(ϕ^ ,−i) =((−2i (−i)^λ )/(−2i)) =(−i)^λ ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ −i^λ +(−i)^λ } =−2iπ{ i^λ −(−i)^λ } =−2iπ 2i Im(i^λ ) =4π Im( cos(((λπ)/2)) +isin(((λπ)/2))} =4π sin(((λπ)/2)) ⇒ I_λ =4π sin(((λπ)/2)) ⇒ 2π a_n = 4πsin((((n+α)π)/2)) +4π sin((((n−α)π)/2))⇒ a_n = (1/2){ sin((((n+α)π)/2)) +sin((((n−α)π)/2))} a_0 =(1/π) ∫_(−π) ^π ((cos(αx))/(cosx))dx =(1/π)4π sin(((απ)/2)) =4 sin((α/2)) ⇒(a_0 /2) = 2 sin((α/2)) ⇒ f(x) = 2sin((α/2)) +(1/2)Σ_(n=1) ^∞ {sin((((n+α)π)/2))+sin((((n−α)π)/2))}](Q39305.png)
$${f}\left({x}\right)\:=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \:{cos}\left({nx}\right)\:{with} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:\:{f}\left({x}\right){cos}\left({nx}\right)\:{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\:\int_{−\pi} ^{\pi} \:\:\frac{{cos}\left(\alpha{x}\right){cos}\left({nx}\right)}{{cosx}}{dx}\:\Rightarrow \\ $$$$\pi\:{a}_{{n}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\pi} ^{\pi} \:\:\frac{{cos}\left({n}+\alpha\right){x}\:+{cos}\left({n}−\alpha\right){x}}{{cosx}}{dx} \\ $$$$\mathrm{2}\pi\:{a}_{{n}} \:=\:\int_{−\pi} ^{\pi} \:\:\frac{{cos}\left({n}+\alpha\right){x}}{{cosx}}{dx}\:\:+\int_{−\pi} ^{\pi} \:\frac{{cos}\left({n}−\alpha\right){x}}{{cosx}}{dx} \\ $$$${let}\:{find}\:{first}\:{I}_{\lambda} \:=\int_{−\pi} ^{\pi} \:\frac{{cos}\left(\lambda{x}\right)}{{cosx}}{dx} \\ $$$${I}_{\lambda} \:\:=\:{Re}\left(\:\int_{−\pi} ^{\pi} \:\:\frac{{e}^{{i}\lambda{x}} }{{cosx}}{dx}\right)\:\:\:{changement}\:{e}^{{ix}} ={z}\:{give} \\ $$$$\int_{−\pi} ^{\pi} \:\:\frac{{e}^{{i}\lambda{x}} }{{cosx}}\:{dx}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{z}^{\lambda} }{{z}+{z}^{−\mathrm{1}} }\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{i}\:{z}^{\lambda} }{{z}^{\mathrm{2}} \:+\mathrm{1}}{dz}\:\:\:{let}\:\varphi\left({z}\right)\:=\:\frac{−\mathrm{2}{iz}^{\lambda} }{{z}^{\mathrm{2}\:} \:+\mathrm{1}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,−{i}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)=\:\frac{−\mathrm{2}{i}\:{i}^{\lambda} }{\mathrm{2}{i}}\:=\:−{i}^{\lambda} \\ $$$${Res}\left(\bar {\varphi},−{i}\right)\:=\frac{−\mathrm{2}{i}\:\left(−{i}\right)^{\lambda} }{−\mathrm{2}{i}}\:=\left(−{i}\right)^{\lambda} \:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:−{i}^{\lambda} \:+\left(−{i}\right)^{\lambda} \:\right\} \\ $$$$=−\mathrm{2}{i}\pi\left\{\:{i}^{\lambda} \:−\left(−{i}\right)^{\lambda} \right\}\:=−\mathrm{2}{i}\pi\:\mathrm{2}{i}\:{Im}\left({i}^{\lambda} \right) \\ $$$$=\mathrm{4}\pi\:{Im}\left(\:{cos}\left(\frac{\lambda\pi}{\mathrm{2}}\right)\:+{isin}\left(\frac{\lambda\pi}{\mathrm{2}}\right)\right\}\:=\mathrm{4}\pi\:{sin}\left(\frac{\lambda\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}_{\lambda} =\mathrm{4}\pi\:{sin}\left(\frac{\lambda\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\mathrm{2}\pi\:{a}_{{n}} =\:\mathrm{4}\pi{sin}\left(\frac{\left({n}+\alpha\right)\pi}{\mathrm{2}}\right)\:+\mathrm{4}\pi\:{sin}\left(\frac{\left({n}−\alpha\right)\pi}{\mathrm{2}}\right)\Rightarrow \\ $$$${a}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{sin}\left(\frac{\left({n}+\alpha\right)\pi}{\mathrm{2}}\right)\:+{sin}\left(\frac{\left({n}−\alpha\right)\pi}{\mathrm{2}}\right)\right\} \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\pi}\:\int_{−\pi} ^{\pi} \:\frac{{cos}\left(\alpha{x}\right)}{{cosx}}{dx}\:=\frac{\mathrm{1}}{\pi}\mathrm{4}\pi\:{sin}\left(\frac{\alpha\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:\Rightarrow\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\:\mathrm{2}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\left\{{sin}\left(\frac{\left({n}+\alpha\right)\pi}{\mathrm{2}}\right)+{sin}\left(\frac{\left({n}−\alpha\right)\pi}{\mathrm{2}}\right)\right\} \\ $$$$ \\ $$$$ \\ $$