Question Number 35620 by abdo mathsup 649 cc last updated on 21/May/18
$${let}\:\:{f}\left({x}\right)\:={e}^{−{x}} \:{sinx}\:\:\:{odd}\:\mathrm{2}\pi\:{periodic}\: \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 25/May/18
![we have f(x)=Σ_(n=1) ^∞ a_n sin(nx) with a_n = (2/T) ∫_([T]) f(x)sin(nx)dx =(2/(2π)) ∫_(−π) ^π e^(−x) sinx sin(nx)dx = (2/π) ∫_0 ^π e^(−x) sin(nx)sin(x)dx cos(a+b) =cosa cosb −sinasinb cos(a−b)=cosa cosb +sina sinb ⇒I sina sinb =(1/2){cos(a−b)−cos(a+b)} π a_n = ∫_0 ^π ( e^(−x) cos(n−1)x −e^(−x) cos(n+1)x) but I(λ) = ∫_0 ^π e^(−x) cos(λx)dx =Re( ∫_0 ^π e^(−x) e^(iλx) dx) =Re( ∫_0 ^π e^((−1+iλ)x) dx) ∫_0 ^π e^((−1+iλ)x) dx = [(1/(−1+iλ)) e^((−1+iλ)x) ]_0 ^π =((−1)/(1−iλ)){ e^((−1+iλ)π) −1} =((−1−iλ)/(1+λ^2 )) { e^(−π) (cos(λπ) +isin(λπ) −1} =−(((1+iλ){ e^(−π) cos(λπ)−1 +i e^(−π) sin(λπ)})/(1+λ^2 )) =−(1/(1+λ^2 )){ e^(−π) cos(λπ) −1 +i e^(−π) sin(λπ) +iλ e^(−π) cos(λπ) −iλ −λ e^(−π) sin(λπ)}−⇒ I(λ) =((−1)/(1+λ^2 )){ e^(−π) cos(λπ) −λ e^(−π) sin(λπ) −1} I(n−1)= ((−1)/(1+(n−1)^2 )){ e^(−π) cos(n−1)π −(n−1) e^(−π) sin(n−1)π −1} =((−1)/(1+(n−1)^2 )){ (−1)^(n−1) e^(−π) −1} = (((−1)^n e^(−π) +1)/(1+(n−1)^2 )) I(n) = ((−1)/(1+n^2 )){ (−1)^n e^(−π) −1}=((1−(−1)^n e^(−π) )/(1+n^2 )) π a_n = I(n−1)−I(n)= ((1+(−1)^n e^(−π) )/(1+(n−1)^2 )) −((1 −(−1)^n e^(−π) )/(1+n^2 )) ⇒ e^(−x) sinx =Σ_(n=1) ^∞ (1/π){ ((1+(−1)^n e^(−π) )/(1+(n−1)^2 )) +(((−1)^n e^(−π) −1)/(1+n^2 ))}sin(nx)](https://www.tinkutara.com/question/Q35917.png)
$${we}\:{have}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \:{sin}\left({nx}\right)\:{with} \\ $$$${a}_{{n}} =\:\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right){sin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:\:{e}^{−{x}} \:{sinx}\:{sin}\left({nx}\right){dx} \\ $$$$=\:\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{x}} \:{sin}\left({nx}\right){sin}\left({x}\right){dx} \\ $$$${cos}\left({a}+{b}\right)\:={cosa}\:{cosb}\:−{sinasinb} \\ $$$${cos}\left({a}−{b}\right)={cosa}\:{cosb}\:+{sina}\:{sinb}\:\Rightarrow{I} \\ $$$${sina}\:{sinb}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left({a}−{b}\right)−{cos}\left({a}+{b}\right)\right\} \\ $$$$\pi\:{a}_{{n}} \:=\:\int_{\mathrm{0}} ^{\pi} \:\:\left(\:{e}^{−{x}} {cos}\left({n}−\mathrm{1}\right){x}\:−{e}^{−{x}} \:{cos}\left({n}+\mathrm{1}\right){x}\right)\:{but} \\ $$$${I}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{x}} \:{cos}\left(\lambda{x}\right){dx} \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{−{x}} \:{e}^{{i}\lambda{x}} {dx}\right)\:={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:{e}^{\left(−\mathrm{1}+{i}\lambda\right){x}} {dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:{e}^{\left(−\mathrm{1}+{i}\lambda\right){x}} {dx}\:=\:\left[\frac{\mathrm{1}}{−\mathrm{1}+{i}\lambda}\:{e}^{\left(−\mathrm{1}+{i}\lambda\right){x}} \right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{1}−{i}\lambda}\left\{\:{e}^{\left(−\mathrm{1}+{i}\lambda\right)\pi} \:−\mathrm{1}\right\} \\ $$$$=\frac{−\mathrm{1}−{i}\lambda}{\mathrm{1}+\lambda^{\mathrm{2}} }\:\left\{\:{e}^{−\pi} \left({cos}\left(\lambda\pi\right)\:+{isin}\left(\lambda\pi\right)\:−\mathrm{1}\right\}\right. \\ $$$$=−\frac{\left(\mathrm{1}+{i}\lambda\right)\left\{\:{e}^{−\pi} \:{cos}\left(\lambda\pi\right)−\mathrm{1}\:+{i}\:{e}^{−\pi} \:{sin}\left(\lambda\pi\right)\right\}}{\mathrm{1}+\lambda^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{1}+\lambda^{\mathrm{2}} }\left\{\:\:{e}^{−\pi} {cos}\left(\lambda\pi\right)\:−\mathrm{1}\:+{i}\:{e}^{−\pi} \:{sin}\left(\lambda\pi\right)\right. \\ $$$$\left.+{i}\lambda\:{e}^{−\pi} \:{cos}\left(\lambda\pi\right)\:−{i}\lambda\:−\lambda\:{e}^{−\pi} \:{sin}\left(\lambda\pi\right)\right\}−\Rightarrow \\ $$$${I}\left(\lambda\right)\:=\frac{−\mathrm{1}}{\mathrm{1}+\lambda^{\mathrm{2}} }\left\{\:{e}^{−\pi} \:{cos}\left(\lambda\pi\right)\:−\lambda\:{e}^{−\pi} \:{sin}\left(\lambda\pi\right)\:−\mathrm{1}\right\} \\ $$$${I}\left({n}−\mathrm{1}\right)=\:\frac{−\mathrm{1}}{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\left\{\:{e}^{−\pi} \:{cos}\left({n}−\mathrm{1}\right)\pi\right. \\ $$$$\left.−\left({n}−\mathrm{1}\right)\:{e}^{−\pi} \:{sin}\left({n}−\mathrm{1}\right)\pi\:−\mathrm{1}\right\} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\left\{\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{e}^{−\pi} \:−\mathrm{1}\right\}\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} \:+\mathrm{1}}{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}\left({n}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{1}+{n}^{\mathrm{2}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} \:−\mathrm{1}\right\}=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\pi\:{a}_{{n}} \:=\:{I}\left({n}−\mathrm{1}\right)−{I}\left({n}\right)=\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}\:−\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\pi} }{\mathrm{1}+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${e}^{−{x}} \:{sinx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\pi}\left\{\:\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\boldsymbol{{e}}^{−\pi} }{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}} {e}^{−\pi} \:−\mathrm{1}}{\mathrm{1}+{n}^{\mathrm{2}} }\right\}{sin}\left({nx}\right) \\ $$