Question Number 28683 by abdo imad last updated on 28/Jan/18
$${developp}\:{f}\left({x}\right)={e}^{−\alpha{x}} \:\:\:\mathrm{2}\pi\:{periodic}\:{at}\:{Fourier}\:{serie}\:{with} \\ $$$$\alpha>\mathrm{0}. \\ $$
Commented by abdo imad last updated on 31/Jan/18
![f(x)=Σ_(n=−∞) ^(+∞) c_n e^(inx) and c_n = (1/T) ∫_([T]) f(x) e^(−inx) dx =(1/(2π)) ∫_(−π) ^π e^(−αx) e^(−inx) dx ⇒2π c_n = ∫_(−π) ^π e^(−(α+in)x) dx =((−1)/(α+in)) [ e^(−(α+in)x) ]_(−π) ^π =((−1)/(α+in)) ( e^(−(α+in)π) −e^((𝛂+in)π) ) =((−1)/(α+in))( (−1)^n e^(−απ) −(−1)^n e^(απ) ) = (((−1)^n )/(α+in))(e^(απ) −e^(−απ) ) =((2(−1)^n )/(α+in)) sh(απ) ⇒ c_n = ((sh(απ))/π) (((−1)^n )/(α +in)) so f(x)= Σ_(n=−∞) ^(+∞) ((sh(απ))/π) (((−1)^n )/(α+in)) e^(inx) .](https://www.tinkutara.com/question/Q28880.png)
$${f}\left({x}\right)=\sum_{{n}=−\infty} ^{+\infty} \:{c}_{{n}} \:{e}^{{inx}} \:\:{and}\:{c}_{{n}} =\:\frac{\mathrm{1}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right)\:{e}^{−{inx}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{e}^{−\alpha{x}} \:{e}^{−{inx}} {dx}\:\Rightarrow\mathrm{2}\pi\:{c}_{{n}} =\:\int_{−\pi} ^{\pi} \:{e}^{−\left(\alpha+{in}\right){x}} {dx} \\ $$$$=\frac{−\mathrm{1}}{\alpha+{in}}\:\left[\:{e}^{−\left(\alpha+{in}\right){x}} \:\right]_{−\pi} ^{\pi} \:=\frac{−\mathrm{1}}{\alpha+{in}}\:\left(\:\:{e}^{−\left(\alpha+{in}\right)\pi} \:−\boldsymbol{{e}}^{\left(\boldsymbol{\alpha}+{in}\right)\pi} \right) \\ $$$$=\frac{−\mathrm{1}}{\alpha+{in}}\left(\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\alpha\pi} \:−\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} \right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha+{in}}\left({e}^{\alpha\pi} \:−{e}^{−\alpha\pi} \right) \\ $$$$=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\alpha+{in}}\:{sh}\left(\alpha\pi\right)\:\Rightarrow\:\:{c}_{{n}} =\:\frac{{sh}\left(\alpha\pi\right)}{\pi}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha\:+{in}}\:{so} \\ $$$${f}\left({x}\right)=\:\:\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{{sh}\left(\alpha\pi\right)}{\pi}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha+{in}}\:{e}^{{inx}} \:\:. \\ $$