Question Number 38209 by prof Abdo imad last updated on 22/Jun/18
$${let}\:{f}\left({x}\right)={e}^{−{x}} {cosx} \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:. \\ $$
Commented by math khazana by abdo last updated on 25/Jun/18
![f(x)=Σ_(n=−∞) ^(+∞) c_n e^(inx) with c_n =(1/T)∫_([T]) f(x)e^(−inx) dx = (1/(2π)) ∫_(−π) ^π e^((−1−in)x) cosx dx 2π c_n = Re( ∫_(−π) ^π e^((−1−in +i)x) dx) but ∫_(−π) ^π e^(−(1+(n−1)i)x) dx=[ (1/(−(1+(n−1)i)))e^(−(1+(n−1)i)x) ]_(−π) ^π =((−1)/(1+(n−1)i)) { e^(−π) (−1)^(n−1) −e^π (−1)^(n−1) } =(((−1)^(n−1) 2i sh(π)(1−(n−1)i))/(1+(n−1)^2 )) =((2ish(π)(−1)^(n−1) +2sh(π)(−1)^(n−1) )/((n−1)^2 +1)) ⇒ 2πc_n = ((2sh(π)(−1)^(n−1) )/((n−1)^2 +1)) ⇒ c_n = (((−1)^(n−1) sh(π))/(π{ (n−1)^2 +1})) ⇒ f(x)= Σ_(n=−∞) ^(+∞) (((−1)^(n−1) sh(π))/(π{ (n−1)^2 +1})) e^(inx) =((sh(π))/π) Σ_(n=−∞) ^(+∞) (((−1)^(n−1) )/((n−1)^2 +1)) e^(inx) =e^(−x) cosx](https://www.tinkutara.com/question/Q38434.png)
$${f}\left({x}\right)=\sum_{{n}=−\infty} ^{+\infty} \:{c}_{{n}} {e}^{{inx}} \:\:{with} \\ $$$${c}_{{n}} =\frac{\mathrm{1}}{{T}}\int_{\left[{T}\right]} {f}\left({x}\right){e}^{−{inx}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:\:{e}^{\left(−\mathrm{1}−{in}\right){x}} {cosx}\:{dx} \\ $$$$\mathrm{2}\pi\:{c}_{{n}} =\:{Re}\left(\:\int_{−\pi} ^{\pi} \:\:{e}^{\left(−\mathrm{1}−{in}\:+{i}\right){x}} {dx}\right)\:{but} \\ $$$$\:\int_{−\pi} ^{\pi} \:\:\:{e}^{−\left(\mathrm{1}+\left({n}−\mathrm{1}\right){i}\right){x}} {dx}=\left[\:\frac{\mathrm{1}}{−\left(\mathrm{1}+\left({n}−\mathrm{1}\right){i}\right)}{e}^{−\left(\mathrm{1}+\left({n}−\mathrm{1}\right){i}\right){x}} \right]_{−\pi} ^{\pi} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{1}+\left({n}−\mathrm{1}\right){i}}\:\left\{\:{e}^{−\pi} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:−{e}^{\pi} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \mathrm{2}{i}\:{sh}\left(\pi\right)\left(\mathrm{1}−\left({n}−\mathrm{1}\right){i}\right)}{\mathrm{1}+\left({n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{ish}\left(\pi\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:+\mathrm{2}{sh}\left(\pi\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{2}\pi{c}_{{n}} =\:\frac{\mathrm{2}{sh}\left(\pi\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${c}_{{n}} =\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{sh}\left(\pi\right)}{\pi\left\{\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}}\:\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{sh}\left(\pi\right)}{\pi\left\{\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}}\:{e}^{{inx}} \\ $$$$=\frac{{sh}\left(\pi\right)}{\pi}\:\sum_{{n}=−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:{e}^{{inx}} \:={e}^{−{x}} {cosx} \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 25/Jun/18
$$\left.\mathrm{2}\right){f}\left(\mathrm{0}\right)=\mathrm{1}\:=\:\frac{{sh}\left(\pi\right)}{\pi}\:\sum_{{n}=−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} }{\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=_{\boldsymbol{{n}}−\mathrm{1}=\boldsymbol{{p}}} \:\:\:\frac{\boldsymbol{{sh}}\left(\pi\right)}{\pi}\:\sum_{{p}=−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{{p}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{p}=−\infty} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{{p}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\frac{\pi}{{sh}\left(\pi\right)}\:. \\ $$$$\left.\mathrm{3}\right)\:{f}\left(\pi\right)\:=\:−{e}^{−\pi} \:=\:−\frac{{sh}\left(\pi\right)}{\pi}\:\sum_{{n}=−\infty} ^{+\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow\sum_{{n}=−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\frac{\pi\:{e}^{−\pi} }{{sh}\left(\pi\right)}\:. \\ $$