Question Number 62455 by aliesam last updated on 21/Jun/19
Commented by mathmax by abdo last updated on 22/Jun/19
 ⇒A_n = e^((π/2)−n) ∫_0 ^π (sin(n)cos(t) +cos(n)sin(t))dt =sin(n) e^((π/2)−n) ∫_0 ^π cost dt+cos(n) e^((π/2)−n) ∫_0 ^π sint dt =0 +cos(n) e^((π/2)−n) [−cost]_0 ^π =2 cos(n) e^((π/2)−n) ⇒ S =2Σ_(n=1) ^∞ cos(n)e^((π/2)−n) =2 e^(π/2) Σ_(n=1) ^∞ e^(−n) cos(n) =2 e^(π/2) (Σ_(n=0) ^∞ e^(−n) cos(n)−1) but Σ_(n=0) ^∞ e^(−n) cosn =Re( Σ_(n=0) ^∞ e^(−n+in) ) =Re(Σ_(n=0) ^∞ (e^(−1+i) )^n ) we have ∣e^(−1 +i) ∣ =(1/e)<1 ⇒ Σ_(n=0) ^∞ (e^(−1+i) )^n =(1/(1−e^(−1+i) )) =(1/(1−e^(−1) (cos(1) +isin(1)))) =(1/(1−e^(−1) cos(1)−ie^(−1) sin(1))) =((1−e^(−1) cos(1)+e^(−1) sin(1))/((1−e^(−1) cos(1))^2 +e^(−2) sin^2 (1))) ⇒ Σ_(n=0) ^∞ e^(−n) cos(n) =((1−e^(−1) cos(1))/((1−e^(−1) cos(1))^2 +e^(−2) sin^2 (1))) ⇒ S =2 e^(π/2) { ((1−e^(−1) cos(1))/(1−2e^(−1) cos(1) +e^(−2) )) −1} .](Q62495.png)
$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} \:\:\:{with}\:{A}_{{n}} =\int_{{n}−\frac{\pi}{\mathrm{2}}} ^{{n}+\frac{\pi}{\mathrm{2}}} \:{e}^{−{x}} \mid{cos}\left({x}\right)\mid{dx}\:\:\:{changement}\:{x}\:={n}−\frac{\pi}{\mathrm{2}}\:+{t}\:{give} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\:{e}^{\left.−\left({n}−\frac{\pi}{\mathrm{2}}+{t}\right)\right)} \:\mid{cos}\left(\frac{\pi}{\mathrm{2}}−\left({n}+{t}\right)\right)\mid{dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$=\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{−{t}} \:\:\mid{sin}\left({n}+{t}\right)\mid\:{dt}\:\:\:\:\:\:{if}\:{we}\:{suppose}\:{sin}\left({n}+{t}\right)\geqslant\mathrm{0}\:{on}\left[\mathrm{0},\pi\right]\left(\:{that}\:{need}\right. \\ $$$$\left.{a}\:{proof}\right)\:\Rightarrow{A}_{{n}} =\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\int_{\mathrm{0}} ^{\pi} \left({sin}\left({n}\right){cos}\left({t}\right)\:+{cos}\left({n}\right){sin}\left({t}\right)\right){dt} \\ $$$$={sin}\left({n}\right)\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\int_{\mathrm{0}} ^{\pi} \:{cost}\:{dt}+{cos}\left({n}\right)\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\int_{\mathrm{0}} ^{\pi} \:{sint}\:{dt} \\ $$$$=\mathrm{0}\:+{cos}\left({n}\right)\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\left[−{cost}\right]_{\mathrm{0}} ^{\pi} \:=\mathrm{2}\:{cos}\left({n}\right)\:{e}^{\frac{\pi}{\mathrm{2}}−{n}} \:\Rightarrow \\ $$$${S}\:=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{cos}\left({n}\right){e}^{\frac{\pi}{\mathrm{2}}−{n}} \:=\mathrm{2}\:{e}^{\frac{\pi}{\mathrm{2}}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{n}} \:{cos}\left({n}\right) \\ $$$$=\mathrm{2}\:{e}^{\frac{\pi}{\mathrm{2}}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{e}^{−{n}} \:{cos}\left({n}\right)−\mathrm{1}\right)\:\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \:{cosn}\:={Re}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}+{in}} \right)\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}+{i}} \right)^{{n}} \right)\:\:{we}\:{have}\:\mid{e}^{−\mathrm{1}\:+{i}} \mid\:=\frac{\mathrm{1}}{{e}}<\mathrm{1}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \left({e}^{−\mathrm{1}+{i}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}+{i}} }\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} \left({cos}\left(\mathrm{1}\right)\:+{isin}\left(\mathrm{1}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)−{ie}^{−\mathrm{1}} {sin}\left(\mathrm{1}\right)}\:=\frac{\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)+{e}^{−\mathrm{1}} \:{sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} \:{sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \:{cos}\left({n}\right)\:=\frac{\mathrm{1}−{e}^{−\mathrm{1}} \:{cos}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{1}} \:{cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:\Rightarrow \\ $$$${S}\:=\mathrm{2}\:{e}^{\frac{\pi}{\mathrm{2}}} \left\{\:\frac{\mathrm{1}−{e}^{−\mathrm{1}} \:{cos}\left(\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{e}^{−\mathrm{1}} \:{cos}\left(\mathrm{1}\right)\:+{e}^{−\mathrm{2}} }\:−\mathrm{1}\right\}\:. \\ $$$$ \\ $$
Commented by aliesam last updated on 22/Jun/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{brilliant}\:\mathrm{solution} \\ $$
Commented by mathmax by abdo last updated on 22/Jun/19
$${you}\:{are}\:{welcome}\:{sir}. \\ $$