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Question Number 87530 by mathmax by abdo last updated on 04/Apr/20
![1) calculate U_n =∫_0 ^∞ e^(−n[x]) sin(((πx)/n))dx nnatural and n≥1 2)determine nature of Σ U_n](Q87530.png)
$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{n}\left[{x}\right]} {sin}\left(\frac{\pi{x}}{{n}}\right){dx}\:\:{nnatural}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right){determine}\:{nature}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 05/Apr/20
![1) U_n =Σ_(p=0) ^∞ ∫_p ^(p+1) e^(−np) sin(((πx)/n))dx =Σ_(p=0) ^∞ e^(−np) ∫_p ^(p+1) sin(((πx)/n))dx =Σ_(p=0) ^∞ e^(−np) [−(n/π)cos(((πx)/n))]_p ^(p+1) =−(n/π)Σ_(p=0) ^∞ e^(−np) { cos((π/n)(p+1)}−cos(((pπ)/n))} =−(n/π)Σ_(p=0) ^∞ e^(−np) cos(((π(p+1))/n))+(n/π)Σ_(p=0) ^∞ e^(−np) cos(((pπ)/n)) we have Σ_(p=0) ^∞ e^(−np) cos(((pπ)/n)) =Re(Σ_(p=0) ^∞ e^(−np+i((pπ)/n)) )=Re(A_n ) and A_n =Σ_(p=0) ^∞ (e^(−n+((iπ)/n)) )^p =(1/(1−e^(−n+((iπ)/n)) )) =(1/(1−e^(−n) {cos((π/n))+isin((π/n))})) =(1/(1−e^(−n) cos((π/n))−ie^(−n) sin((π/n)))) =((1−e^(−n) cos((π/n))+ie^(−n) sin((π/n)))/((1−e^(−n) cos((π/n)))^2 +e^(−2n) sin^2 ((π/n)))) ⇒ A_n =((1−e^(−n) cos((π/n)))/((1−e^(−n) cos((π/n)))^2 +e^(−2n) sin^2 ((π/n)))) Σ_(p=0) ^∞ e^(−np) cos(((π(p+1))/n)) =Σ_(p=1) ^∞ e^(−n(p−1)) cos(((πp)/n)) =e^n Σ_(p=1) ^∞ e^(−np) cos(((πp)/n)) =e^n (Σ_(p=1) ^∞ e^(−np) cos(((πp)/n))−1) =e^n (A_n −1) ⇒ U_n =−(n/π)e^n (A_n −1) +(n/π) A_n ⇒ U_n =(n/π)(1−e^n )A_n +((ne^n )/π)](Q87672.png)
$$\left.\mathrm{1}\right)\:{U}_{{n}} =\sum_{{p}=\mathrm{0}} ^{\infty} \:\int_{{p}} ^{{p}+\mathrm{1}} \:{e}^{−{np}} \:{sin}\left(\frac{\pi{x}}{{n}}\right){dx} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:\int_{{p}} ^{{p}+\mathrm{1}} \:{sin}\left(\frac{\pi{x}}{{n}}\right){dx}\: \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:\left[−\frac{{n}}{\pi}{cos}\left(\frac{\pi{x}}{{n}}\right)\right]_{{p}} ^{{p}+\mathrm{1}} \\ $$$$=−\frac{{n}}{\pi}\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \left\{\:{cos}\left(\frac{\pi}{{n}}\left({p}+\mathrm{1}\right)\right\}−{cos}\left(\frac{{p}\pi}{{n}}\right)\right\} \\ $$$$=−\frac{{n}}{\pi}\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{\pi\left({p}+\mathrm{1}\right)}{{n}}\right)+\frac{{n}}{\pi}\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{{p}\pi}{{n}}\right) \\ $$$${we}\:{have}\:\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{{p}\pi}{{n}}\right)\:={Re}\left(\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}+{i}\frac{{p}\pi}{{n}}} \right)={Re}\left({A}_{{n}} \right) \\ $$$${and}\:{A}_{{n}} =\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\left({e}^{−{n}+\frac{{i}\pi}{{n}}} \right)^{{p}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}+\frac{{i}\pi}{{n}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} \left\{{cos}\left(\frac{\pi}{{n}}\right)+{isin}\left(\frac{\pi}{{n}}\right)\right\}}\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} \:{cos}\left(\frac{\pi}{{n}}\right)−{ie}^{−{n}} \:{sin}\left(\frac{\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{1}−{e}^{−{n}} \:{cos}\left(\frac{\pi}{{n}}\right)+{ie}^{−{n}} \:{sin}\left(\frac{\pi}{{n}}\right)}{\left(\mathrm{1}−{e}^{−{n}} \:{cos}\left(\frac{\pi}{{n}}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{n}} \:{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}}\right)}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}−{e}^{−{n}} \:{cos}\left(\frac{\pi}{{n}}\right)}{\left(\mathrm{1}−{e}^{−{n}} \:{cos}\left(\frac{\pi}{{n}}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{n}} \:{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}}\right)} \\ $$$$\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{\pi\left({p}+\mathrm{1}\right)}{{n}}\right)\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{n}\left({p}−\mathrm{1}\right)} \:{cos}\left(\frac{\pi{p}}{{n}}\right) \\ $$$$={e}^{{n}} \:\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{\pi{p}}{{n}}\right)\:={e}^{{n}} \left(\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{np}} \:{cos}\left(\frac{\pi{p}}{{n}}\right)−\mathrm{1}\right) \\ $$$$={e}^{{n}} \left({A}_{{n}} −\mathrm{1}\right)\:\Rightarrow\:{U}_{{n}} =−\frac{{n}}{\pi}{e}^{{n}} \left({A}_{{n}} −\mathrm{1}\right)\:+\frac{{n}}{\pi}\:{A}_{{n}} \\ $$$$\Rightarrow\:{U}_{{n}} =\frac{{n}}{\pi}\left(\mathrm{1}−{e}^{{n}} \right){A}_{{n}} \:+\frac{{ne}^{{n}} }{\pi} \\ $$