Question Number 49636 by maxmathsup by imad last updated on 08/Dec/18
![1) calculate A_n =∫_0 ^∞ e^(−n[x]) sin(x)dx with n integr and n≥1 2) find nature of Σ_(n=1) ^∞ A_n](https://www.tinkutara.com/question/Q49636.png)
$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{n}\left[{x}\right]} {sin}\left({x}\right){dx}\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{A}_{{n}} \\ $$
Commented by Abdo msup. last updated on 10/Dec/18
![1) A_n =Σ_(p=0) ^∞ ∫_p ^(p+1) e^(−np) sin(x)dx=Σ_(p=0) ^∞ e^(−np) A_p A_p =∫_p ^(p+1) sinx dx =[−cosx]_p ^(p+1) =cosp −cos(p+1) ⇒ A_n =Σ_(p=0) ^∞ e^(−np) cosp −Σ_(p=0) ^∞ e^(−np) cos(p+1) =Σ_(p=0) ^∞ e^(−np) cosp −Σ_(p=1) ^∞ e^(−n(p−1)) cosp =1 +(1−e^n )Σ_(p=1) ^∞ e^(−np) cosp but Σ_(p=1) ^∞ e^(−np) cosp =Re(Σ_(p=1) ^∞ e^(−np+ip) ) Σ_(p=1) ^∞ e^((−n+i)p) =Σ_(p=0) ^∞ e^((−n+i)p) −1 = (1/(1−e^(−n+i) )) −1 =(1/(1−e^(−n) (cos1 +isin1))) −1 =((1−e^(−n) cos(1)+isin(1))/((1−e^(−n) cos(1))^2 +e^(−2n) sin^2 (1))) −1 ⇒ Re(...) =((1−e^(−n) cos(1))/(1−2e^(−n) cos(1) +e^(−2n) )) −1 ⇒ A_n =1+(1−e^(−n) )(((1−e^(−n) cos(1))/(1−2e^(−n) cos(1)+e^(−2n) )) −1) .](https://www.tinkutara.com/question/Q49744.png)
$$\left.\mathrm{1}\right)\:{A}_{{n}} =\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\int_{{p}} ^{{p}+\mathrm{1}} \:{e}^{−{np}} \:{sin}\left({x}\right){dx}=\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{A}_{{p}} \\ $$$${A}_{{p}} =\int_{{p}} ^{{p}+\mathrm{1}} \:{sinx}\:{dx}\:=\left[−{cosx}\right]_{{p}} ^{{p}+\mathrm{1}} ={cosp}\:−{cos}\left({p}+\mathrm{1}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} {cosp}\:−\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} \:{cos}\left({p}+\mathrm{1}\right) \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{−{np}} {cosp}\:−\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{n}\left({p}−\mathrm{1}\right)} \:{cosp} \\ $$$$=\mathrm{1}\:\:+\left(\mathrm{1}−{e}^{{n}} \right)\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{np}} {cosp}\:{but} \\ $$$$\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{np}} {cosp}\:={Re}\left(\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{−{np}+{ip}} \right) \\ $$$$\sum_{{p}=\mathrm{1}} ^{\infty} \:{e}^{\left(−{n}+{i}\right){p}} \:=\sum_{{p}=\mathrm{0}} ^{\infty} \:{e}^{\left(−{n}+{i}\right){p}} \:−\mathrm{1} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}+{i}} }\:−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} \left({cos}\mathrm{1}\:+{isin}\mathrm{1}\right)}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{e}^{−{n}} \:{cos}\left(\mathrm{1}\right)+{isin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−{n}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{n}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:−\mathrm{1}\:\Rightarrow \\ $$$${Re}\left(…\right)\:=\frac{\mathrm{1}−{e}^{−{n}} {cos}\left(\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{e}^{−{n}} {cos}\left(\mathrm{1}\right)\:+{e}^{−\mathrm{2}{n}} }\:−\mathrm{1}\:\Rightarrow \\ $$$${A}_{{n}} =\mathrm{1}+\left(\mathrm{1}−{e}^{−{n}} \right)\left(\frac{\mathrm{1}−{e}^{−{n}} {cos}\left(\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{e}^{−{n}} {cos}\left(\mathrm{1}\right)+{e}^{−\mathrm{2}{n}} }\:−\mathrm{1}\right)\:. \\ $$
Commented by Abdo msup. last updated on 10/Dec/18
$$\left.\mathrm{2}\right)\:{we}\:{hsve}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\mathrm{1}\:\:\Rightarrow\:\Sigma\:{A}_{{n}} \:{diverges}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18
![P_n =∫_0 ^∞ e^(−n[x]) cosxdx Q_n =∫_0 ^∞ e^(−n[x]) sinx A_n =Q_n P_n +iQ_n =∫_0 ^∞ e^(−n[x]) e^(ix) dx =∫_0 ^∞ e^(−n[x]+ix) dx =∫_0 ^1 e^(−n×0+ix) dx+∫_1 ^2 e^(−n×1+ix) dx+...+∫_r ^(r+1) e^(−n×r+ix) dx+..upto ∞ =Σ_(r=0) ^∞ ∫_r ^(r+1) e^(−nr+ix) dx =Σ_(r=0) ^∞ e^(−nr) ∫_r ^(r+1) e^(ix) dx =Σ_(r=0) ^∞ e^(−nr) ×∣(e^(ix) /i)∣_r ^(r+1) =Σ_(r=0) ^∞ e^(−nr) ×(−i)(e^(i(r+1)) −e^(ir) ) =Σ_(r=0) ^∞ e^(−nr) ×i×(e^(ir) −e^(i(r+1)) ) =Σ_(r=0) ^∞ e^(−nr) ×i×[{cosr+isinr}−{cos(r+1)+isin(r+1)}] =Σ_(r=0) ^∞ e^(−nr) ×i×[2sin(r+(1/2))sin((1/2))+i{2cos(r+(1/2))sin((1/2))] =Σ_(r=0) ^∞ e^(−nr) ×2sin((1/2))[−cos(r+(1/2))+isin(r+(1/2)) =(−2)sin((1/2))Σ_(r=0) ^∞ e^(−nr) ×[cos(r+(1/2))−isin(r+(1/2))] Q_n =A_n =2sin((1/2))Σ_(r=0) ^∞ e^(−nr) sin(r+(1/2))](https://www.tinkutara.com/question/Q49711.png)
$${P}_{{n}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{n}\left[{x}\right]} {cosxdx} \\ $$$${Q}_{{n}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{n}\left[{x}\right]} {sinx} \\ $$$${A}_{{n}} ={Q}_{{n}} \\ $$$${P}_{{n}} +{iQ}_{{n}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{n}\left[{x}\right]} {e}^{{ix}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{n}\left[{x}\right]+{ix}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{n}×\mathrm{0}+{ix}} {dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {e}^{−{n}×\mathrm{1}+{ix}} {dx}+…+\int_{{r}} ^{{r}+\mathrm{1}} {e}^{−{n}×{r}+{ix}} {dx}+..{upto}\:\infty \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{{r}} ^{{r}+\mathrm{1}} {e}^{−{nr}+{ix}} {dx} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} \int_{{r}} ^{{r}+\mathrm{1}} {e}^{{ix}} {dx} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×\mid\frac{{e}^{{ix}} }{{i}}\mid_{{r}} ^{{r}+\mathrm{1}} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×\left(−{i}\right)\left({e}^{{i}\left({r}+\mathrm{1}\right)} −{e}^{{ir}} \right) \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×{i}×\left({e}^{{ir}} −{e}^{{i}\left({r}+\mathrm{1}\right)} \right) \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×{i}×\left[\left\{{cosr}+{isinr}\right\}−\left\{{cos}\left({r}+\mathrm{1}\right)+{isin}\left({r}+\mathrm{1}\right)\right\}\right] \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×{i}×\left[\mathrm{2}{sin}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{i}\left\{\mathrm{2}{cos}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right]\right. \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×\mathrm{2}{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left[−{cos}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{isin}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right. \\ $$$$=\left(−\mathrm{2}\right){sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} ×\left[{cos}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)−{isin}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$${Q}_{{n}} ={A}_{{n}} \\ $$$$=\mathrm{2}{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{nr}} {sin}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$