Question Number 45976 by maxmathsup by imad last updated on 19/Oct/18
![find u_n = ∫_0 ^∞ e^(−n[x]) cos(nx)dx and v_n =∫_0 ^∞ e^(n[x]) sin(nx)dx 2) find nature of Σ u_n v_n and Σ (u_n /v_n )](Q45976.png)
$${find}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{n}\left[{x}\right]} {cos}\left({nx}\right){dx}\:{and}\:{v}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{{n}\left[{x}\right]} {sin}\left({nx}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} {v}_{{n}} \:\:{and}\:\Sigma\:\frac{{u}_{{n}} }{{v}_{{n}} } \\ $$
Commented by maxmathsup by imad last updated on 20/Oct/18
![we have u_n +iv_n = ∫_0 ^∞ e^(−n[x]) e^(inx) dx = ∫_0 ^∞ e^(−n[x] +inx) dx = Σ_(k=0) ^∞ ∫_k ^(k+1) e^(−kn +inx) dx =Σ_(k=0) ^∞ e^(−nk) ∫_k ^(k+1) e^(inx) dx =Σ_(k=0) ^∞ e^(−nk) (1/(in))[e^(inx) ]_k ^(k+1) =(1/(in)) Σ_(k=0) ^∞ e^(−nk) { e^(in(k+1)) −e^(ink) } =(1/(in)) e^(in) Σ_(k=0) ^∞ e^(−nk+ink) −(1/(in)) Σ_(k=0) ^∞ e^(−nk+ink) =(1/(in)) e^(in) Σ_(k=0) ^∞ (e^(−n+in) )^k −(1/(in)) Σ_(k=0) ^∞ (e^(−n+in) )^k =(1/(in)) e^(in) (1/(1−e^(−n+in) )) −(1/(in)) (1/(1−e^(−n+in) )) =((e^(in) −1)/(in)) (1/(1−e^(−n+in) )) =((cos(n)+isin(n)−1)/(in)) (1/(1−e^(−n) (cosn +isin(n)))) =((−i(cos(n)+isin(n)−1))/n) (1/(1−e^(−n) cos(n)−ie^(−n) sin(n))) =((sin(n)−i cos(n) +i)/n) ((1−e^(−n) cos(n)+i e^(−n) sin(n))/((1−e^(−n) cos(n))^2 +e^(−2n) sin^2 (n))) af ter we separate Re (..) and Im(...)...](Q46082.png)
$${we}\:{have}\:{u}_{{n}} \:+{iv}_{{n}} \:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{n}\left[{x}\right]} \:{e}^{{inx}} {dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{n}\left[{x}\right]\:+{inx}} {dx} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:{e}^{−{kn}\:+{inx}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:{e}^{{inx}} {dx}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{e}^{−{nk}} \:\frac{\mathrm{1}}{{in}}\left[{e}^{{inx}} \right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{in}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}} \:\left\{\:{e}^{{in}\left({k}+\mathrm{1}\right)} \:−{e}^{{ink}} \right\} \\ $$$$=\frac{\mathrm{1}}{{in}}\:{e}^{{in}} \:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}+{ink}} \:\:−\frac{\mathrm{1}}{{in}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{nk}+{ink}} \\ $$$$=\frac{\mathrm{1}}{{in}}\:{e}^{{in}} \:\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\left({e}^{−{n}+{in}} \right)^{{k}} \:−\frac{\mathrm{1}}{{in}}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−{n}+{in}} \right)^{{k}} \\ $$$$=\frac{\mathrm{1}}{{in}}\:{e}^{{in}} \frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}+{in}} }\:−\frac{\mathrm{1}}{{in}}\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}+{in}} }\:=\frac{{e}^{{in}} −\mathrm{1}}{{in}}\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}+{in}} } \\ $$$$=\frac{{cos}\left({n}\right)+{isin}\left({n}\right)−\mathrm{1}}{{in}}\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} \left({cosn}\:+{isin}\left({n}\right)\right)} \\ $$$$=\frac{−{i}\left({cos}\left({n}\right)+{isin}\left({n}\right)−\mathrm{1}\right)}{{n}}\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{n}} {cos}\left({n}\right)−{ie}^{−{n}} {sin}\left({n}\right)} \\ $$$$=\frac{{sin}\left({n}\right)−{i}\:{cos}\left({n}\right)\:+{i}}{{n}}\:\frac{\mathrm{1}−{e}^{−{n}} {cos}\left({n}\right)+{i}\:{e}^{−{n}} \:{sin}\left({n}\right)}{\left(\mathrm{1}−{e}^{−{n}} {cos}\left({n}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{n}} \:{sin}^{\mathrm{2}} \left({n}\right)} \\ $$$${af}\:{ter}\:\:{we}\:{separate}\:{Re}\:\left(..\right)\:{and}\:{Im}\left(...\right)... \\ $$