Question Number 59282 by Mr X pcx last updated on 07/May/19
![calculate f(x)=∫_0 ^∞ e^(−x[t]) sin([t])dt with x>0 2) calculate ∫_0 ^∞ e^(−3[t]) sin([t])dt .](https://www.tinkutara.com/question/Q59282.png)
$${calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}\left[{t}\right]} {sin}\left(\left[{t}\right]\right){dt} \\ $$$${with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{3}\left[{t}\right]} {sin}\left(\left[{t}\right]\right){dt}\:. \\ $$
Commented by maxmathsup by imad last updated on 08/May/19
![1) we have f(x) =Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−nx) sin(n) dt =Σ_(n=0) ^∞ sin(n)e^(−nx) (n+1−n) =Σ_(n=0) ^∞ e^(−nx) sin(n) =Im( Σ_(n=0) ^∞ e^(−nx+in) ) but Σ_(n=0) ^∞ e^(−nx +in) =Σ_(n=0) ^∞ (e^(i−x) )^n =(1/(1−e^(i−x) )) =(1/(1−e^(−x) (cos(1)+isin(1))) =(1/(1−e^(−x) cos(1)−i e^(−x) sin(1))) =((1−e^(−x) cos(1)+ie^(−x) sin(1))/((1−e^(−x) cos(1))^2 +e^(−2x) sin^2 (1))) ⇒ f(x) =((e^(−x) sin(1))/((1−e^(−x) cos(1))^2 +e^(−2x) sin^2 (1))) =((e^(−x) sin(1))/(1−2 e^(−x) cos(1) +e^(−2x) )) 2) ∫_0 ^∞ e^(−3[t]) sin([t])dt =f(3) =((e^(−3) sin(1))/(e^(−6) −2e^(−3) cos(1)+1)) =((e^3 sin(1))/(1−2e^3 cos(1) +e^6 )) .](https://www.tinkutara.com/question/Q59320.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:{e}^{−{nx}} \:{sin}\left({n}\right)\:{dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{sin}\left({n}\right){e}^{−{nx}} \left({n}+\mathrm{1}−{n}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}} \:{sin}\left({n}\right)\:={Im}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}+{in}} \right)\:\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}\:+{in}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({e}^{{i}−{x}} \right)^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{{i}−{x}} }\:=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} \left({cos}\left(\mathrm{1}\right)+{isin}\left(\mathrm{1}\right)\right.} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} {cos}\left(\mathrm{1}\right)−{i}\:{e}^{−{x}} {sin}\left(\mathrm{1}\right)}\:=\frac{\mathrm{1}−{e}^{−{x}} {cos}\left(\mathrm{1}\right)+{ie}^{−{x}} {sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−{x}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{x}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{{e}^{−{x}} {sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−{x}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{x}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:=\frac{{e}^{−{x}} {sin}\left(\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}\:{e}^{−{x}} {cos}\left(\mathrm{1}\right)\:+{e}^{−\mathrm{2}{x}} } \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\mathrm{3}\left[{t}\right]} {sin}\left(\left[{t}\right]\right){dt}\:={f}\left(\mathrm{3}\right)\:=\frac{{e}^{−\mathrm{3}} {sin}\left(\mathrm{1}\right)}{{e}^{−\mathrm{6}} \:−\mathrm{2}{e}^{−\mathrm{3}} {cos}\left(\mathrm{1}\right)+\mathrm{1}}\:=\frac{{e}^{\mathrm{3}} {sin}\left(\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{e}^{\mathrm{3}} {cos}\left(\mathrm{1}\right)\:+{e}^{\mathrm{6}} }\:. \\ $$
Answered by tanmay last updated on 07/May/19
![f(x)=∫_0 ^1 e^(−x×0) sin(0)dt+∫_1 ^2 e^(−x×1) sin(1)dt+ ∫_2 ^3 e^(−x×2) sin(2)dt +...+∫_r ^(r+1) e^(−x×r) sin(r)dt+... =0+e^(−x) sin1+e^(−2x) sin2+e^(−3x) sin3+... =Σ_(r=0) ^∞ e^(−rx) ×sinr Q=e^(−x) sin1+e^(−2x) sin2+e^(−3x) sin3+... P=e^(−x) cos1+e^(−2x) cos2+e^(−3x) cos3+... P+iQ=e^(−x) ×e^i +e^(−2x) ×e^(i2) +e^(−3x) ×e^(i3) +... S=t+t^2 +t^3 +...∞ [t=(e^i /e^x )] S=(t/(1−t))=(e^i /(e^x (1−(e^i /e^x ))))=((cos1+isin1)/(e^x −cos1−isin1)) S=(((cos1+isin1))/((e^x −cos1)^2 +sin^2 1))×(e^x −cos1+isin1) S=((cos1(e^x −cos1)+isin1cos1+isin1(e^x −cos1)−sin^2 1)/(e^(2x) −2e^x cos1+1)) =((e^x cos1−1+i(sin1cos1+e^x sin1−sin1cos1))/(e^(2x) −2e^x cos1+1)) so reauired ans is Q =(complex part) =((e^x sin1)/(e^(2x) −2e^x cos1+1)) pls check ... 2)((e^3 sin1)/(e^6 −2e^3 cos1+1))](https://www.tinkutara.com/question/Q59292.png)
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}×\mathrm{0}} {sin}\left(\mathrm{0}\right){dt}+\int_{\mathrm{1}} ^{\mathrm{2}} {e}^{−{x}×\mathrm{1}} {sin}\left(\mathrm{1}\right){dt}+ \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} {e}^{−{x}×\mathrm{2}} {sin}\left(\mathrm{2}\right){dt}\:+…+\int_{{r}} ^{{r}+\mathrm{1}} {e}^{−{x}×{r}} {sin}\left({r}\right){dt}+… \\ $$$$=\mathrm{0}+{e}^{−{x}} {sin}\mathrm{1}+{e}^{−\mathrm{2}{x}} {sin}\mathrm{2}+{e}^{−\mathrm{3}{x}} {sin}\mathrm{3}+… \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{rx}} ×{sinr} \\ $$$$ \\ $$$${Q}={e}^{−{x}} {sin}\mathrm{1}+{e}^{−\mathrm{2}{x}} {sin}\mathrm{2}+{e}^{−\mathrm{3}{x}} {sin}\mathrm{3}+… \\ $$$${P}={e}^{−{x}} {cos}\mathrm{1}+{e}^{−\mathrm{2}{x}} {cos}\mathrm{2}+{e}^{−\mathrm{3}{x}} {cos}\mathrm{3}+… \\ $$$${P}+{iQ}={e}^{−{x}} ×{e}^{{i}} +{e}^{−\mathrm{2}{x}} ×{e}^{{i}\mathrm{2}} +{e}^{−\mathrm{3}{x}} ×{e}^{{i}\mathrm{3}} +… \\ $$$${S}={t}+{t}^{\mathrm{2}} +{t}^{\mathrm{3}} +…\infty\:\:\:\:\left[{t}=\frac{{e}^{{i}} }{{e}^{{x}} }\right] \\ $$$${S}=\frac{{t}}{\mathrm{1}−{t}}=\frac{{e}^{{i}} }{{e}^{{x}} \left(\mathrm{1}−\frac{{e}^{{i}} }{{e}^{{x}} }\right)}=\frac{{cos}\mathrm{1}+{isin}\mathrm{1}}{{e}^{{x}} −{cos}\mathrm{1}−{isin}\mathrm{1}} \\ $$$${S}=\frac{\left({cos}\mathrm{1}+{isin}\mathrm{1}\right)}{\left({e}^{{x}} −{cos}\mathrm{1}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \mathrm{1}}×\left({e}^{{x}} −{cos}\mathrm{1}+{isin}\mathrm{1}\right) \\ $$$${S}=\frac{{cos}\mathrm{1}\left({e}^{{x}} −{cos}\mathrm{1}\right)+{isin}\mathrm{1}{cos}\mathrm{1}+{isin}\mathrm{1}\left({e}^{{x}} −{cos}\mathrm{1}\right)−{sin}^{\mathrm{2}} \mathrm{1}}{{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} {cos}\mathrm{1}+\mathrm{1}} \\ $$$$=\frac{{e}^{{x}} {cos}\mathrm{1}−\mathrm{1}+{i}\left({sin}\mathrm{1}{cos}\mathrm{1}+{e}^{{x}} {sin}\mathrm{1}−{sin}\mathrm{1}{cos}\mathrm{1}\right)}{{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} {cos}\mathrm{1}+\mathrm{1}} \\ $$$${so}\:{reauired}\:{ans}\:{is}\:{Q}\:=\left({complex}\:{part}\right) \\ $$$$=\frac{{e}^{{x}} {sin}\mathrm{1}}{{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} {cos}\mathrm{1}+\mathrm{1}} \\ $$$${pls}\:{check}\:… \\ $$$$\left.\mathrm{2}\right)\frac{{e}^{\mathrm{3}} {sin}\mathrm{1}}{{e}^{\mathrm{6}} −\mathrm{2}{e}^{\mathrm{3}} {cos}\mathrm{1}+\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 08/May/19
$${sir}\:{Tanmay}\:{your}\:{answer}\:{is}\:{correct}\:{thanks}… \\ $$
Commented by tanmay last updated on 08/May/19
$${most}\:{welcome}\:{sir} \\ $$