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Question Number 59282 by Mr X pcx last updated on 07/May/19

$$calculatef\left(x\right)={\int }_0^{\mathrm{\infty }}{e}^{-x\left[t\right]}sin\left(\left[t\right]\right)dt$$ $$withx>0$$ $$2)calculate{\int }_0^{\mathrm{\infty }}{e}^{-3\left[t\right]}sin\left(\left[t\right]\right)dt.$$

Commented bymaxmathsup by imad last updated on 08/May/19

$$1)wehavef\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{\int }_n^{n+1}{e}^{-nx}sin\left(n\right)dt=\sum _{n=0}^{\mathrm{\infty }}sin\left(n\right)e^{-nx}(n+1-n)$$ $$=\sum _{n=0}^{\mathrm{\infty }}{e}^{-nx}sin\left(n\right)=Im\left(\sum _{n=0}^{\mathrm{\infty }}{e}^{-nx+in}\right)but$$ $$\sum _{n=0}^{\mathrm{\infty }}{e}^{-nx+in}=\sum _{n=0}^{\mathrm{\infty }}{\left(e^{i-x}\right)}^n=\frac{1}{1-e^{i-x}}=\frac{1}{1-e^{-x}(cos\left(1\right)+isin\left(1\right)}$$ $$=\frac{1}{1-e^{-x}cos\left(1\right)-i{e}^{-x}sin\left(1\right)}=\frac{1-e^{-x}cos\left(1\right)+{ie}^{-x}sin\left(1\right)}{{(1-e^{-x}cos\left(1\right))}^2+e^{-2x}{sin}^2\left(1\right)}\Rightarrow$$ $$f\left(x\right)=\frac{e^{-x}sin\left(1\right)}{{(1-e^{-x}cos\left(1\right))}^2+e^{-2x}{sin}^2\left(1\right)}=\frac{e^{-x}sin\left(1\right)}{1-2e^{-x}cos\left(1\right)+e^{-2x}}$$ $$2){\int }_0^{\mathrm{\infty }}{e}^{-3\left[t\right]}sin\left(\left[t\right]\right)dt=f\left(3\right)=\frac{e^{-3}sin\left(1\right)}{e^{-6}-2e^{-3}cos\left(1\right)+1}=\frac{e^{3}sin\left(1\right)}{1-2e^{3}cos\left(1\right)+e^6}.$$

Answered by tanmay last updated on 07/May/19

$$f\left(x\right)={\int }_0^1{e}^{-x\times 0}sin\left(0\right)dt+{\int }_1^2{e}^{-x\times 1}sin\left(1\right)dt+$$ $${\int }_2^3{e}^{-x\times 2}sin\left(2\right)dt+...+{\int }_r^{r+1}{e}^{-x\times r}sin\left(r\right)dt+...$$ $$=0+e^{-x}sin1+e^{-2x}sin2+e^{-3x}sin3+...$$ $$=\underset{r=0}{\sum ^{\mathrm{\infty }}}{e}^{-rx}\times sinr$$ $$$$ $$Q=e^{-x}sin1+e^{-2x}sin2+e^{-3x}sin3+...$$ $$P=e^{-x}cos1+e^{-2x}cos2+e^{-3x}cos3+...$$ $$P+iQ=e^{-x}\times e^i+e^{-2x}\times e^{i2}+e^{-3x}\times e^{i3}+...$$ $$S=t+t^2+t^3+...\mathrm{\infty }[t=\frac{e^i}{e^x}]$$ $$S=\frac{t}{1-t}=\frac{e^i}{e^x(1-\frac{e^i}{e^x})}=\frac{cos1+isin1}{e^x-cos1-isin1}$$ $$S=\frac{(cos1+isin1)}{{(e^x-cos1)}^2+{sin}^{2}1}\times (e^x-cos1+isin1)$$ $$S=\frac{cos1(e^x-cos1)+isin1cos1+isin1(e^x-cos1)-{sin}^{2}1}{e^{2x}-2e^{x}cos1+1}$$ $$=\frac{e^{x}cos1-1+i(sin1cos1+e^{x}sin1-sin1cos1)}{e^{2x}-2e^{x}cos1+1}$$ $$soreauiredansisQ=\left(complexpart\right)$$ $$=\frac{e^{x}sin1}{e^{2x}-2e^{x}cos1+1}$$ $$plscheck...$$ $$2)\frac{e^{3}sin1}{e^6-2e^{3}cos1+1}$$ $$$$ $$$$

Commented bymaxmathsup by imad last updated on 08/May/19

$$sirTanmayyouransweriscorrectthanks...$$

Commented bytanmay last updated on 08/May/19

$$mostwelcomesir$$

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