Question Number 190420 by mnjuly1970 last updated on 02/Apr/23
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{laplace}\:\:\mathrm{transform} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathcal{L}_{{t}} \:\left\{\:\:\frac{\:\mathrm{sin}\left({t}\:\right)}{{t}}\:\:\right\}\:=\:\mathcal{F}\:\left({s}\:\right)\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathcal{F}\:\left({s}\:\right)=\:\:?\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:{then}\:\:{calculate}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{\:−\mathrm{2}{t}} {sin}\left({t}\:\right)}{{t}}\:{dt}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$
Commented by mahdipoor last updated on 02/Apr/23
$$ \\ $$$${L}\left(\frac{{sin}\left({t}\right)}{{t}}\right)={L}\left(\frac{{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}!}−…}{{t}}\right)={L}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{5}!}−…\right) \\ $$$$={L}\left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}!}{L}\left({t}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{5}!}{L}\left({t}^{\mathrm{4}} \right)−…= \\ $$$$\frac{\mathrm{1}}{{s}}−\frac{\mathrm{2}!}{\mathrm{3}!{s}^{\mathrm{3}} }+\frac{\mathrm{4}!}{\mathrm{5}!{s}^{\mathrm{5}} }−….=\frac{\mathrm{1}}{{s}}−\frac{\mathrm{1}}{\mathrm{3}{s}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}{s}^{\mathrm{5}} }−…={G}\left({s}\right) \\ $$$${G}\left({s}\right)=?\Rightarrow\Omega={G}\left(\mathrm{2}\right)=? \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
$$\:\:{thx}\:…\:{G}\:\left({s}\:\right)=\:{arctan}\left(\frac{\mathrm{1}}{{s}}\right) \\ $$$${G}\left(\mathrm{2}\right)=\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right)… \\ $$
Answered by witcher3 last updated on 04/Apr/23
$$\mathrm{F}\left(\mathrm{s}\right)=\mathrm{Im}\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{−\mathrm{1}} \mathrm{e}^{−\mathrm{t}\left(\mathrm{s}−\mathrm{i}\right)} \mathrm{dt},\mathrm{s}>\mathrm{0} \\ $$$$\mathrm{F}'\left(\mathrm{s}\right)=\mathrm{Im}\int_{\mathrm{0}} ^{\infty} −\mathrm{e}^{−\mathrm{ts}+\mathrm{it}} \mathrm{dt} \\ $$$$=\mathrm{Im}\frac{\mathrm{1}}{−\mathrm{s}+\mathrm{i}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} +\mathrm{1}}\Rightarrow\mathrm{F}\left(\mathrm{s}\right)=−\mathrm{arctan}\left(\mathrm{s}\right)+\mathrm{c} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}F}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{c}=\frac{\pi}{\mathrm{2}} \\ $$$$\Omega=−\mathrm{arctan}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{2}}=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$