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Question Number 190420 by mnjuly1970 last updated on 02/Apr/23

laplace transform −−−−−−− L_t { (( sin(t ))/t) } = F (s ) F (s )= ? then calculate . Ω=∫_0 ^( ∞) ((e^( −2t) sin(t ))/t) dt =? −−−−−−−−

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{laplace}\:\:\mathrm{transform} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathcal{L}_{{t}} \:\left\{\:\:\frac{\:\mathrm{sin}\left({t}\:\right)}{{t}}\:\:\right\}\:=\:\mathcal{F}\:\left({s}\:\right)\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathcal{F}\:\left({s}\:\right)=\:\:?\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:{then}\:\:{calculate}\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{\:−\mathrm{2}{t}} {sin}\left({t}\:\right)}{{t}}\:{dt}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$

Commented by mahdipoor last updated on 02/Apr/23

L(((sin(t))/t))=L(((t−(t^3 /(3!))+(t^5 /(5!))−...)/t))=L(1−(t^2 /(3!))+(t^4 /(5!))−...) =L(1)−(1/(3!))L(t^2 )+(1/(5!))L(t^4 )−...= (1/s)−((2!)/(3!s^3 ))+((4!)/(5!s^5 ))−....=(1/s)−(1/(3s^3 ))+(1/(5s^5 ))−...=G(s) G(s)=?⇒Ω=G(2)=?

$$ \\ $$$${L}\left(\frac{{sin}\left({t}\right)}{{t}}\right)={L}\left(\frac{{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}!}−...}{{t}}\right)={L}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{5}!}−...\right) \\ $$$$={L}\left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{3}!}{L}\left({t}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{5}!}{L}\left({t}^{\mathrm{4}} \right)−...= \\ $$$$\frac{\mathrm{1}}{{s}}−\frac{\mathrm{2}!}{\mathrm{3}!{s}^{\mathrm{3}} }+\frac{\mathrm{4}!}{\mathrm{5}!{s}^{\mathrm{5}} }−....=\frac{\mathrm{1}}{{s}}−\frac{\mathrm{1}}{\mathrm{3}{s}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}{s}^{\mathrm{5}} }−...={G}\left({s}\right) \\ $$$${G}\left({s}\right)=?\Rightarrow\Omega={G}\left(\mathrm{2}\right)=? \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 02/Apr/23

thx ... G (s )= arctan((1/s)) G(2)= arctan((1/2) )...

$$\:\:{thx}\:...\:{G}\:\left({s}\:\right)=\:{arctan}\left(\frac{\mathrm{1}}{{s}}\right) \\ $$$${G}\left(\mathrm{2}\right)=\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right)... \\ $$

Answered by witcher3 last updated on 04/Apr/23

F(s)=Im∫_0 ^∞ t^(−1) e^(−t(s−i)) dt,s>0 F′(s)=Im∫_0 ^∞ −e^(−ts+it) dt =Im(1/(−s+i)) =−(1/(s^2 +1))⇒F(s)=−arctan(s)+c lim_(x→∞) F(x)=0⇒c=(π/2) Ω=−arctan(2)+(π/2)=arctan((1/2))

$$\mathrm{F}\left(\mathrm{s}\right)=\mathrm{Im}\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{−\mathrm{1}} \mathrm{e}^{−\mathrm{t}\left(\mathrm{s}−\mathrm{i}\right)} \mathrm{dt},\mathrm{s}>\mathrm{0} \\ $$$$\mathrm{F}'\left(\mathrm{s}\right)=\mathrm{Im}\int_{\mathrm{0}} ^{\infty} −\mathrm{e}^{−\mathrm{ts}+\mathrm{it}} \mathrm{dt} \\ $$$$=\mathrm{Im}\frac{\mathrm{1}}{−\mathrm{s}+\mathrm{i}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} +\mathrm{1}}\Rightarrow\mathrm{F}\left(\mathrm{s}\right)=−\mathrm{arctan}\left(\mathrm{s}\right)+\mathrm{c} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}F}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{c}=\frac{\pi}{\mathrm{2}} \\ $$$$\Omega=−\mathrm{arctan}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{2}}=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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