Question Number 219871 by SdC355 last updated on 03/May/25
$${F}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} +\alpha^{\mathrm{2}} }{e}^{−{st}} \:\mathrm{d}{t} \\ $$$$\mathrm{F}\left(\mathrm{3}\right)=?? \\ $$
Answered by MrGaster last updated on 04/May/25
![F(3)=∫_0 ^∞ ((sin(t))/(t^2 +α^2 ))e^(−3t) dt F(3)=Im[∫_0 ^∞ (e^(−(3−i)t) /(t^2 +α^2 ))dt] ∫_0 ^∞ (e^(−kt) /(t^2 +α^2 ))dt=(π/(2e))e^(−kα) (via contour integration) k=3−i⇒Im[(π/(2α))e^(−(3−i)α) ] e^(−(3−i)α) =e^(−3α) (cos(α)+i sin(α)) Im[e^(−3α) (cos(α)+i sin(α))]=e^(−3α) sin(α) F(3)=(π/(2α))e^(−3α) sinh(α)](https://www.tinkutara.com/question/Q219980.png)
$${F}\left(\mathrm{3}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({t}\right)}{{t}^{\mathrm{2}} +\alpha^{\mathrm{2}} }{e}^{−\mathrm{3}{t}} {dt} \\ $$$${F}\left(\mathrm{3}\right)=\mathrm{Im}\left[\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\left(\mathrm{3}−{i}\right){t}} }{{t}^{\mathrm{2}} +\alpha^{\mathrm{2}} }{dt}\right] \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{kt}} }{{t}^{\mathrm{2}} +\alpha^{\mathrm{2}} }{dt}=\frac{\pi}{\mathrm{2}{e}}{e}^{−{k}\alpha} \left(\mathrm{via}\:\mathrm{contour}\:\mathrm{integration}\right) \\ $$$${k}=\mathrm{3}−{i}\Rightarrow\mathrm{Im}\left[\frac{\pi}{\mathrm{2}\alpha}{e}^{−\left(\mathrm{3}−{i}\right)\alpha} \right] \\ $$$${e}^{−\left(\mathrm{3}−{i}\right)\alpha} ={e}^{−\mathrm{3}\alpha} \left(\mathrm{cos}\left(\alpha\right)+{i}\:\mathrm{sin}\left(\alpha\right)\right) \\ $$$$\mathrm{Im}\left[{e}^{−\mathrm{3}\alpha} \left(\mathrm{cos}\left(\alpha\right)+{i}\:\mathrm{sin}\left(\alpha\right)\right)\right]={e}^{−\mathrm{3}\alpha} \mathrm{sin}\left(\alpha\right) \\ $$$${F}\left(\mathrm{3}\right)=\frac{\pi}{\mathrm{2}\alpha}{e}^{−\mathrm{3}\alpha} \mathrm{sinh}\left(\alpha\right) \\ $$