Question Number 60681 by maxmathsup by imad last updated on 24/May/19
$${calculate}\:\:{L}\left({e}^{−\mathrm{2}{x}} {sin}\left(\alpha{x}\right)\right)\:\:\:\:\alpha\:{real}\:\:\:{and}\:{L}\:{laplace}\:{transform} \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
![L(e^(−2x) sin(αx)) =∫_0 ^∞ f(t)e^(−xt) dt =∫_0 ^∞ e^(−2t) sin(αt) e^(−xt) dt =∫_0 ^∞ e^(−(x+2)t) sin(αt)dt =Im(∫_0 ^∞ e^(−(x+2)t) e^(iαt) dt) ∫_0 ^∞ e^(−(x+2)t) e^(iαt) dt =∫_0 ^∞ e^((−(x+2) +iα)t) dt =[(1/(−(x+2) +iα)) e^(−{(x+2)+iα}) ]_0 ^∞ =(1/(x+2−iα)) =((x+2 +iα)/((x+2)^2 +α^2 )) ⇒ L (e^(−2x) sin(αx)) = (α/((x+2)^2 +α^2 )) .](https://www.tinkutara.com/question/Q60803.png)
$${L}\left({e}^{−\mathrm{2}{x}} {sin}\left(\alpha{x}\right)\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{f}\left({t}\right){e}^{−{xt}} \:{dt}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{2}{t}} \:{sin}\left(\alpha{t}\right)\:{e}^{−{xt}} \:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}+\mathrm{2}\right){t}} \:{sin}\left(\alpha{t}\right){dt}\:={Im}\left(\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}+\mathrm{2}\right){t}} \:{e}^{{i}\alpha{t}} {dt}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}+\mathrm{2}\right){t}} \:{e}^{{i}\alpha{t}} \:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(−\left({x}+\mathrm{2}\right)\:+{i}\alpha\right){t}} {dt} \\ $$$$=\left[\frac{\mathrm{1}}{−\left({x}+\mathrm{2}\right)\:+{i}\alpha}\:{e}^{−\left\{\left({x}+\mathrm{2}\right)+{i}\alpha\right\}} \right]_{\mathrm{0}} ^{\infty} \:=\frac{\mathrm{1}}{{x}+\mathrm{2}−{i}\alpha}\:=\frac{{x}+\mathrm{2}\:+{i}\alpha}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$${L}\:\left({e}^{−\mathrm{2}{x}} \:{sin}\left(\alpha{x}\right)\right)\:=\:\frac{\alpha}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:. \\ $$