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Question Number 219519 by OmoloyeMichael last updated on 27/Apr/25
find the laplace transform of ∫_0 ^∞ te^(−2t) sintdt
$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}}\:\boldsymbol{{of}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{te}}^{−\mathrm{2}\boldsymbol{{t}}} \boldsymbol{{sintdt}} \\ $$
Answered by SdC355 last updated on 27/Apr/25
First idea..Let′s define F(s) as F(s)=∫_0 ^( ∞) sin(t)e^(−st) dt −((d )/ds)F(s)=−((d )/ds)∫_0 ^( ∞) sin(t)e^(−st) dt=∫_0 ^( ∞) t∙sin(t)e^(−st) dt F(s)=(1/(s^2 +1)) , s∈[0,∞) −((d )/ds) (1/(s^2 +1))=((2s)/((s^2 +1)^2 )) ∴ ∫_0 ^( ∞) t∙sin(t)e^(−st) dt=((2s)/((s^2 +1)^2 )) ∴∫_0 ^( ∞) t∙sin(t)e^(−2t) dt=(4/(25))
$$\mathrm{First}\:\mathrm{idea}..\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:{F}\left({s}\right)\:\mathrm{as}\: \\ $$$${F}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\mathrm{sin}\left({t}\right){e}^{−{st}} \mathrm{d}{t} \\ $$$$−\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}{F}\left({s}\right)=−\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}\left({t}\right){e}^{−{st}} \mathrm{d}{t}=\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot\mathrm{sin}\left({t}\right){e}^{−{st}} \mathrm{d}{t} \\ $$$${F}\left({s}\right)=\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}}\:,\:{s}\in\left[\mathrm{0},\infty\right) \\ $$$$−\frac{\mathrm{d}\:\:}{\mathrm{d}{s}}\:\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}{s}}{\left({s}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot\mathrm{sin}\left({t}\right){e}^{−{st}} \:\mathrm{d}{t}=\frac{\mathrm{2}{s}}{\left({s}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot\mathrm{sin}\left({t}\right){e}^{−\mathrm{2}{t}} \:\mathrm{d}{t}=\frac{\mathrm{4}}{\mathrm{25}} \\ $$
Commented by OmoloyeMichael last updated on 28/Apr/25
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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