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Find $\displaystyle \mathcal{L}\left (\int\limits_0^t e^{-2\tau}\sin(3\tau)\;d\tau \right )$. I know we can solve this problem by solving the integral and then finding the Laplace Transformation. Is this can be solved using convolution? I tried to manipulate $$ \int\limits_0^t e^{-2\tau}\sin(3\tau)\;d\tau =e^{-2t}\int\limits_0^t e^{2(t-\tau)}\sin(3\tau)\;d\tau,$$ but I have no idea how to solve using the convolution theorem to find the Laplace Transformation.

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    $\begingroup$ Why would you want to use convolutions here? Doesn't seem intuitive or make it simpler. Maybe you could use the fact that if $f(t)\rightarrow F(s)$, then $\int_0^t f(t)dt\rightarrow F(s)\dfrac1s$ $\endgroup$ Commented Nov 21, 2023 at 15:40
  • $\begingroup$ Moreover, this integral can be computed exactly and the result is not harder to Laplace transform. $\endgroup$
    – Abezhiko
    Commented Nov 21, 2023 at 15:57

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$$ \begin{aligned} \mathcal{L}\left(\int_0^t e^{-2\tau}\sin(3\tau)d\tau\right) &= \mathcal{L}\left(e^{-2t}\int_0^t e^{2(t-τ)}\sin(3\tau)d\tau\right) = \\ &= \left[ \begin{array}{l} \text{by using: }\mathcal{L}\left(e^{at}y(t)\right) = Y(s-a), \\ \text{where }a = -2,\text{ and }Y(s)\text{ is a Laplace transform of the convolution:} \\ \begin{aligned} \mathcal{L}\left(\int_0^t e^{2(t-τ)}\sin(3\tau)d\tau\right) &= \mathcal{L}\left(e^{2t}\right)\cdot\mathcal{L}\left(\sin(3t)\right) = \\ &= \frac{1}{s-2}\cdot\frac{3}{s^2+3^2} \end{aligned} \end{array} \right] = \\ &= \frac{1}{(s-(-2))-2}\cdot\frac{3}{(s-(-2))^2+9} = \\ &= \frac{3}{s(s^2+4s + 13)} = \\ &= \frac{3}{s^3+4s^2 + 13s} \end{aligned} $$

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