Laplace transform of $\,\sin\left(t\right) \,\frac{d^2y}{{dt}^2}$

Question

I would like to know what is, and how to work out the Laplace transform with respect to $t$ of:

$$\sin\left(t\right)\,\dfrac{d^2y}{{dt}^2}$$

I know that the transform of $\sin\left(t\right)$ is $\,\dfrac{a}{s^2+a^2}$, and transform of $\,\dfrac{d^2y}{{dt}^2}\,$ is $s^2F\left(s\right) - s\,f\left(0\right) - s\,f'\left(0\right)$

Answer 1

I assume $y$ and its derivatives are exponentially bounded. Then using integration by parts twice, I get (for $\text{Re}(s)$ sufficiently large)

$$ y(0) + (s^2-1) F(s) - 2 s G(s) $$ where $F(s)$ and $G(s)$ are the Laplace transforms of $y(t) \sin(t)$ and $y(t) \cos(t)$ respectively.

Answer 2

Explicit computation yiels

$$\int_{0}^{\infty} y''(t) \> \sin t \> e^{-st} \> dt$$ $$=\sin t \> e^{-st} \> y'(t) \> |_{0}^{\infty} + \int_{0}^{\infty} y'(t) (se^{-st} \sin t - e^{-st} \cos t) \> dt $$

We see the first term is $0$ if y is of exponential order. Assuming this we continue

$$= se^{-st} \sin t \> y \> |_0^{\infty} - \int_{0}^{\infty} y (se^{-st} \cos t + \sin t (-s^2 e^{-st})) \> dt - y \> e^{-st} \cos t \> |_0^{\infty} + \int_{0}^{\infty} y (-e^{-st} \sin t + \cos t(-se^{-st})) \> dt$$

We have

$$\mathscr{L}\{y''(t) \sin t\}= y(0) + (s^2 - 1) \int_{0}^{\infty} y \> e^{-st} \sin t \> dt - 2s\int_{0}^{\infty} y \> \cos t e^{-st} \> dt$$