Laplace transform of $t \cos(t)$ by definition I want to find the Laplace transform of $t \cos(t)$ by the definition $$\int e^{-st} t \cos(2t)dt$$
The solution manual just say try the $$u = t, dv = e^{-st} \cos(2t)$$
I use the integration by parts, but still cannot delete any function. Where is the problem?
 A: First note that it is sufficient to find $$I(s) = \int e^{-st} \cos(2t) dt$$ since $$I'(s) = \int - t e^{-st} \cos(2t) dt$$ which gives us the desired integral we are looking for. Now note that
$$I(s) = \int e^{-st} \cos(2t) dt = -\dfrac1s \int \cos(2t) d\left( e^{-st} \right)$$
Hence, $$\int \cos(2t) d\left( e^{-st} \right) = \cos(2t) e^{-st} - \int e^{-st} d\left( \cos(2t) \right) = \cos(2t) e^{-st} + 2 \int e^{-st} \sin(2t) dt$$
Now
$$\int e^{-st} \sin(2t) dt = -\dfrac1s \int \sin(2t) d \left( e^{-st}\right)$$
$$\int \sin(2t) d \left( e^{-st}\right) = \sin(2t) e^{-st} - \int e^{-st} d(\sin(2t)) = \sin(2t) e^{-st} - 2 \int e^{-st} \cos(2t) dt$$
Hence, putting all this together, we get that
$$I(s) = - \dfrac1s \left(\cos(2t) e^{-st}- \dfrac2s \left(\sin(2t) e^{-st} - 2 I(s) \right) \right)$$
This gives us
$$s^2 I(s) = -\left(s \cos(2t) e^{-st} - 2\sin(2t)e^{-st} + 4 I(s) \right)$$
Hence, we get that
$$\left(s^2 + 4\right) I(s) = \left(2 \sin(2t) - s \cos(2t) \right) e^{-st}$$
This gives us that $$I(s) = \dfrac{2 \sin(2t) - s \cos(2t)}{s^2 + 4} e^{-st} + c$$
Now $$J(s) = \int_0^{\infty} e^{-st} \cos(2t) dt = \dfrac{s}{s^2+4}$$
The Laplace transform is given by
$$L(t\cos(2t)) = \int t e^{-st} \cos(2t) dt = - J'(s) = \dfrac{s^2-4}{\left(s^2+4 \right)^2}$$
A: Hint:
Integrate $e^{-st}\cos{2t}$ by parts twice.
