Laplace transform using the definition

Question

Find the Laplace of the given function using the definition

$$f(t)=tsin(t)$$

I know what the answer is according to a sheet that I have of common transforms but I am not 100% on how to get there using the definition.

I know that if the question were to be: $$f(t)= sin(t)$$ then it's a simple $$\int_0^{\infty}e^{-st}*sin(t)$$ which by integration by parts give you $$1/((s-1)^2-1)$$

But how do I do it with the additional t??

Answer 1

Hint: $$ \int_0^\infty t\sin(t)\mathrm e^{-st}\mathrm dt=-\frac{\mathrm d}{\mathrm ds}\int_0^\infty \sin(t)\mathrm e^{-st}\mathrm dt$$ Beware though that the formula you give for $$ \int_0^\infty \sin(t)\mathrm e^{-st}\mathrm dt$$ is inaccurate.

Answer 2

Another way is to write $\sin t = {1 \over 2 i } (e^{it} - e^{-it})$. If we let $r(t) = t 1_{(0,1)}(t)$, it is straightforward to establish that ${\cal L} r(s) = {1 \over s^2}$ (that is, the Laplace transform of a ramp).

Then \begin{eqnarray} {\cal L} f(s) &=& \int_0^\infty e^{-st} {1 \over 2 i } (e^{it} - e^{-it}) t dt \\ &=& {1 \over 2 i } ( \int_0^\infty e^{-(s-i)t}t dt - \int_0^\infty e^{-(s+i)t}t dt ) \\ &=& {1 \over 2 i } ( {\cal L} r(s-i) - {\cal L} r(s+i) ) \\ &=& \cdots \end{eqnarray}