How do I determine the (bilateral) Laplace transform of $t\theta(t-1)$, where the $\theta$ is the Heaviside step function.
I can solve it "manualy" (i.e. solving $\int_1^\infty tdt$), but I would like to understand how to do it using a table of standard Laplace transformations. I don't get how to deal with the $t-1$ in the Heaviside function.
Sorry if I use the wrong terms, I am taking the course in Swedish and it's kinda hard finding the translations to some of the terms.
For the Heavisde step function (Heaviside theta function or unit step function):
$$\theta\left(t\right)=\begin{cases}0\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space t<0\\\\1\space\space\space\space\space\space\space\space\space\space\space\text{when}\space t\ge0\end{cases}\tag1$$
So, we get that:
$$\theta\left(t-1\right)=\begin{cases}0\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space t<1\\\\1\space\space\space\space\space\space\space\space\space\space\space\text{when}\space t\ge1\end{cases}\tag2$$
And we also get:
$$t\cdot\theta\left(t-1\right)=\begin{cases}0\space\space\space\space\space\space\space\space\space\space\space\text{when}\space\space t<1\\\\t\space\space\space\space\space\space\space\space\space\space\space\text{when}\space t\ge1\end{cases}\tag3$$
So, we need to solve:
$$\mathcal{B}_t\left[t\cdot\theta\left(t-1\right)\right]_{\left(\text{s}\right)}:=\int_{-\infty}^\infty e^{-\text{s}t}\cdot t\cdot\theta\left(t-1\right)\space\text{d}t=\int_1^\infty te^{-\text{s}t}\space\text{d}t\tag4$$
Now, use:
$$\int te^{-\text{s}t}\space\text{d}t=\text{C}-\frac{e^{-\text{s}t}\cdot\left(1+\text{s}\cdot t\right)}{\text{s}^2}\tag5$$