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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.

1 Answers 1

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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:

$$\int te^{-\text{s}t}\space\text{d}t=\text{C}-\frac{e^{-\text{s}t}\cdot\left(1+\text{s}\cdot t\right)}{\text{s}^2}\tag5$$

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    That's the part I understand. As I said, I can solve it manually, what I don't understand is how to solve it using a "standard Laplace transformations"-table, like this one: http://www.intmath.com/laplace-transformation/table-laplace-transforms.php – 2017-01-29
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    @EAminy Actually you can not use that table, because it gives not the bilateral Laplace transform, but the standard Laplace transform; defined as: $$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t$$ Instead of the bilateral Laplace transfrom, which is defined as: $$\text{Y}\left(\text{s}\right)=\mathcal{B}_t\left[\text{y}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_{-\infty}^\infty\text{y}\left(t\right)e^{-\text{s}t}\space\text{d}t$$ – 2017-02-01