# Using definition of Laplace transform in determining Laplace of a step function

#### shorty

##### New member
I have a question that has stumped me a bit, i am not sure how to use the definition to calculate it, i can use the tables, but i don't think that's what is needed.

Using the definition of the Laplace transform, determine the Laplace transform of I can do it with the table but i am not sure how to to this using the definition. #### Chris L T521

##### Well-known member
Staff member
I have a question that has stumped me a bit, i am not sure how to use the definition to calculate it, i can use the tables, but i don't think that's what is needed.

Using the definition of the Laplace transform, View attachment 153 determine the Laplace transform of

View attachment 154

I can do it with the table but i am not sure how to to this using the definition. We can break up the integral into two parts since $f(t)$ is a piecewise function:

$\mathcal{L}[f(t)] = \int_0^{\infty}e^{-st}f(t)\,dt=\int_0^2 e^{-st}0\,dt + \int_2^{\infty}e^{-st}t\,dt = \int_2^{\infty}te^{-st}\,dt.$

This should now be a relatively simple improper integral to compute.

Can you take it from here?

#### shorty

##### New member
Thank you, but I got that far into the separation, but I wasn't sure how to proceed from there, my integrals kept repeating when I tried it by parts, and I wasn't getting anything to substitute to use that wasn't still leaving me with multiple variables to integrate. ...

#### Chris L T521

##### Well-known member
Staff member
Thank you, but I got that far into the separation, but I wasn't sure how to proceed from there, my integrals kept repeating when I tried it by parts, and I wasn't getting anything to substitute to use that wasn't still leaving me with multiple variables to integrate. ...
In this case, you only need to apply integration by parts once. Let $u=t$, $dv=e^{-st}dt$; thus $du=dt$ and $v=-\dfrac{e^{-st}}{s}$. Plugging this into the integration by parts formula, we have

$\int_2^{\infty}te^{-st}\,dt = \lim\limits_{b\to\infty}\left.\left[-\frac{te^{-st}}{s}\right]\right|_2^b + \frac{1}{s}\int_2^{\infty}e^{-st}\,dt=\ldots$

Can you take it from here?