To calculate the Laplace transform of the linear function f(t)=t, we should follow these steps:

1. Apply Laplace transform

To find the Laplace transform of f(t)=1, Let’s start by the Laplace integral:

\displaystyle\mathscr{L}\{f(t)\}=\int_{t=0}^{\infty}f(t)e^{-st}dt

Put f(t)=t in above equation, we get

\displaystyle \mathscr{L}\{t\}=\int_{t=0}^{\infty}t\cdot e^{-st}dt

2. Use integration by parts

Let’s apply the integration by parts forumula to the above equation. We have:

\displaystyle\int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)dx,

Let f(t)=t and g'(t)=e^{-st}. The integration formula for exponential function is:

\displaystyle\int e^{ax}dx=\frac{e^{ax}}{a}+c

Hence, we have:

\displaystyle\mathcal{L}\{t\}=\left|\frac{te^{-st}}{-s}\right|_{0}^{\infty}-\int_{t=0}^{\infty}\frac{e^{-st}}{-s}\cdot\frac{d}{dt}(t) dt

Using the limits of t in the first part of right hand side and simplifying the derivative in second part, we get

\displaystyle \mathscr{L}\{t\}=\frac{-1}{s}\left[te^{-st}\right]_{0}^{\infty}+\frac{1}{s}\int_{t=0}^{\infty}e^{-st}\cdot1 dt

Here s is positive and large enough. So, the product te^{-st} converge to zero as t\to\infty.

The first part of right hand side will be zero and using the integral of exponential in the second part, we get:

\displaystyle\mathscr{L}\{t\}=0+\frac{1}{s}\mathcal{L}\{1\}

3. Use table of Laplace transforms

Using the Laplace transform of a constant in the above equation, we get:

\displaystyle\mathscr{L}\{t\}=\frac{1}{s}\left(\frac{1}{s}\right)

Simplifying above expression, the Laplace transform of t is given by:

\displaystyle \mathcal{L}\{t\}=\frac{1}{s^2}

That’s all!

We reached the end of this exercise about calculating the Laplace transform of t using the definition. The Laplace transform of the general case, t^{n}, is discussed in this post.