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# System prepended metadata

title: PID 915

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The Laplace transform of a function $f(t)$ is given by the integral [[1]](https://byjus.com/maths/laplace-transform/)

$$\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt $$

For $f(t) = t^2$, the Laplace transform is:

$$\begin{split} \mathcal{L}\{t^2\} & = \int_{0}^{\infty} e^{-st} t^2 \, dt \\ 
& = \int_{0}^{\infty} t^2 e^{-st} \, dt \end{split}$$

To solve this integral, we will use integration by parts. The formula for integration by parts is [[2]](https://www.cuemath.com/calculus/integration-by-parts/):

$$\int u \, dv = uv - \int v \, du\tag1$$

Letting $u = t^2$ and $dv = e^{-st} \, dt$, then $du=2t~dt~\text{and}~v=-\frac{1}{s}e^{-st}$, we get:

$$\begin{split} \mathcal{L}\{t^2\} & = \left. -\frac{t^2}{s} e^{-st} \right|_{0}^{\infty} + \frac{2}{s} \int_{0}^{\infty} t e^{-st} \, dt \end{split}$$

The first term evaluates to zero when $t$ goes to infinity, and it's also zero when $t$ is zero. So, we're left with the integral:

$$\mathcal{L}\{t^2\}=\frac{2}{s} \int_{0}^{\infty} t e^{-st} \, dt\tag2$$

Again, Letting $u = t$ and $dv = e^{-st} \, dt$, then $du=dt~\text{and}~v=-\frac{1}{s}e^{-st}$, we get from (1) and (2):

$$\begin{split} \mathcal{L}\{t^2\} & = \frac{2}{s}\left(\left. -\frac{t}{s} e^{-st} \right|_{0}^{\infty} + \frac{1}{s} \int_{0}^{\infty} e^{-st} \, dt \right)\\
  &= \frac{2}{s}\left(0 + \frac{1}{s} \left(\frac{e^{-st}}{-s}\right)_0^{\infty} \, dt \right)\\
  &= -\frac{2}{s^3} \left(e^{-st}\right)_0^{\infty} \, dt\\
  &= -\frac{2}{s^3} \left(0-1\right) \, dt\\
  &= \frac{2}{s^3}
\end{split}$$

Therefore, the Laplace transform of $t^2$ is also $\frac{2}{s^3}$.