Question

Prove the laplace transform of t is 1/(s^2)

Answer 1

$$\mathcal{L}\{t\}=\int_0^\infty te^{-st}\,\mathrm{d}t$$Consider the antiderivative separately first and integrate by parts:$$\begin{align*}\int\underbrace{t}_u\underbrace{e^{-st}\,\mathrm{d}s}_{\mathrm{d}v}&=-\frac1ste^{-st}+\frac1s\int e^{-st}\,\mathrm{d}t\\&=-\frac1ste^{-st}-\frac1{s^2}e^{-st}\\&=-\frac1s\left(te^{-st}+\frac1se^{-st}\right)\end{align*}$$Now consider our improper integral as a limit of integrals:$$\begin{align*}\int_0^\infty te^{-st}\,\mathrm{d}t&=\lim_{a\to\infty}\int_0^ate^{-st}\,\mathrm{d}t\\&=\lim_{a\to\infty}\left[-\frac1s\left(te^{-st}+\frac1se^{-st}\right)\right]_0^a\\&=\frac1s\lim_{a\to\infty}\left(-ae^{-at}-\frac1se^{-at}+0+\frac1s\right)\\&=\frac1s\cdot\frac1s\\&=\frac1{s^2}\end{align*}$$

Answer 2

Omfg that was perfect