Laplace transform of time integration

In the previous lesson, we discussed the Laplace transform of the convolution which is part of the Integrals transformation property. The next section presents another Laplace transformation property that corresponds to time integration!

1. So, what is the Laplace transform of integration?

Proof:

To prove this theorem, let g(t)=1, so

G(s)=\mathscr{L}\{g(t)\}=\frac{1}{s}

And from Laplace transform of convultion, we have:

\displaystyle (f\ast g)(t)=\int_{0}^{t}f(\tau)g(t-\tau)d\tau=\int_{0}^{t}f(\tau)1d\tau=\int_{0}^{t}f(\tau)d\tau

Hence:

\displaystyle \mathscr{L}\{(f\ast g)(t)\}=\mathscr{L}\left\{\int_{0}^{t}f(\tau)d\tau\right\}=F(s)G(s)

i.e.,

\displaystyle \mathscr{L}\left\{\int_{0}^{t}f(\tau)d\tau\right\}=\frac{1}{s}F(s)

This ends the proof.

Illustrative example of the Laplace transform of time integration

From the laplace transform of an exponential, and since a=-4 at f(t)=e^{-4t}, we have:

\displaystyle F(s)=\mathscr{L}\{e^{-4t}\}=\frac{1}{s-(-4)}=\frac{1}{s+4}

Using the above theorem, we get:

\displaystyle \mathscr{L}\left\{\int_{0}^{t}e^{-4\tau}d\tau\right\}=\frac{1}{s}\cdot\frac{1}{s+4}=\frac{1}{s(s+4)}

That’s all!

We reached the end of this short lesson about the Laplace transform of integration. For a full list of Laplace transform properties, check this post!