So I'm doing the maths tutorials for a second year now and finally realized how awesome the Laplace transform is. Despite having understood Fourier transform and the analysis in my DSP classes, I can only now say (after some years of working...) that Laplace is pretty f***ing awesome! And just because I can, I want to put down some proofs that I find particularly cool...

Well, the basic Laplace transform, as shown in many textbooks, is:

\[ F(s)=\mathcal{L}\left(f(t)\right)=\int_0^{\infty}f(t)e^{-st}dt\]

Whether we were dealing with feedback systems or multi-order differentials, we simply substituted \(\frac{d}{dt}\) for \(s\) and \(\int_0^tdt\) with \(\frac{1}{s}\), since that is what the lookup table told us. So I never fully understood why we could do that - in our engineering classes we rarely did mathematical proofs. But in my moments of spare time, I decided to give proving the Laplace transform of derivatives and integrals a try. And this is what I concluded.

Laplace transform of a derivative

So let's start with the differential and write out the full equation.

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = \int_0^\infty \frac{d}{dt}f(t)e^{-st}dt\]

We can solve this integral by using integration by parts...

Integration by parts

This kind of integration is used when integrating a product of two functions that both contain the to term that is be integrated over. The theorem is pretty straight forward:

\[ \int_a^bu(x)\frac{d}{dx}v(x) = \left[u(x)v(x)\right]\Big|_a^b - \int_a^bv(x)\frac{d}{dx}u(x)dx\]

Or a shorter version:

\[ \int u\,dv = uv - \int v\,du\]

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = \lim_{R \to \infty} \left[f(t)e^{-st}\right]\Big|_{t=0}^{t=R} + s\int_0^\infty f(t)e^{-st}dt\]

Expanding the left summand results in the following.

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = \lim_{R \to \infty} \left[f(R)e^{-sR} - f(0)e^{-s0}\right] + s\int_0^\infty f(t)e^{-st}dt\]

Now we can simplify! First if all, as \(\lim_{R \to \infty} f(R)e^{-sR} = 0\) due to the negative exponential. All that's left is \(-f(0)e^{-s0}\) or, since \(e^{-s0} = 1\), \(-f(0)\).

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = -f(0) + s\int_0^\infty f(t)e^{-st}dt\]

And finally, \(s\int_0^\infty f(t)e^{-st}dt\) can be rewritten as \(sF(s)\), hence proving that:

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = -f(0) + sF(s)\]

In most systems, the initial conditions are assumed to be zero, hence you'll often find the simplified formula:

\[ \mathcal{L}\left(\frac{d}{dt}f(t)\right) = sF(s)\]

How awesome was that!

Laplace transform of an integral

In the 1st proof we've already done most of the work that allows us to work our way through the integration problem. First, let's simplify a bit by stating that \(g(t)=\int_0^tf(\tau)d\tau\) because then \(\frac{d}{dt}g(t) = f(t)\).

\[ \mathcal{L}\left(f(t)\right) = \mathcal{L}\left(\frac{d}{dt}g(t)\right)\]

We already know how to solve the differential problem, so we can simply substitute (assuming \(g(0) = 0\)).

\[ \mathcal{L}\left(f(t)\right) = s\mathcal{L}\left(g(t)\right)\]

If we now divide by \(s\) and substitute \(g(t)\) we find the following.

\[ \frac{1}{s}\mathcal{L}\left(f(t)\right) = \mathcal{L}\left(\int_0^tf(\tau)d\tau\right)\]

And look at that! We've shown that (I flipped the sides):

\[ \mathcal{L}\left(\int_0^tf(\tau)d\tau\right) = \frac{1}{s}F(s)\]

Conclusion

So from these two short proofs it can be seen that differentiation is literally the inverse of integration, and vice-versa! Although that's pretty obvious, I still like the maths work out...